Electric Charges and Fields

NEET Ug . NEET Physics . Physics . Electric Charges and Fields

Chapter 16: Electric Charges and Fields

Ch 16

NEET Physics

NEET Ug

Introduction

Static Electricity

Electric shock sensation occurs due to discharge of static electric charges through the body.
Static electricity is generated by accumulation of electric charges on surfaces.
Electrostatics studies forces, fields, and potentials from static charges.

Solved Examples

Ex 1: Calculate force between two charges 2 μC and 3 μC separated by 0.5 m in vacuum.
Use F = 1/(4 π ε₀) |q₁ q₂|/r²
F = 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/(0.5²) = 216 N

Practice Set

Q1: Two charges +1 μC and +2 μC are 1 m apart. Force magnitude is?
- A) 18 N
- B) 9 N
- C) 36 N
- D) 0 N
Q2: If distance between charges is doubled, force becomes?
- A) Half
- B) Quarter
- C) Double
- D) Four times
Q3: Two charges q₁ = 3 μC, q₂ = -3 μC separated by 0.3 m in dielectric medium K=3. Force magnitude?
- A) 90 N
- B) 30 N
- C) 10 N
- D) 270 N

Answer Key

Q1: A) 18 N - F = 9 × 10⁹ × (1 × 10^-6 × 2 × 10^-6)/(1²) = 18 N
Q2: B) Quarter - Force varies as 1/r², doubling r reduces force by 4.
Q3: C) 10 N - Force in vacuum = (9 × 10⁹ × 3 × 10^-6 × 3 × 10^-6)/(0.3²) = 30 N, in medium F = 30/3 = 10 N

Memory Aids

Force F ∝ |q₁ q₂|/r²
Like charges repel, unlike attract
Force in dielectric = Force in vacuum/K
k = 9 × 10⁹, ε₀ = 8.854 × 10^-12

Static Electricity

Electric shock sensation occurs due to discharge of static electric charges through the body.
Static electricity is generated by accumulation of electric charges on surfaces.
Electrostatics studies forces, fields, and potentials from static charges.

Solved Examples

Ex 1: Why do we feel a shock when touching a car door after sliding on a bus seat?
Step 1: Sliding causes charge accumulation on body.
Step 2: Touching metal door discharges static electricity.
Step 3: Discharge causes shock sensation.

Practice Set

Q1: What causes the sensation of electric shock when touching a metal object after sliding on a seat?
A. Current flow
B. Discharge of static electricity
C. Magnetic field
D. Heat generation
Q2: Electrostatics primarily deals with:
A. Moving charges
B. Static charges
C. Magnetic fields
D. Electric currents
Q3: Which of the following is NOT related to static electricity?
A. Charge accumulation
B. Electric discharge
C. Electric current
D. Electrostatic force

Answer Key

Q1: B - Shock is due to discharge of static electric charges.
Q2: B - Electrostatics studies static charges and their effects.
Q3: C - Electric current involves moving charges, not static electricity.

Memory Aids

Static electricity = charges at rest.
Shock = sudden discharge of static charge.
Electrostatics = study of static charges and forces.

Electric Charge

Types and Properties

Neutral body becomes charged by gaining or losing electrons.
Two types of charges: positive (+) and negative (−).
Losing electrons → positive charge; gaining electrons → negative charge.
Like charges repel; unlike charges attract.
Glass rod or cat’s fur rubbed with silk → positive charge.
Plastic rod or silk rubbed → negative charge.
Charged object = electrified; no net charge = neutral.
Gold-leaf electroscope detects presence and type of charge.

Diagram 1 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Identify charge types on rods:
Step 1: Glass rod rubbed with silk → positive charge.
Step 2: Plastic rod rubbed with silk → negative charge.
Step 3: Like charges repel; unlike charges attract.

Practice Set

Q1: What happens when a neutral object loses electrons?
A. Becomes negatively charged
B. Becomes positively charged
C. Remains neutral
D. Becomes magnetic
Q2: Which pair of charges will repel each other?
A. Positive and negative
B. Positive and positive
C. Negative and positive
D. Neutral and positive
Q3: What device is used to detect electric charge?
A. Ammeter
B. Voltmeter
C. Gold-leaf electroscope
D. Galvanometer

Answer Key

Q1: B - Losing electrons makes object positively charged.
Q2: B - Like charges (positive-positive) repel each other.
Q3: C - Gold-leaf electroscope detects electric charge.

Memory Aids

Positive charge = loss of electrons.
Negative charge = gain of electrons.
Like charges repel; unlike charges attract.
Gold-leaf electroscope detects charge.

Solved Examples

Ex 1: Identify charge types on rods:
Step 1: Glass rod rubbed with silk → positive charge.
Step 2: Plastic rod rubbed with silk → negative charge.
Step 3: Like charges repel; unlike charges attract.

Practice Set

Q1: What happens when a neutral object loses electrons?
A. Becomes negatively charged
B. Becomes positively charged
C. Remains neutral
D. Becomes magnetic
Q2: Which pair of charges will repel each other?
A. Positive and negative
B. Positive and positive
C. Negative and positive
D. Neutral and positive
Q3: What device is used to detect electric charge?
A. Ammeter
B. Voltmeter
C. Gold-leaf electroscope
D. Galvanometer

Answer Key

Q1: B - Losing electrons makes object positively charged.
Q2: B - Like charges (positive-positive) repel each other.
Q3: C - Gold-leaf electroscope detects electric charge.

Memory Aids

Positive charge = loss of electrons.
Negative charge = gain of electrons.
Like charges repel; unlike charges attract.
Gold-leaf electroscope detects charge.

Solved Examples

Ex 1: Identify charge types on rods:
Step 1: Glass rod rubbed with silk → positive charge.
Step 2: Plastic rod rubbed with silk → negative charge.
Step 3: Like charges repel; unlike charges attract.

Practice Set

Q1: What happens when a neutral object loses electrons?
A. Becomes negatively charged
B. Becomes positively charged
C. Remains neutral
D. Becomes magnetic
Q2: Which pair of charges will repel each other?
A. Positive and negative
B. Positive and positive
C. Negative and positive
D. Neutral and positive
Q3: What device is used to detect electric charge?
A. Ammeter
B. Voltmeter
C. Gold-leaf electroscope
D. Galvanometer

Answer Key

Q1: B - Losing electrons makes object positively charged.
Q2: B - Like charges (positive-positive) repel each other.
Q3: C - Gold-leaf electroscope detects electric charge.

Memory Aids

Positive charge = loss of electrons.
Negative charge = gain of electrons.
Like charges repel; unlike charges attract.
Gold-leaf electroscope detects charge.

Solved Examples

Ex 1: Identify charge types on rods:
Step 1: Glass rod rubbed with silk → positive charge.
Step 2: Plastic rod rubbed with silk → negative charge.
Step 3: Like charges repel; unlike charges attract.

Practice Set

Q1: What happens when a neutral object loses electrons?
A. Becomes negatively charged
B. Becomes positively charged
C. Remains neutral
D. Becomes magnetic
Q2: Which pair of charges will repel each other?
A. Positive and negative
B. Positive and positive
C. Negative and positive
D. Neutral and positive
Q3: What device is used to detect electric charge?
A. Ammeter
B. Voltmeter
C. Gold-leaf electroscope
D. Galvanometer

Answer Key

Q1: B - Losing electrons makes object positively charged.
Q2: B - Like charges (positive-positive) repel each other.
Q3: C - Gold-leaf electroscope detects electric charge.

Memory Aids

Positive charge = loss of electrons.
Negative charge = gain of electrons.
Like charges repel; unlike charges attract.
Gold-leaf electroscope detects charge.

Types and Properties

Neutral body becomes charged by gaining or losing electrons.
Two types of charges: positive (+) and negative (−).
Losing electrons → positive charge; gaining electrons → negative charge.
Like charges repel; unlike charges attract.
Glass rod or cat’s fur rubbed with silk → positive charge.
Plastic rod or silk rubbed → negative charge.
Charged object = electrified; no net charge = neutral.
Gold-leaf electroscope detects presence and type of charge.

Diagram 2 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Identify charge types on rods:
Step 1: Glass rod rubbed with silk → positive charge.
Step 2: Plastic rod rubbed with silk → negative charge.
Step 3: Like charges repel; unlike charges attract.

Practice Set

Q1: What happens when a neutral object loses electrons?
A. Becomes negatively charged
B. Becomes positively charged
C. Remains neutral
D. Becomes magnetic
Q2: Which pair of charges will repel each other?
A. Positive and negative
B. Positive and positive
C. Negative and positive
D. Neutral and positive
Q3: What device is used to detect electric charge?
A. Ammeter
B. Voltmeter
C. Gold-leaf electroscope
D. Galvanometer

Answer Key

Q1: B - Losing electrons makes object positively charged.
Q2: B - Like charges (positive-positive) repel each other.
Q3: C - Gold-leaf electroscope detects electric charge.

Memory Aids

Positive charge = loss of electrons.
Negative charge = gain of electrons.
Like charges repel; unlike charges attract.
Gold-leaf electroscope detects charge.

Conductors and Insulators

Properties of Conductors

Allow electric current to pass easily due to free electrons.
Examples: Metals (copper, aluminum), human and animal bodies, earth.
Charge distributes uniformly over the surface when transferred.

Properties of Insulators

Offer high resistance to electric current.
Examples: Glass, porcelain, plastic, nylon, wood.
Charge remains localized at the point of contact.

Electric Wiring in Houses

Three wires: Live, Neutral, Earth.
Live and Neutral carry current from power station.
Earth wire connected to ground for safety.
Metallic parts of appliances connected to earth wire to discharge extra charge safely.

Solved Examples

Ex 1: Charge distribution on a conductor
- Given charge Q on conductor.
- Charge spreads uniformly on surface.
- Surface charge density σ = Q/A, where A is surface area.
Ex 2: Identifying conductor or insulator
- Material allows current easily → Conductor.
- Material resists current → Insulator.

Practice Set

Q1: Which of the following is a good conductor?
- A. Glass
- B. Copper
- C. Plastic
- D. Wood
Q2: What happens to charge on an insulator when charged?
- A. Spreads over surface
- B. Moves freely inside
- C. Remains localized
- D. Flows to earth
Q3: In household wiring, the earth wire is used to:
- A. Carry current to appliances
- B. Connect live and neutral wires
- C. Discharge extra charge safely to ground
- D. Increase resistance

Answer Key

Q1: B - Copper is a metal and a good conductor of electricity.
Q2: C - Charge on insulators remains localized and does not spread.
Q3: C - Earth wire safely discharges extra charge to the ground to prevent shocks.

Memory Aids

Conductors: "Free electrons flow" → easy current flow.
Insulators: "Charge stays put" → no free electron movement.
Earth wire: "Safety path" for extra charge.

Properties of Insulators

Offer high resistance to electric current.
Examples: Glass, porcelain, plastic, nylon, wood.
Charge remains localized at the point of contact.

Electric Wiring in Houses

Three wires: Live, Neutral, Earth.
Live and Neutral carry current from power station.
Earth wire connected to ground for safety.
Metallic parts of appliances connected to earth wire to discharge extra charge safely.

Solved Examples

Ex 1: Charge distribution on a conductor
- Given charge Q on conductor.
- Charge spreads uniformly on surface.
- Surface charge density σ = Q/A, where A is surface area.
Ex 2: Identifying conductor or insulator
- Material allows current easily → Conductor.
- Material resists current → Insulator.

Practice Set

Q1: Which of the following is a good conductor?
- A. Glass
- B. Copper
- C. Plastic
- D. Wood
Q2: What happens to charge on an insulator when charged?
- A. Spreads over surface
- B. Moves freely inside
- C. Remains localized
- D. Flows to earth
Q3: In household wiring, the earth wire is used to:
- A. Carry current to appliances
- B. Connect live and neutral wires
- C. Discharge extra charge safely to ground
- D. Increase resistance

Answer Key

Q1: B - Copper is a metal and a good conductor of electricity.
Q2: C - Charge on insulators remains localized and does not spread.
Q3: C - Earth wire safely discharges extra charge to the ground to prevent shocks.

Memory Aids

Conductors: "Free electrons flow" → easy current flow.
Insulators: "Charge stays put" → no free electron movement.
Earth wire: "Safety path" for extra charge.

Electric Wiring in Houses

Three wires: Live, Neutral, Earth.
Live and Neutral carry current from power station.
Earth wire connected to ground for safety.
Metallic parts of appliances connected to earth wire to discharge extra charge safely.

Solved Examples

Ex 1: Charge distribution on a conductor
- Given charge Q on conductor.
- Charge spreads uniformly on surface.
- Surface charge density σ = Q/A, where A is surface area.
Ex 2: Identifying conductor or insulator
- Material allows current easily → Conductor.
- Material resists current → Insulator.

Practice Set

Q1: Which of the following is a good conductor?
- A. Glass
- B. Copper
- C. Plastic
- D. Wood
Q2: What happens to charge on an insulator when charged?
- A. Spreads over surface
- B. Moves freely inside
- C. Remains localized
- D. Flows to earth
Q3: In household wiring, the earth wire is used to:
- A. Carry current to appliances
- B. Connect live and neutral wires
- C. Discharge extra charge safely to ground
- D. Increase resistance

Answer Key

Q1: B - Copper is a metal and a good conductor of electricity.
Q2: C - Charge on insulators remains localized and does not spread.
Q3: C - Earth wire safely discharges extra charge to the ground to prevent shocks.

Memory Aids

Conductors: "Free electrons flow" → easy current flow.
Insulators: "Charge stays put" → no free electron movement.
Earth wire: "Safety path" for extra charge.

Properties of Insulators

Offer high resistance to electric current.
Examples: Glass, porcelain, plastic, nylon, wood.
Charge remains localized at the point of contact.

Electric Wiring in Houses

Three wires: Live, Neutral, Earth.
Live and Neutral carry current from power station.
Earth wire connected to ground for safety.
Metallic parts of appliances connected to earth wire to discharge extra charge safely.

Solved Examples

Ex 1: Charge distribution on a conductor
- Given charge Q on conductor.
- Charge spreads uniformly on surface.
- Surface charge density σ = Q/A, where A is surface area.
Ex 2: Identifying conductor or insulator
- Material allows current easily → Conductor.
- Material resists current → Insulator.

Practice Set

Q1: Which of the following is a good conductor?
- A. Glass
- B. Copper
- C. Plastic
- D. Wood
Q2: What happens to charge on an insulator when charged?
- A. Spreads over surface
- B. Moves freely inside
- C. Remains localized
- D. Flows to earth
Q3: In household wiring, the earth wire is used to:
- A. Carry current to appliances
- B. Connect live and neutral wires
- C. Discharge extra charge safely to ground
- D. Increase resistance

Answer Key

Q1: B - Copper is a metal and a good conductor of electricity.
Q2: C - Charge on insulators remains localized and does not spread.
Q3: C - Earth wire safely discharges extra charge to the ground to prevent shocks.

Memory Aids

Conductors: "Free electrons flow" → easy current flow.
Insulators: "Charge stays put" → no free electron movement.
Earth wire: "Safety path" for extra charge.

Properties of Insulators

Offer high resistance to electric current.
Examples: Glass, porcelain, plastic, nylon, wood.
Charge remains localized at the point of contact.

Electric Wiring in Houses

Three wires: Live, Neutral, Earth.
Live and Neutral carry current from power station.
Earth wire connected to ground for safety.
Metallic parts of appliances connected to earth wire to discharge extra charge safely.

Solved Examples

Ex 1: Charge distribution on a conductor
- Given charge Q on conductor.
- Charge spreads uniformly on surface.
- Surface charge density σ = Q/A, where A is surface area.
Ex 2: Identifying conductor or insulator
- Material allows current easily → Conductor.
- Material resists current → Insulator.

Practice Set

Q1: Which of the following is a good conductor?
- A. Glass
- B. Copper
- C. Plastic
- D. Wood
Q2: What happens to charge on an insulator when charged?
- A. Spreads over surface
- B. Moves freely inside
- C. Remains localized
- D. Flows to earth
Q3: In household wiring, the earth wire is used to:
- A. Carry current to appliances
- B. Connect live and neutral wires
- C. Discharge extra charge safely to ground
- D. Increase resistance

Answer Key

Q1: B - Copper is a metal and a good conductor of electricity.
Q2: C - Charge on insulators remains localized and does not spread.
Q3: C - Earth wire safely discharges extra charge to the ground to prevent shocks.

Memory Aids

Conductors: "Free electrons flow" → easy current flow.
Insulators: "Charge stays put" → no free electron movement.
Earth wire: "Safety path" for extra charge.

Properties of Conductors

Allow electric current to pass easily due to free electrons.
Examples: Metals (copper, aluminum), human and animal bodies, earth.
Charge distributes uniformly over the surface when transferred.

Properties of Insulators

Offer high resistance to electric current.
Examples: Glass, porcelain, plastic, nylon, wood.
Charge remains localized at the point of contact.

Electric Wiring in Houses

Three wires: Live, Neutral, Earth.
Live and Neutral carry current from power station.
Earth wire connected to ground for safety.
Metallic parts of appliances connected to earth wire to discharge extra charge safely.

Solved Examples

Ex 1: Charge distribution on a conductor
- Given charge Q on conductor.
- Charge spreads uniformly on surface.
- Surface charge density \( \sigma = \frac{Q}{A} \), where A is surface area.
Ex 2: Identifying conductor or insulator
- Material allows current easily → Conductor.
- Material resists current → Insulator.

Practice Set

Q1: Which of the following is a good conductor?
- A. Glass
- B. Copper
- C. Plastic
- D. Wood
Q2: What happens to charge on an insulator when charged?
- A. Spreads over surface
- B. Moves freely inside
- C. Remains localized
- D. Flows to earth
Q3: In household wiring, the earth wire is used to:
- A. Carry current to appliances
- B. Connect live and neutral wires
- C. Discharge extra charge safely to ground
- D. Increase resistance

Answer Key

Q1: B - Copper is a metal and a good conductor of electricity.
Q2: C - Charge on insulators remains localized and does not spread.
Q3: C - Earth wire safely discharges extra charge to the ground to prevent shocks.

Memory Aids

Conductors: "Free electrons flow" → easy current flow.
Insulators: "Charge stays put" → no free electron movement.
Earth wire: "Safety path" for extra charge.

Basic Properties of Electric Charge

Addition of Charges

Total charge in a system with discrete charges q₁, q₂, q₃, ..., q_n is q = q₁ + q₂ + q₃ + ... + q_n.

Conservation of Charge

Total charge in an isolated system remains constant.
Charge can neither be created nor destroyed, only transferred.

Quantisation of Charge

Charge is quantised: Q = ±ne, where n = 1, 2, 3, ... and e = 1.6 × 10^-19 C (charge of one electron).
1 Coulomb charge contains approximately 6 × 10¹⁸ electrons.

Solved Examples

Ex 1: Find total charge if a system has charges 2 × 10^-6 C, -3 × 10^-6 C, and 5 × 10^-6 C.

Total charge q = 2 × 10^-6 - 3 × 10^-6 + 5 × 10^-6 = 4 × 10^-6 C.

Ex 2: Calculate number of electrons in 3 C charge.

Number of electrons = 3/(1.6 × 10^-19) ≈ 1.875 × 10¹⁹.

Practice Set

Q1: What is the total charge of charges 1 μC, -2 μC, 3 μC?
A) 2 μC B) 6 μC C) 1 μC D) 0 μC>

Q2: Which statement is true about charge conservation?
A) Charge can be created
B) Charge can be destroyed
C) Total charge in isolated system is constant
D) Charge is not quantised

Q3: If Q = 4.8 × 10^-19 C, what is n in Q = ne?
A) 2 B) 3 C) 4 D) 5

Answer Key

Q1: A) 2 μC - Sum of charges: 1 - 2 + 3 = 2 μC.

Q2: C) Total charge in isolated system is constant - Conservation law states charge is neither created nor destroyed.

Q3: B) 3 - n = Q/e = (4.8 × 10^-19)/(1.6 × 10^-19) = 3.

Memory Aids

Charge addition: Sum all charges algebraically.
Conservation: Charge constant in isolated system.
Quantisation: Charge = integer × electron charge.

Solved Examples

Ex 1: Find total charge if a system has charges 2 × 10^-6 C, -3 × 10^-6 C, and 5 × 10^-6 C.

Total charge q = 2 × 10^-6 - 3 × 10^-6 + 5 × 10^-6 = 4 × 10^-6 C.

Ex 2: Calculate number of electrons in 3 C charge.

Number of electrons = 3/(1.6 × 10^-19) ≈ 1.875 × 10¹⁹.

Practice Set

Q1: What is the total charge of charges 1 μC, -2 μC, 3 μC?
A) 2 μC B) 6 μC C) 1 μC D) 0 μC>

Q2: Which statement is true about charge conservation?
A) Charge can be created
B) Charge can be destroyed
C) Total charge in isolated system is constant
D) Charge is not quantised

Q3: If Q = 4.8 × 10^-19 C, what is n in Q = ne?
A) 2 B) 3 C) 4 D) 5

Answer Key

Q1: A) 2 μC - Sum of charges: 1 - 2 + 3 = 2 μC.

Q2: C) Total charge in isolated system is constant - Conservation law states charge is neither created nor destroyed.

Q3: B) 3 - n = Q/e = (4.8 × 10^-19)/(1.6 × 10^-19) = 3.

Memory Aids

Charge addition: Sum all charges algebraically.
Conservation: Charge constant in isolated system.
Quantisation: Charge = integer × electron charge.

Quantisation of Charge

Charge is quantised: Q = ±ne, where n = 1, 2, 3, ... and e = 1.6 × 10^-19 C (charge of one electron).
1 Coulomb charge contains approximately 6 × 10¹⁸ electrons.

Solved Examples

Ex 1: Find total charge if a system has charges 2 × 10^-6 C, -3 × 10^-6 C, and 5 × 10^-6 C.

Total charge q = 2 × 10^-6 - 3 × 10^-6 + 5 × 10^-6 = 4 × 10^-6 C.

Ex 2: Calculate number of electrons in 3 C charge.

Number of electrons = 3/(1.6 × 10^-19) ≈ 1.875 × 10¹⁹.

Practice Set

Q1: What is the total charge of charges 1 μC, -2 μC, 3 μC?
A) 2 μC B) 6 μC C) 1 μC D) 0 μC>

Q2: Which statement is true about charge conservation?
A) Charge can be created
B) Charge can be destroyed
C) Total charge in isolated system is constant
D) Charge is not quantised

Q3: If Q = 4.8 × 10^-19 C, what is n in Q = ne?
A) 2 B) 3 C) 4 D) 5

Answer Key

Q1: A) 2 μC - Sum of charges: 1 - 2 + 3 = 2 μC.

Q2: C) Total charge in isolated system is constant - Conservation law states charge is neither created nor destroyed.

Q3: B) 3 - n = Q/e = (4.8 × 10^-19)/(1.6 × 10^-19) = 3.

Memory Aids

Charge addition: Sum all charges algebraically.
Conservation: Charge constant in isolated system.
Quantisation: Charge = integer × electron charge.

Conservation of Charge

Total charge in an isolated system remains constant.
Charge can neither be created nor destroyed, only transferred.

Quantisation of Charge

Charge is quantised: Q = ±ne, where n = 1, 2, 3, ... and e = 1.6 × 10^-19 C (charge of one electron).
1 Coulomb charge contains approximately 6 × 10¹⁸ electrons.

Solved Examples

Ex 1: Find total charge if a system has charges 2 × 10^-6 C, -3 × 10^-6 C, and 5 × 10^-6 C.

Total charge q = 2 × 10^-6 - 3 × 10^-6 + 5 × 10^-6 = 4 × 10^-6 C.

Ex 2: Calculate number of electrons in 3 C charge.

Number of electrons = 3/(1.6 × 10^-19) ≈ 1.875 × 10¹⁹.

Practice Set

Q1: What is the total charge of charges 1 μC, -2 μC, 3 μC?
A) 2 μC B) 6 μC C) 1 μC D) 0 μC>

Q2: Which statement is true about charge conservation?
A) Charge can be created
B) Charge can be destroyed
C) Total charge in isolated system is constant
D) Charge is not quantised

Q3: If Q = 4.8 × 10^-19 C, what is n in Q = ne?
A) 2 B) 3 C) 4 D) 5

Answer Key

Q1: A) 2 μC - Sum of charges: 1 - 2 + 3 = 2 μC.

Q2: C) Total charge in isolated system is constant - Conservation law states charge is neither created nor destroyed.

Q3: B) 3 - n = Q/e = (4.8 × 10^-19)/(1.6 × 10^-19) = 3.

Memory Aids

Charge addition: Sum all charges algebraically.
Conservation: Charge constant in isolated system.
Quantisation: Charge = integer × electron charge.

Conservation of Charge

Total charge in an isolated system remains constant.
Charge can neither be created nor destroyed, only transferred.

Quantisation of Charge

Charge is quantised: Q = ±ne, where n = 1, 2, 3, ... and e = 1.6 × 10^-19 C (charge of one electron).
1 Coulomb charge contains approximately 6 × 10¹⁸ electrons.

Solved Examples

Ex 1: Find total charge if a system has charges 2 × 10^-6 C, -3 × 10^-6 C, and 5 × 10^-6 C.

Total charge q = 2 × 10^-6 - 3 × 10^-6 + 5 × 10^-6 = 4 × 10^-6 C.

Ex 2: Calculate number of electrons in 3 C charge.

Number of electrons = 3/(1.6 × 10^-19) ≈ 1.875 × 10¹⁹.

Practice Set

Q1: What is the total charge of charges 1 μC, -2 μC, 3 μC?
A) 2 μC B) 6 μC C) 1 μC D) 0 μC>

Q2: Which statement is true about charge conservation?
A) Charge can be created
B) Charge can be destroyed
C) Total charge in isolated system is constant
D) Charge is not quantised

Q3: If Q = 4.8 × 10^-19 C, what is n in Q = ne?
A) 2 B) 3 C) 4 D) 5

Answer Key

Q1: A) 2 μC - Sum of charges: 1 - 2 + 3 = 2 μC.

Q2: C) Total charge in isolated system is constant - Conservation law states charge is neither created nor destroyed.

Q3: B) 3 - n = Q/e = (4.8 × 10^-19)/(1.6 × 10^-19) = 3.

Memory Aids

Charge addition: Sum all charges algebraically.
Conservation: Charge constant in isolated system.
Quantisation: Charge = integer × electron charge.

Addition of Charges

Total charge in a system with discrete charges \(q_1, q_2, q_3, \ldots, q_n\) is \(q = q_1 + q_2 + q_3 + \ldots + q_n\).

Conservation of Charge

Total charge in an isolated system remains constant.
Charge can neither be created nor destroyed, only transferred.

Quantisation of Charge

Charge is quantised: \(Q = \pm ne\), where \(n = 1, 2, 3, \ldots\) and \(e = 1.6 \times 10^{-19}\) C (charge of one electron).
1 Coulomb charge contains approximately \(6 \times 10^{18}\) electrons.

Solved Examples

Ex 1: Find total charge if a system has charges \(2 \times 10^{-6}\) C, \(-3 \times 10^{-6}\) C, and \(5 \times 10^{-6}\) C.

Total charge \(q = 2 \times 10^{-6} - 3 \times 10^{-6} + 5 \times 10^{-6} = 4 \times 10^{-6}\) C.

Ex 2: Calculate number of electrons in \(3\) C charge.

Number of electrons \(= \frac{3}{1.6 \times 10^{-19}} \approx 1.875 \times 10^{19}\).

Practice Set

Q1: What is the total charge of charges \(1 \mu C, -2 \mu C, 3 \mu C\)?
A) \(2 \mu C\) B) \(6 \mu C\) C) \(1 \mu C\) D) \(0 \mu C\>

Q2: Which statement is true about charge conservation?
A) Charge can be created
B) Charge can be destroyed
C) Total charge in isolated system is constant
D) Charge is not quantised

Q3: If \(Q = 4.8 \times 10^{-19}\) C, what is \(n\) in \(Q = ne\)?
A) 2 B) 3 C) 4 D) 5

Answer Key

Q1: A) \(2 \mu C\) - Sum of charges: \(1 - 2 + 3 = 2 \mu C\).

Q2: C) Total charge in isolated system is constant - Conservation law states charge is neither created nor destroyed.

Q3: B) 3 - \(n = \frac{Q}{e} = \frac{4.8 \times 10^{-19}}{1.6 \times 10^{-19}} = 3\).

Memory Aids

Charge addition: Sum all charges algebraically.
Conservation: Charge constant in isolated system.
Quantisation: Charge = integer × electron charge.

Coulomb's Law

Fundamental Concept

Force between two point charges q₁ and q₂ separated by distance r in vacuum:
[ F = k |q₁ q₂|/r² = 1/(4 π ε₀) |q₁ q₂|/r² ]
ε₀ = permittivity of free space = 8.854 × 10^-12 C² N^-1 m^-2
k = 1/(4 π ε₀) = 9 × 10⁹ N m² C^-2
Force is attractive if charges are unlike, repulsive if like.

Vector Form

Force on q₂ due to q₁:
[ F₂₁ = 1/(4 π ε₀) q₁ q₂/|r₂ - r₁|³ (r₂ - r₁) = 1/(4 π ε₀) q₁ q₂/|r₂₁|² r̂₂₁ = - F₁₂ ]
Obeys Newton's third law: forces are equal in magnitude and opposite in direction.
r₂₁ = r₂ - r₁ is the vector from q₁ to q₂.

Force Direction for Charges

For like charges (q₁ q₂ > 0), force vectors point away (repulsion).
For unlike charges (q₁ q₂ < 0), force vectors point towards each other (attraction).

Effect of Dielectric Medium

In a medium with dielectric constant K, force reduces by factor K:
[ F = F₀/K = 1/(4 π ε₀ K) q₁ q₂/r² ]
F₀ is force in vacuum.

Solved Examples

Ex 1: Calculate force between two charges 2 μC and 3 μC separated by 0.5 m in vacuum.
Use F = 1/(4 π ε₀) |q₁ q₂|/r²
F = 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/(0.5²) = 216 N

Practice Set

Q1: Two charges +1 μC and +2 μC are 1 m apart. Force magnitude is?
- A) 18 N
- B) 9 N
- C) 36 N
- D) 0 N
Q2: If distance between charges is doubled, force becomes?
- A) Half
- B) Quarter
- C) Double
- D) Four times
Q3: Two charges q₁ = 3 μC, q₂ = -3 μC separated by 0.3 m in dielectric medium K=3. Force magnitude?
- A) 90 N
- B) 30 N
- C) 10 N
- D) 270 N

Answer Key

Q1: A) 18 N - F = 9 × 10⁹ × (1 × 10^-6 × 2 × 10^-6)/(1²) = 18 N
Q2: B) Quarter - Force varies as 1/r², doubling r reduces force by 4.
Q3: C) 10 N - Force in vacuum = (9 × 10⁹ × 3 × 10^-6 × 3 × 10^-6)/(0.3²) = 30 N, in medium F = 30/3 = 10 N

Memory Aids

Force F ∝ |q₁ q₂|/r²
Like charges repel, unlike attract
Force in dielectric = Force in vacuum/K
k = 9 × 10⁹, ε₀ = 8.854 × 10^-12

Vector Form

Force on q₂ due to q₁:
[ F₂₁ = 1/(4 π ε₀) q₁ q₂/|r₂ - r₁|³ (r₂ - r₁) = 1/(4 π ε₀) q₁ q₂/|r₂₁|² r̂₂₁ = - F₁₂ ]
Obeys Newton's third law: forces are equal in magnitude and opposite in direction.
r₂₁ = r₂ - r₁ is the vector from q₁ to q₂.

Diagram 3 – NEET Physics Electric Charges and Fields

Force Direction for Charges

For like charges (q₁ q₂ > 0), force vectors point away (repulsion).
For unlike charges (q₁ q₂ < 0), force vectors point towards each other (attraction).

Diagram 4 – NEET Physics Electric Charges and Fields

Effect of Dielectric Medium

In a medium with dielectric constant K, force reduces by factor K:
[ F = F₀/K = 1/(4 π ε₀ K) q₁ q₂/r² ]
F₀ is force in vacuum.

Solved Examples

Ex 1: Calculate force between two charges 2 μC and 3 μC separated by 0.5 m in vacuum.
Use F = 1/(4 π ε₀) |q₁ q₂|/r²
F = 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/(0.5²) = 216 N

Practice Set

Q1: Two charges +1 μC and +2 μC are 1 m apart. Force magnitude is?
- A) 18 N
- B) 9 N
- C) 36 N
- D) 0 N
Q2: If distance between charges is doubled, force becomes?
- A) Half
- B) Quarter
- C) Double
- D) Four times
Q3: Two charges q₁ = 3 μC, q₂ = -3 μC separated by 0.3 m in dielectric medium K=3. Force magnitude?
- A) 90 N
- B) 30 N
- C) 10 N
- D) 270 N

Answer Key

Q1: A) 18 N - F = 9 × 10⁹ × (1 × 10^-6 × 2 × 10^-6)/(1²) = 18 N
Q2: B) Quarter - Force varies as 1/r², doubling r reduces force by 4.
Q3: C) 10 N - Force in vacuum = (9 × 10⁹ × 3 × 10^-6 × 3 × 10^-6)/(0.3²) = 30 N, in medium F = 30/3 = 10 N

Memory Aids

Force F ∝ |q₁ q₂|/r²
Like charges repel, unlike attract
Force in dielectric = Force in vacuum/K
k = 9 × 10⁹, ε₀ = 8.854 × 10^-12

Force Direction for Charges

For like charges (q₁ q₂ > 0), force vectors point away (repulsion).
For unlike charges (q₁ q₂ < 0), force vectors point towards each other (attraction).

Diagram 5 – NEET Physics Electric Charges and Fields

Effect of Dielectric Medium

In a medium with dielectric constant K, force reduces by factor K:
[ F = F₀/K = 1/(4 π ε₀ K) q₁ q₂/r² ]
F₀ is force in vacuum.

Solved Examples

Ex 1: Calculate force between two charges 2 μC and 3 μC separated by 0.5 m in vacuum.
Use F = 1/(4 π ε₀) |q₁ q₂|/r²
F = 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/(0.5²) = 216 N

Practice Set

Q1: Two charges +1 μC and +2 μC are 1 m apart. Force magnitude is?
- A) 18 N
- B) 9 N
- C) 36 N
- D) 0 N
Q2: If distance between charges is doubled, force becomes?
- A) Half
- B) Quarter
- C) Double
- D) Four times
Q3: Two charges q₁ = 3 μC, q₂ = -3 μC separated by 0.3 m in dielectric medium K=3. Force magnitude?
- A) 90 N
- B) 30 N
- C) 10 N
- D) 270 N

Answer Key

Q1: A) 18 N - F = 9 × 10⁹ × (1 × 10^-6 × 2 × 10^-6)/(1²) = 18 N
Q2: B) Quarter - Force varies as 1/r², doubling r reduces force by 4.
Q3: C) 10 N - Force in vacuum = (9 × 10⁹ × 3 × 10^-6 × 3 × 10^-6)/(0.3²) = 30 N, in medium F = 30/3 = 10 N

Memory Aids

Force F ∝ |q₁ q₂|/r²
Like charges repel, unlike attract
Force in dielectric = Force in vacuum/K
k = 9 × 10⁹, ε₀ = 8.854 × 10^-12

Vector Form

Force on q₂ due to q₁:
[ F₂₁ = 1/(4 π ε₀) q₁ q₂/|r₂ - r₁|³ (r₂ - r₁) = 1/(4 π ε₀) q₁ q₂/|r₂₁|² r̂₂₁ = - F₁₂ ]
Obeys Newton's third law: forces are equal in magnitude and opposite in direction.
r₂₁ = r₂ - r₁ is the vector from q₁ to q₂.

Diagram 6 – NEET Physics Electric Charges and Fields

Force Direction for Charges

For like charges (q₁ q₂ > 0), force vectors point away (repulsion).
For unlike charges (q₁ q₂ < 0), force vectors point towards each other (attraction).

Diagram 7 – NEET Physics Electric Charges and Fields

Effect of Dielectric Medium

In a medium with dielectric constant K, force reduces by factor K:
[ F = F₀/K = 1/(4 π ε₀ K) q₁ q₂/r² ]
F₀ is force in vacuum.

Solved Examples

Ex 1: Calculate force between two charges 2 μC and 3 μC separated by 0.5 m in vacuum.
Use F = 1/(4 π ε₀) |q₁ q₂|/r²
F = 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/(0.5²) = 216 N

Practice Set

Q1: Two charges +1 μC and +2 μC are 1 m apart. Force magnitude is?
- A) 18 N
- B) 9 N
- C) 36 N
- D) 0 N
Q2: If distance between charges is doubled, force becomes?
- A) Half
- B) Quarter
- C) Double
- D) Four times
Q3: Two charges q₁ = 3 μC, q₂ = -3 μC separated by 0.3 m in dielectric medium K=3. Force magnitude?
- A) 90 N
- B) 30 N
- C) 10 N
- D) 270 N

Answer Key

Q1: A) 18 N - F = 9 × 10⁹ × (1 × 10^-6 × 2 × 10^-6)/(1²) = 18 N
Q2: B) Quarter - Force varies as 1/r², doubling r reduces force by 4.
Q3: C) 10 N - Force in vacuum = (9 × 10⁹ × 3 × 10^-6 × 3 × 10^-6)/(0.3²) = 30 N, in medium F = 30/3 = 10 N

Memory Aids

Force F ∝ |q₁ q₂|/r²
Like charges repel, unlike attract
Force in dielectric = Force in vacuum/K
k = 9 × 10⁹, ε₀ = 8.854 × 10^-12

Vector Form

Force on q₂ due to q₁:
[ F₂₁ = 1/(4 π ε₀) q₁ q₂/|r₂ - r₁|³ (r₂ - r₁) = 1/(4 π ε₀) q₁ q₂/|r₂₁|² r̂₂₁ = - F₁₂ ]
Obeys Newton's third law: forces are equal in magnitude and opposite in direction.
r₂₁ = r₂ - r₁ is the vector from q₁ to q₂.

Diagram 8 – NEET Physics Electric Charges and Fields

Force Direction for Charges

For like charges (q₁ q₂ > 0), force vectors point away (repulsion).
For unlike charges (q₁ q₂ < 0), force vectors point towards each other (attraction).

Diagram 9 – NEET Physics Electric Charges and Fields

Effect of Dielectric Medium

In a medium with dielectric constant K, force reduces by factor K:
[ F = F₀/K = 1/(4 π ε₀ K) q₁ q₂/r² ]
F₀ is force in vacuum.

Solved Examples

Ex 1: Calculate force between two charges 2 μC and 3 μC separated by 0.5 m in vacuum.
Use F = 1/(4 π ε₀) |q₁ q₂|/r²
F = 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/(0.5²) = 216 N

Practice Set

Q1: Two charges +1 μC and +2 μC are 1 m apart. Force magnitude is?
- A) 18 N
- B) 9 N
- C) 36 N
- D) 0 N
Q2: If distance between charges is doubled, force becomes?
- A) Half
- B) Quarter
- C) Double
- D) Four times
Q3: Two charges q₁ = 3 μC, q₂ = -3 μC separated by 0.3 m in dielectric medium K=3. Force magnitude?
- A) 90 N
- B) 30 N
- C) 10 N
- D) 270 N

Answer Key

Q1: A) 18 N - F = 9 × 10⁹ × (1 × 10^-6 × 2 × 10^-6)/(1²) = 18 N
Q2: B) Quarter - Force varies as 1/r², doubling r reduces force by 4.
Q3: C) 10 N - Force in vacuum = (9 × 10⁹ × 3 × 10^-6 × 3 × 10^-6)/(0.3²) = 30 N, in medium F = 30/3 = 10 N

Memory Aids

Force F ∝ |q₁ q₂|/r²
Like charges repel, unlike attract
Force in dielectric = Force in vacuum/K
k = 9 × 10⁹, ε₀ = 8.854 × 10^-12

Fundamental Concept

Force between two point charges \(q_1\) and \(q_2\) separated by distance \(r\) in vacuum:
\[ F = k \frac{|q_1 q_2|}{r^2} = \frac{1}{4 \pi \epsilon_0} \frac{|q_1 q_2|}{r^2} \]
\(\epsilon_0\) = permittivity of free space = \(8.854 \times 10^{-12} \text{ C}^2 \text{N}^{-1} \text{m}^{-2}\)
\(k = \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \text{ Nm}^2 \text{C}^{-2}\)
Force is attractive if charges are unlike, repulsive if like.

Vector Form

Force on \(q_2\) due to \(q_1\):
\[ \vec{F}_{21} = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{|\vec{r}_2 - \vec{r}_1|^3} (\vec{r}_2 - \vec{r}_1) = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{|\vec{r}_{21}|^2} \hat{r}_{21} = - \vec{F}_{12} \]
Obeys Newton's third law: forces are equal in magnitude and opposite in direction.
\(\vec{r}_{21} = \vec{r}_2 - \vec{r}_1\) is the vector from \(q_1\) to \(q_2\).

Diagram 10 – NEET Physics Electric Charges and Fields

Force Direction for Charges

For like charges (\(q_1 q_2 > 0\)), force vectors point away (repulsion).
For unlike charges (\(q_1 q_2 < 0\)), force vectors point towards each other (attraction).

Diagram 11 – NEET Physics Electric Charges and Fields

Effect of Dielectric Medium

In a medium with dielectric constant \(K\), force reduces by factor \(K\):
\[ F = \frac{F_0}{K} = \frac{1}{4 \pi \epsilon_0 K} \frac{q_1 q_2}{r^2} \]
\(F_0\) is force in vacuum.

Solved Examples

Ex 1: Calculate force between two charges \(2 \mu C\) and \(3 \mu C\) separated by 0.5 m in vacuum.
Use \(F = \frac{1}{4 \pi \epsilon_0} \frac{|q_1 q_2|}{r^2}\)
\(F = 9 \times 10^9 \times \frac{2 \times 10^{-6} \times 3 \times 10^{-6}}{0.5^2} = 216 \text{ N}\)

Practice Set

Q1: Two charges \(+1 \mu C\) and \(+2 \mu C\) are 1 m apart. Force magnitude is?
- A) 18 N
- B) 9 N
- C) 36 N
- D) 0 N
Q2: If distance between charges is doubled, force becomes?
- A) Half
- B) Quarter
- C) Double
- D) Four times
Q3: Two charges \(q_1 = 3 \mu C\), \(q_2 = -3 \mu C\) separated by 0.3 m in dielectric medium \(K=3\). Force magnitude?
- A) 90 N
- B) 30 N
- C) 10 N
- D) 270 N

Answer Key

Q1: A) 18 N - \(F = 9 \times 10^9 \times \frac{1 \times 10^{-6} \times 2 \times 10^{-6}}{1^2} = 18 \text{ N}\)
Q2: B) Quarter - Force varies as \(1/r^2\), doubling \(r\) reduces force by 4.
Q3: C) 10 N - Force in vacuum \(= \frac{9 \times 10^9 \times 3 \times 10^{-6} \times 3 \times 10^{-6}}{0.3^2} = 30 \text{ N}\), in medium \(F = 30/3 = 10 \text{ N}\)

Memory Aids

Force \(F \propto \frac{|q_1 q_2|}{r^2}\)
Like charges repel, unlike attract
Force in dielectric \(= \frac{\text{Force in vacuum}}{K}\)
\(k = 9 \times 10^9\), \(\epsilon_0 = 8.854 \times 10^{-12}\)

Forces between Multiple Charges

Principle of Superposition

Force on a charge due to multiple charges is the vector sum of individual forces.
Individual forces remain unaffected by other charges.
Mathematically, vec{F}₁ = vec{F}₁₂ + vec{F}₁₃ + ... + vec{F}_1n.
Each force vec{F}_1i is calculated using Coulomb's law: vec{F}_1i = 1/(4πϵ₀) q₁q_i/r_1i² r̂_1i.
Unit vector r̂_1i points from charge q₁ to q_i.

Vector Addition of Forces

Charges q₁, q₂, q₃ have position vectors r₁, r₂, r₃ from reference point O.
Vectors r₁₂ = r₂ - r₁ and r₁₃ = r₃ - r₁ represent directions and distances.
Forces vec{F}₁₂ and vec{F}₁₃ act on q₁ due to q₂ and q₃ respectively.
Total force on q₁ is vec{F}₁ = vec{F}₁₂ + vec{F}₁₃.

Generalization for n Charges

For n charges, net force on q₁ is sum of forces from all other charges.
Formula: vec{F}₁ = ∑_i=2ⁿ vec{F}_1i = 1/(4πϵ₀) ∑_i=2ⁿ q₁q_i/r_1i² r̂_1i.
Each force vector is calculated independently and then summed vectorially.

Electric Field from Multiple Charges

Net force on test charge q₀ near charges q₁, q₂, ..., q_n:
vec{F} = vec{F}₁ + vec{F}₂ + ... + vec{F}_n
Net electric field at test charge:
vec{E} = vec{F}₁/q₀ + vec{F}₂/q₀ + ... + vec{F}_n/q₀ = vec{E}₁ + vec{E}₂ + ... + vec{E}_n
Formula:
[ vec{E}(vec{r}) = 1/(4 π ϵ₀) ∑_i=1ⁿ q_i/r_i² r̂_{p_i} ]
r̂_{p_i} points from the i^th charge to point P.

Physical Significance

Electric field is a property of the charge system, independent of the test charge.
Electric field is a vector quantity (like force).
Changes in electric field propagate at the speed of light due to electromagnetic waves.
Electric field is detected by forces on charges and can transport energy.

Solved Examples

Ex 1: Calculate electric field at distance r = 0.5 m from a point charge q = 2 × 10^-6 C.
Step 1: Use formula E = 1/(4 π ϵ₀) q/r²
Step 2: Substitute 1/(4 π ϵ₀) = 9 × 10⁹ N m²/C²
Step 3: E = 9 × 10⁹ × (2 × 10^-6)/(0.5)² = 7.2 × 10⁴ N/C

Practice Set

Q1: Electric field due to a charge +3 μC at 2 m is:
A) 6.75 × 10³ N/C
B) 2.25 × 10³ N/C
C) 1.5 × 10³ N/C
D) 9 × 10³ N/C
Q2: Net electric field at point P due to two charges +q and -q placed 4 m apart, P is midpoint. The field magnitude is:
A) Zero
B) 1/(4 π ϵ₀) 2q/(2)²
C) 1/(4 π ϵ₀) q/4
D) 1/(4 π ϵ₀) q/2
Q3: Four charges q₁ = +1 μC, q₂ = -1 μC, q₃ = +2 μC, q₄ = -2 μC are placed at equal distances 1 m from point P. Net electric field at P is:
A) Zero
B) 9 × 10³ N/C
C) 18 × 10³ N/C
D) 36 × 10³ N/C

Answer Key

Q1: A
Calculation: E = 9 × 10⁹ × (3 × 10^-6)/(2²) = 6.75 × 10³ N/C
Q2: A
Fields due to equal and opposite charges at midpoint cancel out.
Q3: A
Equal positive and negative charges symmetrically placed produce zero net field.

Memory Aids

Electric field formula: E = k q/r², where k = 9 × 10⁹ N m²/C².
Electric field vectors add like forces (vector sum).
Field direction: away from positive, toward negative charges.

Solved Examples

Ex 1

Given charges q₁ = 2μC, q₂ = 3μC at distance r₁₂ = 0.5m.
Calculate force vec{F}₁₂ = 1/(4πϵ₀) q₁q₂/r₁₂² r̂₁₂.
Use k = 9 × 10⁹ Nm²/C², F = 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/(0.5²) = 0.216 N.
Direction along r̂₁₂.

Ex 2

Three charges q₁, q₂, q₃ at known positions.
Calculate vec{F}₁₂ and vec{F}₁₃ separately.
Add vectors to find net force vec{F}₁ = vec{F}₁₂ + vec{F}₁₃.

Practice Set

Q1: Two charges +2μC and +3μC are 1 m apart. Force magnitude is?
- A) 5.4 N
- B) 6.0 N
- C) 10.8 N
- D) 12.0 N
Q2: If q₁ = +1μC, q₂ = -1μC, r = 0.5 m, force is?
- A) Attractive, 36 N
- B) Repulsive, 36 N
- C) Attractive, 18 N
- D) Repulsive, 18 N
Q3: Three charges q₁ = 1μC, q₂ = 2μC, q₃ = -1μC at 1 m apart. Net force on q₁ is?
- A) Zero
- B) Directed towards q₃
- C) Directed towards q₂
- D) Directed away from both

Answer Key

Q1: C) 10.8 N - Use Coulomb's law: F = k q₁q₂/r² = 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/(1²) = 10.8 N.
Q2: A) Attractive, 36 N - Opposite charges attract; force magnitude calculated similarly.
Q3: B) Directed towards q₃ - Negative charge q₃ attracts q₁ stronger than q₂ repels due to charge values.

Memory Aids

Superposition: Add forces vectorially, independent of other charges.
Coulomb's law: F = k q₁q₂/r².
Unit vector r̂ shows force direction.
Opposite charges attract, like charges repel.

Principle of Superposition

Force on a charge due to multiple charges is the vector sum of individual forces.
Individual forces remain unaffected by other charges.
Mathematically, \( \vec{F}_1 = \vec{F}_{12} + \vec{F}_{13} + ... + \vec{F}_{1n} \).
Each force \( \vec{F}_{1i} \) is calculated using Coulomb's law: \( \vec{F}_{1i} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_i}{r_{1i}^2} \hat{r}_{1i} \).
Unit vector \( \hat{r}_{1i} \) points from charge \( q_1 \) to \( q_i \).

Vector Addition of Forces

Charges \( q_1, q_2, q_3 \) have position vectors \( \vec{r}_1, \vec{r}_2, \vec{r}_3 \) from reference point \( O \).
Vectors \( \vec{r}_{12} = \vec{r}_2 - \vec{r}_1 \) and \( \vec{r}_{13} = \vec{r}_3 - \vec{r}_1 \) represent directions and distances.
Forces \( \vec{F}_{12} \) and \( \vec{F}_{13} \) act on \( q_1 \) due to \( q_2 \) and \( q_3 \) respectively.
Total force on \( q_1 \) is \( \vec{F}_1 = \vec{F}_{12} + \vec{F}_{13} \).

Generalization for n Charges

For \( n \) charges, net force on \( q_1 \) is sum of forces from all other charges.
Formula: \( \vec{F}_1 = \sum_{i=2}^n \vec{F}_{1i} = \frac{1}{4\pi\epsilon_0} \sum_{i=2}^n \frac{q_1 q_i}{r_{1i}^2} \hat{r}_{1i} \).
Each force vector is calculated independently and then summed vectorially.

Electric Field from Multiple Charges

Electric field at a point due to multiple charges is vector sum of fields from each charge.
Electric field due to charge \( q_i \) at distance \( r_{1i} \) is \( \vec{E}_{1i} = \frac{1}{4\pi\epsilon_0} \frac{q_i}{r_{1i}^2} \hat{r}_{1i} \).
Total electric field: \( \vec{E} = \sum_{i=1}^n \vec{E}_i \).
Direction given by unit vectors \( \hat{r}_{1i} \).

Solved Examples

Ex 1

Given charges \( q_1 = 2\mu C, q_2 = 3\mu C \) at distance \( r_{12} = 0.5m \).
Calculate force \( \vec{F}_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}^2} \hat{r}_{12} \).
Use \( k = 9 \times 10^9 Nm^2/C^2 \), \( F = 9 \times 10^9 \times \frac{2 \times 10^{-6} \times 3 \times 10^{-6}}{0.5^2} = 0.216 N \).
Direction along \( \hat{r}_{12} \).

Ex 2

Three charges \( q_1, q_2, q_3 \) at known positions.
Calculate \( \vec{F}_{12} \) and \( \vec{F}_{13} \) separately.
Add vectors to find net force \( \vec{F}_1 = \vec{F}_{12} + \vec{F}_{13} \).

Practice Set

Q1: Two charges \( +2\mu C \) and \( +3\mu C \) are 1 m apart. Force magnitude is?
- A) 5.4 N
- B) 6.0 N
- C) 10.8 N
- D) 12.0 N
Q2: If \( q_1 = +1\mu C, q_2 = -1\mu C, r = 0.5 m \), force is?
- A) Attractive, 36 N
- B) Repulsive, 36 N
- C) Attractive, 18 N
- D) Repulsive, 18 N
Q3: Three charges \( q_1 = 1\mu C, q_2 = 2\mu C, q_3 = -1\mu C \) at 1 m apart. Net force on \( q_1 \) is?
- A) Zero
- B) Directed towards \( q_3 \)
- C) Directed towards \( q_2 \)
- D) Directed away from both

Answer Key

Q1: C) 10.8 N - Use Coulomb's law: \( F = k \frac{q_1 q_2}{r^2} = 9 \times 10^9 \times \frac{2 \times 10^{-6} \times 3 \times 10^{-6}}{1^2} = 10.8 N \).
Q2: A) Attractive, 36 N - Opposite charges attract; force magnitude calculated similarly.
Q3: B) Directed towards \( q_3 \) - Negative charge \( q_3 \) attracts \( q_1 \) stronger than \( q_2 \) repels due to charge values.

Memory Aids

Superposition: Add forces vectorially, independent of other charges.
Coulomb's law: \( F = k \frac{q_1 q_2}{r^2} \).
Unit vector \( \hat{r} \) shows force direction.
Opposite charges attract, like charges repel.

Vector Addition of Forces

Charges q₁, q₂, q₃ have position vectors r₁, r₂, r₃ from reference point O.
Vectors r₁₂ = r₂ - r₁ and r₁₃ = r₃ - r₁ represent directions and distances.
Forces F₁₂ and F₁₃ act on q₁ due to q₂ and q₃ respectively.
Total force on q₁ is F₁ = F₁₂ + F₁₃.

Generalization for n Charges

For n charges, net force on q₁ is sum of forces from all other charges.
Formula: F₁ = ∑_i=2ⁿ F_1i = 1/(4π ε₀) ∑_i=2ⁿ q₁ q_i/r_1i² r̂_1i.
Each force vector is calculated independently and then summed vectorially.

Electric Field from Multiple Charges

Electric field at a point due to multiple charges is vector sum of fields from each charge.
Electric field due to charge q_i at distance r_1i is E_1i = 1/(4π ε₀) q_i/r_1i² r̂_1i.
Total electric field: E = ∑_i=1ⁿ E_i.
Direction given by unit vectors r̂_1i.

Solved Examples

Ex 1

Given charges q₁ = 2μC, q₂ = 3μC at distance r₁₂ = 0.5m.
Calculate force F₁₂ = 1/(4π ε₀) q₁ q₂/r₁₂² r̂₁₂.
Use k = 9 × 10⁹ Nm²/C², F = 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/(0.5²) = 0.216 N.
Direction along r̂₁₂.

Ex 2

Three charges q₁, q₂, q₃ at known positions.
Calculate F₁₂ and F₁₃ separately.
Add vectors to find net force F₁ = F₁₂ + F₁₃.

Practice Set

Q1: Two charges +2μC and +3μC are 1 m apart. Force magnitude is?
- A) 5.4 N
- B) 6.0 N
- C) 10.8 N
- D) 12.0 N
Q2: If q₁ = +1μC, q₂ = -1μC, r = 0.5 m, force is?
- A) Attractive, 36 N
- B) Repulsive, 36 N
- C) Attractive, 18 N
- D) Repulsive, 18 N
Q3: Three charges q₁ = 1μC, q₂ = 2μC, q₃ = -1μC at 1 m apart. Net force on q₁ is?
- A) Zero
- B) Directed towards q₃
- C) Directed towards q₂
- D) Directed away from both

Answer Key

Q1: C) 10.8 N - Use Coulomb's law: F = k q₁ q₂/r² = 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/(1²) = 10.8 N.
Q2: A) Attractive, 36 N - Opposite charges attract; force magnitude calculated similarly.
Q3: B) Directed towards q₃ - Negative charge q₃ attracts q₁ stronger than q₂ repels due to charge values.

Memory Aids

Superposition: Add forces vectorially, independent of other charges.
Coulomb's law: F = k q₁ q₂/r².
Unit vector r̂ shows force direction.
Opposite charges attract, like charges repel.

Generalization for n Charges

For [32m[32m [0m[0mn [32m[32m[0m[0mcharges, net force on [32m[32m [0m[0mq_1 [32m[32m[0m[0mis sum of forces from all other charges.
Formula: [32m[32m [0m[0mvec{F}_1 = [32m[32m [0m[0msum_{i=2}^n [32m[32m [0m[0mvec{F}_{1i} = [32m[32m [0m[0mfrac{1}{4pi epsilon_0} [32m[32m [0m[0msum_{i=2}^n [32m[32m [0m[0mfrac{q_1 q_i}{r_{1i}^2} hat{r}_{1i} [32m[32m[0m[0m.
Each force vector is calculated independently and then summed vectorially.

Electric Field from Multiple Charges

Electric field at a point due to multiple charges is vector sum of fields from each charge.
Electric field due to charge [32m[32m [0m[0mq_i [32m[32m[0m[0mat distance [32m[32m [0m[0mr_{1i} [32m[32m[0m[0mis [32m[32m [0m[0mE_{1i} = [32m[32m [0m[0mfrac{1}{4pi epsilon_0} [32m[32m [0m[0mfrac{q_i}{r_{1i}^2} hat{r}_{1i} [32m[32m[0m[0m.
Total electric field: [32m[32m [0m[0mE = [32m[32m [0m[0msum_{i=1}^n [32m[32m [0m[0mE_i [32m[32m[0m[0m.
Direction given by unit vectors [32m[32m [0m[0mhat{r}_{1i} [32m[32m[0m[0m.

Solved Examples

Ex 1

Given charges [32m[32m [0m[0mq_1 = 2μC, q_2 = 3μC [32m[32m[0m[0mat distance [32m[32m [0m[0mr_{12} = 0.5m[32m[32m[0m[0m.
Calculate force [32m[32m [0m[0mvec{F}_{12} = [32m[32m [0m[0mfrac{1}{4pi epsilon_0} [32m[32m [0m[0mfrac{q_1 q_2}{r_{12}^2} hat{r}_{12} [32m[32m[0m[0m.
Use k = 9 × 10⁹ Nm²/C², F = 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/(0.5²) = 0.216 N.
Direction along [32m[32m [0m[0mhat{r}_{12} [32m[32m[0m[0m.

Ex 2

Three charges [32m[32m [0m[0mq_1, q_2, q_3 [32m[32m[0m[0mat known positions.
Calculate [32m[32m [0m[0mvec{F}_{12} [32m[32m [0m[0mand [32m[32m [0m[0mvec{F}_{13} [32m[32m[0m[0mseparately.
Add vectors to find net force [32m[32m [0m[0mvec{F}_1 = [32m[32m [0m[0mvec{F}_{12} + [32m[32m [0m[0mvec{F}_{13} [32m[32m[0m[0m.

Practice Set

Q1: Two charges +2μC and +3μC are 1 m apart. Force magnitude is?
- A) 5.4 N
- B) 6.0 N
- C) 10.8 N
- D) 12.0 N
Q2: If q₁ = +1μC, q₂ = -1μC, r = 0.5 m, force is?
- A) Attractive, 36 N
- B) Repulsive, 36 N
- C) Attractive, 18 N
- D) Repulsive, 18 N
Q3: Three charges q₁ = 1μC, q₂ = 2μC, q₃ = -1μC at 1 m apart. Net force on q₁ is?
- A) Zero
- B) Directed towards q₃
- C) Directed towards q₂
- D) Directed away from both

Answer Key

Q1: C) 10.8 N - Use Coulomb's law: F = k q₁ q₂/r² = 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/(1²) = 10.8 N.
Q2: A) Attractive, 36 N - Opposite charges attract; force magnitude calculated similarly.
Q3: B) Directed towards q₃ - Negative charge q₃ attracts q₁ stronger than q₂ repels due to charge values.

Memory Aids

Superposition: Add forces vectorially, independent of other charges.
Coulomb's law: F = k q₁ q₂/r².
Unit vector r̂ shows force direction.
Opposite charges attract, like charges repel.

Vector Addition of Forces

Charges q₁, q₂, q₃ have position vectors r₁, r₂, r₃ from reference point O.
Vectors r₁₂ = r₂ - r₁ and r₁₃ = r₃ - r₁ represent directions and distances.
Forces F₁₂ and F₁₃ act on q₁ due to q₂ and q₃ respectively.
Total force on q₁ is F₁ = F₁₂ + F₁₃.

Generalization for n Charges

For n charges, net force on q₁ is sum of forces from all other charges.
Formula: F₁ = ∑_i=2ⁿ F_1i = 1/(4π ε₀) ∑_i=2ⁿ q₁ q_i/r_1i² r̂_1i.
Each force vector is calculated independently and then summed vectorially.

Electric Field from Multiple Charges

Electric field at a point due to multiple charges is vector sum of fields from each charge.
Electric field due to charge q_i at distance r_1i is E_1i = 1/(4π ε₀) q_i/r_1i² r̂_1i.
Total electric field: E = ∑_i=1ⁿ E_i.
Direction given by unit vectors r̂_1i.

Solved Examples

Ex 1

Given charges q₁ = 2μC, q₂ = 3μC at distance r₁₂ = 0.5m.
Calculate force F₁₂ = 1/(4π ε₀) q₁ q₂/r₁₂² r̂₁₂.
Use k = 9 × 10⁹ Nm²/C², F = 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/(0.5²) = 0.216 N.
Direction along r̂₁₂.

Ex 2

Three charges q₁, q₂, q₃ at known positions.
Calculate F₁₂ and F₁₃ separately.
Add vectors to find net force F₁ = F₁₂ + F₁₃.

Practice Set

Q1: Two charges +2μC and +3μC are 1 m apart. Force magnitude is?
- A) 5.4 N
- B) 6.0 N
- C) 10.8 N
- D) 12.0 N
Q2: If q₁ = +1μC, q₂ = -1μC, r = 0.5 m, force is?
- A) Attractive, 36 N
- B) Repulsive, 36 N
- C) Attractive, 18 N
- D) Repulsive, 18 N
Q3: Three charges q₁ = 1μC, q₂ = 2μC, q₃ = -1μC at 1 m apart. Net force on q₁ is?
- A) Zero
- B) Directed towards q₃
- C) Directed towards q₂
- D) Directed away from both

Answer Key

Q1: C) 10.8 N - Use Coulomb's law: F = k q₁ q₂/r² = 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/(1²) = 10.8 N.
Q2: A) Attractive, 36 N - Opposite charges attract; force magnitude calculated similarly.
Q3: B) Directed towards q₃ - Negative charge q₃ attracts q₁ stronger than q₂ repels due to charge values.

Memory Aids

Superposition: Add forces vectorially, independent of other charges.
Coulomb's law: F = k q₁ q₂/r².
Unit vector r̂ shows force direction.
Opposite charges attract, like charges repel.

Vector Addition of Forces

Charges q₁, q₂, q₃ have position vectors r₁, r₂, r₃ from reference point O.
Vectors r₁₂ = r₂ - r₁ and r₁₃ = r₃ - r₁ represent directions and distances.
Forces F₁₂ and F₁₃ act on q₁ due to q₂ and q₃ respectively.
Total force on q₁ is F₁ = F₁₂ + F₁₃.

Generalization for n Charges

For n charges, net force on q₁ is sum of forces from all other charges.
Formula: F₁ = ∑_i=2ⁿ F_1i = 1/(4π ε₀) ∑_i=2ⁿ q₁ q_i/r_1i² r̂_1i.
Each force vector is calculated independently and then summed vectorially.

Electric Field from Multiple Charges

Electric field at a point due to multiple charges is vector sum of fields from each charge.
Electric field due to charge q_i at distance r_1i is E_1i = 1/(4π ε₀) q_i/r_1i² r̂_1i.
Total electric field: E = ∑_i=1ⁿ E_i.
Direction given by unit vectors r̂_1i.

Solved Examples

Ex 1

Given charges q₁ = 2μC, q₂ = 3μC at distance r₁₂ = 0.5m.
Calculate force F₁₂ = 1/(4π ε₀) q₁ q₂/r₁₂² r̂₁₂.
Use k = 9 × 10⁹ Nm²/C², F = 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/(0.5²) = 0.216 N.
Direction along r̂₁₂.

Ex 2

Three charges q₁, q₂, q₃ at known positions.
Calculate F₁₂ and F₁₃ separately.
Add vectors to find net force F₁ = F₁₂ + F₁₃.

Practice Set

Q1: Two charges +2μC and +3μC are 1 m apart. Force magnitude is?
- A) 5.4 N
- B) 6.0 N
- C) 10.8 N
- D) 12.0 N
Q2: If q₁ = +1μC, q₂ = -1μC, r = 0.5 m, force is?
- A) Attractive, 36 N
- B) Repulsive, 36 N
- C) Attractive, 18 N
- D) Repulsive, 18 N
Q3: Three charges q₁ = 1μC, q₂ = 2μC, q₃ = -1μC at 1 m apart. Net force on q₁ is?
- A) Zero
- B) Directed towards q₃
- C) Directed towards q₂
- D) Directed away from both

Answer Key

Q1: C) 10.8 N - Use Coulomb's law: F = k q₁ q₂/r² = 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/(1²) = 10.8 N.
Q2: A) Attractive, 36 N - Opposite charges attract; force magnitude calculated similarly.
Q3: B) Directed towards q₃ - Negative charge q₃ attracts q₁ stronger than q₂ repels due to charge values.

Memory Aids

Superposition: Add forces vectorially, independent of other charges.
Coulomb's law: F = k q₁ q₂/r².
Unit vector r̂ shows force direction.
Opposite charges attract, like charges repel.

Principle of Superposition

Force on a charge due to multiple charges is the vector sum of individual forces.
Individual forces remain unaffected by other charges.
Mathematically, \( \vec{F}_1 = \vec{F}_{12} + \vec{F}_{13} + ... + \vec{F}_{1n} \).
Each force \( \vec{F}_{1i} \) is calculated using Coulomb's law: \( \vec{F}_{1i} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_i}{r_{1i}^2} \hat{r}_{1i} \).
Unit vector \( \hat{r}_{1i} \) points from charge \( q_1 \) to \( q_i \).

Vector Addition of Forces

Charges \( q_1, q_2, q_3 \) have position vectors \( \vec{r}_1, \vec{r}_2, \vec{r}_3 \) from reference point \( O \).
Vectors \( \vec{r}_{12} = \vec{r}_2 - \vec{r}_1 \) and \( \vec{r}_{13} = \vec{r}_3 - \vec{r}_1 \) represent directions and distances.
Forces \( \vec{F}_{12} \) and \( \vec{F}_{13} \) act on \( q_1 \) due to \( q_2 \) and \( q_3 \) respectively.
Total force on \( q_1 \) is \( \vec{F}_1 = \vec{F}_{12} + \vec{F}_{13} \).

Generalization for n Charges

For \( n \) charges, net force on \( q_1 \) is sum of forces from all other charges.
Formula: \( \vec{F}_1 = \sum_{i=2}^n \vec{F}_{1i} = \frac{1}{4\pi\epsilon_0} \sum_{i=2}^n \frac{q_1 q_i}{r_{1i}^2} \hat{r}_{1i} \).
Each force vector is calculated independently and then summed vectorially.

Electric Field from Multiple Charges

Electric field at a point due to multiple charges is vector sum of fields from each charge.
Electric field due to charge \( q_i \) at distance \( r_{1i} \) is \( \vec{E}_{1i} = \frac{1}{4\pi\epsilon_0} \frac{q_i}{r_{1i}^2} \hat{r}_{1i} \).
Total electric field: \( \vec{E} = \sum_{i=1}^n \vec{E}_i \).
Direction given by unit vectors \( \hat{r}_{1i} \).

Solved Examples

Ex 1

Given charges \( q_1 = 2\mu C, q_2 = 3\mu C \) at distance \( r_{12} = 0.5m \).
Calculate force \( \vec{F}_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}^2} \hat{r}_{12} \).
Use \( k = 9 \times 10^9 Nm^2/C^2 \), \( F = 9 \times 10^9 \times \frac{2 \times 10^{-6} \times 3 \times 10^{-6}}{0.5^2} = 0.216 N \).
Direction along \( \hat{r}_{12} \).

Ex 2

Three charges \( q_1, q_2, q_3 \) at known positions.
Calculate \( \vec{F}_{12} \) and \( \vec{F}_{13} \) separately.
Add vectors to find net force \( \vec{F}_1 = \vec{F}_{12} + \vec{F}_{13} \).

Practice Set

Q1: Two charges \( +2\mu C \) and \( +3\mu C \) are 1 m apart. Force magnitude is?
- A) 5.4 N
- B) 6.0 N
- C) 10.8 N
- D) 12.0 N
Q2: If \( q_1 = +1\mu C, q_2 = -1\mu C, r = 0.5 m \), force is?
- A) Attractive, 36 N
- B) Repulsive, 36 N
- C) Attractive, 18 N
- D) Repulsive, 18 N
Q3: Three charges \( q_1 = 1\mu C, q_2 = 2\mu C, q_3 = -1\mu C \) at 1 m apart. Net force on \( q_1 \) is?
- A) Zero
- B) Directed towards \( q_3 \)
- C) Directed towards \( q_2 \)
- D) Directed away from both

Answer Key

Q1: C) 10.8 N - Use Coulomb's law: \( F = k \frac{q_1 q_2}{r^2} = 9 \times 10^9 \times \frac{2 \times 10^{-6} \times 3 \times 10^{-6}}{1^2} = 10.8 N \).
Q2: A) Attractive, 36 N - Opposite charges attract; force magnitude calculated similarly.
Q3: B) Directed towards \( q_3 \) - Negative charge \( q_3 \) attracts \( q_1 \) stronger than \( q_2 \) repels due to charge values.

Memory Aids

Superposition: Add forces vectorially, independent of other charges.
Coulomb's law: \( F = k \frac{q_1 q_2}{r^2} \).
Unit vector \( \hat{r} \) shows force direction.
Opposite charges attract, like charges repel.

Electric Field

Definition and Formula

Electric field is the region around a charge where its force is felt by a unit positive test charge.
Electric field vec{E} due to a point charge q at position vec{r}:
[ vec{E} = 1/(4 π ϵ₀) q/|vec{r}|² r̂ = vec{F}/q₀ ]
q₀ is the positive test charge.

Electric Field from Multiple Charges

Net force on test charge q₀ near charges q₁, q₂, ..., q_n:
F = F₁ + F₂ + ... + F_n
Net electric field at test charge:
E = F₁/q₀ + F₂/q₀ + ... + F_n/q₀ = E₁ + E₂ + ... + E_n
Formula:
[ E(r) = 1/(4 π ε₀) ∑_i=1ⁿ q_i/r_i² r̂_{p_i} ]
r̂_{p_i} points from the i^th charge to point P.

Physical Significance

Electric field is a property of the charge system, independent of the test charge.
Electric field is a vector quantity (like force).
Changes in electric field propagate at the speed of light due to electromagnetic waves.
Electric field is detected by forces on charges and can transport energy.

Solved Examples

Ex 1: Calculate electric field at distance r = 0.5 m from a point charge q = 2 × 10^-6 C.
Step 1: Use formula E = 1/(4 π ε₀) q/r²
Step 2: Substitute 1/(4 π ε₀) = 9 × 10⁹ N m²/C²
Step 3: E = 9 × 10⁹ × (2 × 10^-6)/(0.5)² = 7.2 × 10⁴ N/C

Practice Set

Q1: Electric field due to a charge +3 μC at 2 m is:
A) 6.75 × 10³ N/C
B) 2.25 × 10³ N/C
C) 1.5 × 10³ N/C
D) 9 × 10³ N/C
Q2: Net electric field at point P due to two charges +q and -q placed 4 m apart, P is midpoint. The field magnitude is:
A) Zero
B) 1/(4 π ε₀) 2q/(2)²
C) 1/(4 π ε₀) q/4
D) 1/(4 π ε₀) q/2
Q3: Four charges q₁ = +1 μC, q₂ = -1 μC, q₃ = +2 μC, q₄ = -2 μC are placed at equal distances 1 m from point P. Net electric field at P is:
A) Zero
B) 9 × 10³ N/C
C) 18 × 10³ N/C
D) 36 × 10³ N/C

Answer Key

Q1: A
Calculation: E = 9 × 10⁹ × (3 × 10^-6)/(2²) = 6.75 × 10³ N/C
Q2: A
Fields due to equal and opposite charges at midpoint cancel out.
Q3: A
Equal positive and negative charges symmetrically placed produce zero net field.

Memory Aids

Electric field formula: E = k q/r², where k = 9 × 10⁹ N m²/C².
Electric field vectors add like forces (vector sum).
Field direction: away from positive, toward negative charges.

Definition and Formula

Electric field is the region around a charge where its force is felt by a unit positive test charge.
Electric field E due to a point charge q at position vector r:
E = (1/4 π ε₀) q/|r|² r̂ = F/q₀
q₀ is the positive test charge.

Electric Field from Multiple Charges

Net force on test charge q₀ near charges q₁, q₂, ..., q_n:
F = F₁ + F₂ + ... + F_n
Net electric field at test charge:
E = F₁/q₀ + F₂/q₀ + ... + F_n/q₀ = E₁ + E₂ + ... + E_n
Formula:
[ E(r) = 1/(4 π ε₀) ∑_i=1ⁿ q_i/r_i² r̂_{p_i} ]
r̂_{p_i} points from the i^th charge to point P.

Physical Significance

Electric field is a property of the charge system, independent of the test charge.
Electric field is a vector quantity (like force).
Changes in electric field propagate at the speed of light due to electromagnetic waves.
Electric field is detected by forces on charges and can transport energy.

Solved Examples

Ex 1: Calculate electric field at distance r = 0.5 m from a point charge q = 2 × 10^-6 C.
Step 1: Use formula E = 1/(4 π ε₀) q/r²
Step 2: Substitute 1/(4 π ε₀) = 9 × 10⁹ N m²/C²
Step 3: E = 9 × 10⁹ × (2 × 10^-6)/(0.5)² = 7.2 × 10⁴ N/C

Practice Set

Q1: Electric field due to a charge +3 μC at 2 m is:
A) 6.75 × 10³ N/C
B) 2.25 × 10³ N/C
C) 1.5 × 10³ N/C
D) 9 × 10³ N/C
Q2: Net electric field at point P due to two charges +q and -q placed 4 m apart, P is midpoint. The field magnitude is:
A) Zero
B) 1/(4 π ε₀) 2q/(2)²
C) 1/(4 π ε₀) q/4
D) 1/(4 π ε₀) q/2
Q3: Four charges q₁ = +1 μC, q₂ = -1 μC, q₃ = +2 μC, q₄ = -2 μC are placed at equal distances 1 m from point P. Net electric field at P is:
A) Zero
B) 9 × 10³ N/C
C) 18 × 10³ N/C
D) 36 × 10³ N/C

Answer Key

Q1: A
Calculation: E = 9 × 10⁹ × (3 × 10^-6)/(2²) = 6.75 × 10³ N/C
Q2: A
Fields due to equal and opposite charges at midpoint cancel out.
Q3: A
Equal positive and negative charges symmetrically placed produce zero net field.

Memory Aids

Electric field formula: E = k q/r², where k = 9 × 10⁹ N m²/C².
Electric field vectors add like forces (vector sum).
Field direction: away from positive, toward negative charges.

Electric Field from Multiple Charges

Net force on test charge q₀ near charges q₁, q₂, ..., q_n:
F = F₁ + F₂ + ... + F_n
Net electric field at test charge:
E = F₁/q₀ + F₂/q₀ + ... + F_n/q₀ = E₁ + E₂ + ... + E_n
Formula:
[ E(r) = 1/(4 π ε₀) ∑_i=1ⁿ q_i/r_i² r̂_{p_i} ]
r̂_{p_i} points from the i^th charge to point P.

Physical Significance

Electric field is a property of the charge system, independent of the test charge.
Electric field is a vector quantity (like force).
Changes in electric field propagate at the speed of light due to electromagnetic waves.
Electric field is detected by forces on charges and can transport energy.

Solved Examples

Ex 1: Calculate electric field at distance r = 0.5 m from a point charge q = 2 × 10^-6 C.
Step 1: Use formula E = 1/(4 π ε₀) q/r²
Step 2: Substitute 1/(4 π ε₀) = 9 × 10⁹ N m²/C²
Step 3: E = 9 × 10⁹ × (2 × 10^-6)/(0.5)² = 7.2 × 10⁴ N/C

Practice Set

Q1: Electric field due to a charge +3 μC at 2 m is:
A) 6.75 × 10³ N/C
B) 2.25 × 10³ N/C
C) 1.5 × 10³ N/C
D) 9 × 10³ N/C
Q2: Net electric field at point P due to two charges +q and -q placed 4 m apart, P is midpoint. The field magnitude is:
A) Zero
B) 1/(4 π ε₀) 2q/(2)²
C) 1/(4 π ε₀) q/4
D) 1/(4 π ε₀) q/2
Q3: Four charges q₁ = +1 μC, q₂ = -1 μC, q₃ = +2 μC, q₄ = -2 μC are placed at equal distances 1 m from point P. Net electric field at P is:
A) Zero
B) 9 × 10³ N/C
C) 18 × 10³ N/C
D) 36 × 10³ N/C

Answer Key

Q1: A
Calculation: E = 9 × 10⁹ × (3 × 10^-6)/(2²) = 6.75 × 10³ N/C
Q2: A
Fields due to equal and opposite charges at midpoint cancel out.
Q3: A
Equal positive and negative charges symmetrically placed produce zero net field.

Memory Aids

Electric field formula: E = k q/r², where k = 9 × 10⁹ N m²/C².
Electric field vectors add like forces (vector sum).
Field direction: away from positive, toward negative charges.

Definition and Formula

Electric field is the region around a charge where its force is felt by a unit positive test charge.
Electric field \( \vec{E} \) due to a point charge \( q \) at position \( \vec{r} \):
\[ \vec{E} = \frac{1}{4 \pi \epsilon_0} \frac{q}{|\vec{r}|^2} \hat{r} = \frac{\vec{F}}{q_0} \]
\( q_0 \) is the positive test charge.

Electric Field from Multiple Charges

Net force on test charge \( q_0 \) near charges \( q_1, q_2, ..., q_n \):
\( \vec{F} = \vec{F}_1 + \vec{F}_2 + ... + \vec{F}_n \)
Net electric field at test charge:
\( \vec{E} = \frac{\vec{F}_1}{q_0} + \frac{\vec{F}_2}{q_0} + ... + \frac{\vec{F}_n}{q_0} = \vec{E}_1 + \vec{E}_2 + ... + \vec{E}_n \)
Formula:
\[ \vec{E}(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \sum_{i=1}^n \frac{q_i}{r_i^2} \hat{r}_{p_i} \]
\( \hat{r}_{p_i} \) points from the \( i^{th} \) charge to point \( P \).

Physical Significance

Electric field is a property of the charge system, independent of the test charge.
Electric field is a vector quantity (like force).
Changes in electric field propagate at the speed of light due to electromagnetic waves.
Electric field is detected by forces on charges and can transport energy.

Solved Examples

Ex 1: Calculate electric field at distance \( r = 0.5 \text{ m} \) from a point charge \( q = 2 \times 10^{-6} \text{ C} \).
Step 1: Use formula \( E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} \)
Step 2: Substitute \( \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2 \)
Step 3: \( E = 9 \times 10^9 \times \frac{2 \times 10^{-6}}{(0.5)^2} = 7.2 \times 10^4 \text{ N/C} \)

Practice Set

Q1: Electric field due to a charge \( +3 \mu C \) at 2 m is:
A) \( 6.75 \times 10^3 \text{ N/C} \)
B) \( 2.25 \times 10^3 \text{ N/C} \)
C) \( 1.5 \times 10^3 \text{ N/C} \)
D) \( 9 \times 10^3 \text{ N/C} \)
Q2: Net electric field at point P due to two charges \( +q \) and \( -q \) placed 4 m apart, P is midpoint. The field magnitude is:
A) Zero
B) \( \frac{1}{4 \pi \epsilon_0} \frac{2q}{(2)^2} \)
C) \( \frac{1}{4 \pi \epsilon_0} \frac{q}{4} \)
D) \( \frac{1}{4 \pi \epsilon_0} \frac{q}{2} \)
Q3: Four charges \( q_1 = +1 \mu C, q_2 = -1 \mu C, q_3 = +2 \mu C, q_4 = -2 \mu C \) are placed at equal distances 1 m from point P. Net electric field at P is:
A) Zero
B) \( 9 \times 10^3 \text{ N/C} \)
C) \( 18 \times 10^3 \text{ N/C} \)
D) \( 36 \times 10^3 \text{ N/C} \)

Answer Key

Q1: A
Calculation: \( E = 9 \times 10^9 \times \frac{3 \times 10^{-6}}{2^2} = 6.75 \times 10^3 \text{ N/C} \)
Q2: A
Fields due to equal and opposite charges at midpoint cancel out.
Q3: A
Equal positive and negative charges symmetrically placed produce zero net field.

Memory Aids

Electric field formula: \( E = k \frac{q}{r^2} \), where \( k = 9 \times 10^9 \text{ Nm}^2/\text{C}^2 \).
Electric field vectors add like forces (vector sum).
Field direction: away from positive, toward negative charges.

Electric Field from Multiple Charges

Net force on test charge q₀ near charges q₁, q₂, ..., q_n:
F = F₁ + F₂ + ... + F_n
Net electric field at test charge:
E = F₁/q₀ + F₂/q₀ + ... + F_n/q₀ = E₁ + E₂ + ... + E_n
Formula:
[ E(r) = 1/(4 π ε₀) ∑_i=1ⁿ q_i/r_i² r̂_{p_i} ]
r̂_{p_i} points from the i^th charge to point P.

Physical Significance

Electric field is a property of the charge system, independent of the test charge.
Electric field is a vector quantity (like force).
Changes in electric field propagate at the speed of light due to electromagnetic waves.
Electric field is detected by forces on charges and can transport energy.

Solved Examples

Ex 1: Calculate electric field at distance r = 0.5 m from a point charge q = 2 × 10^-6 C.
Step 1: Use formula E = 1/(4 π ε₀) q/r²
Step 2: Substitute 1/(4 π ε₀) = 9 × 10⁹ N m²/C²
Step 3: E = 9 × 10⁹ × (2 × 10^-6)/(0.5)² = 7.2 × 10⁴ N/C

Practice Set

Q1: Electric field due to a charge +3 μC at 2 m is:
A) 6.75 × 10³ N/C
B) 2.25 × 10³ N/C
C) 1.5 × 10³ N/C
D) 9 × 10³ N/C
Q2: Net electric field at point P due to two charges +q and -q placed 4 m apart, P is midpoint. The field magnitude is:
A) Zero
B) 1/(4 π ε₀) 2q/(2)²
C) 1/(4 π ε₀) q/4
D) 1/(4 π ε₀) q/2
Q3: Four charges q₁ = +1 μC, q₂ = -1 μC, q₃ = +2 μC, q₄ = -2 μC are placed at equal distances 1 m from point P. Net electric field at P is:
A) Zero
B) 9 × 10³ N/C
C) 18 × 10³ N/C
D) 36 × 10³ N/C

Answer Key

Q1: A
Calculation: E = 9 × 10⁹ × (3 × 10^-6)/(2²) = 6.75 × 10³ N/C
Q2: A
Fields due to equal and opposite charges at midpoint cancel out.
Q3: A
Equal positive and negative charges symmetrically placed produce zero net field.

Memory Aids

Electric field formula: E = k q/r², where k = 9 × 10⁹ N m²/C².
Electric field vectors add like forces (vector sum).
Field direction: away from positive, toward negative charges.

Electric Field from Multiple Charges

Net force on test charge q₀ near charges q₁, q₂, ..., q_n:
F = F₁ + F₂ + ... + F_n
Net electric field at test charge:
E = F₁/q₀ + F₂/q₀ + ... + F_n/q₀ = E₁ + E₂ + ... + E_n
Formula:
[ E(r) = 1/(4 π ε₀) ∑_i=1ⁿ q_i/r_i² r̂_{p_i} ]
r̂_{p_i} points from the i^th charge to point P.

Physical Significance

Electric field is a property of the charge system, independent of the test charge.
Electric field is a vector quantity (like force).
Changes in electric field propagate at the speed of light due to electromagnetic waves.
Electric field is detected by forces on charges and can transport energy.

Solved Examples

Ex 1: Calculate electric field at distance r = 0.5 m from a point charge q = 2 × 10^-6 C.
Step 1: Use formula E = 1/(4 π ε₀) q/r²
Step 2: Substitute 1/(4 π ε₀) = 9 × 10⁹ N m²/C²
Step 3: E = 9 × 10⁹ × (2 × 10^-6)/(0.5)² = 7.2 × 10⁴ N/C

Practice Set

Q1: Electric field due to a charge +3 μC at 2 m is:
A) 6.75 × 10³ N/C
B) 2.25 × 10³ N/C
C) 1.5 × 10³ N/C
D) 9 × 10³ N/C
Q2: Net electric field at point P due to two charges +q and -q placed 4 m apart, P is midpoint. The field magnitude is:
A) Zero
B) 1/(4 π ε₀) 2q/(2)²
C) 1/(4 π ε₀) q/4
D) 1/(4 π ε₀) q/2
Q3: Four charges q₁ = +1 μC, q₂ = -1 μC, q₃ = +2 μC, q₄ = -2 μC are placed at equal distances 1 m from point P. Net electric field at P is:
A) Zero
B) 9 × 10³ N/C
C) 18 × 10³ N/C
D) 36 × 10³ N/C

Answer Key

Q1: A
Calculation: E = 9 × 10⁹ × (3 × 10^-6)/(2²) = 6.75 × 10³ N/C
Q2: A
Fields due to equal and opposite charges at midpoint cancel out.
Q3: A
Equal positive and negative charges symmetrically placed produce zero net field.

Memory Aids

Electric field formula: E = k q/r², where k = 9 × 10⁹ N m²/C².
Electric field vectors add like forces (vector sum).
Field direction: away from positive, toward negative charges.

Definition and Formula

Electric field is the region around a charge where its force is felt by a unit positive test charge.
Electric field \( \vec{E} \) due to a point charge \( q \) at position \( \vec{r} \):
\[ \vec{E} = \frac{1}{4 \pi \epsilon_0} \frac{q}{|\vec{r}|^2} \hat{r} = \frac{\vec{F}}{q_0} \]
\( q_0 \) is the positive test charge.

Electric Field from Multiple Charges

Net force on test charge \( q_0 \) near charges \( q_1, q_2, ..., q_n \):
\( \vec{F} = \vec{F}_1 + \vec{F}_2 + ... + \vec{F}_n \)
Net electric field at test charge:
\( \vec{E} = \frac{\vec{F}_1}{q_0} + \frac{\vec{F}_2}{q_0} + ... + \frac{\vec{F}_n}{q_0} = \vec{E}_1 + \vec{E}_2 + ... + \vec{E}_n \)
Formula:
\[ \vec{E}(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \sum_{i=1}^n \frac{q_i}{r_i^2} \hat{r}_{p_i} \]
\( \hat{r}_{p_i} \) points from the \( i^{th} \) charge to point \( P \).

Physical Significance

Electric field is a property of the charge system, independent of the test charge.
Electric field is a vector quantity (like force).
Changes in electric field propagate at the speed of light due to electromagnetic waves.
Electric field is detected by forces on charges and can transport energy.

Solved Examples

Ex 1: Calculate electric field at distance \( r = 0.5 \text{ m} \) from a point charge \( q = 2 \times 10^{-6} \text{ C} \).
Step 1: Use formula \( E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} \)
Step 2: Substitute \( \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2 \)
Step 3: \( E = 9 \times 10^9 \times \frac{2 \times 10^{-6}}{(0.5)^2} = 7.2 \times 10^4 \text{ N/C} \)

Practice Set

Q1: Electric field due to a charge \( +3 \mu C \) at 2 m is:
A) \( 6.75 \times 10^3 \text{ N/C} \)
B) \( 2.25 \times 10^3 \text{ N/C} \)
C) \( 1.5 \times 10^3 \text{ N/C} \)
D) \( 9 \times 10^3 \text{ N/C} \)
Q2: Net electric field at point P due to two charges \( +q \) and \( -q \) placed 4 m apart, P is midpoint. The field magnitude is:
A) Zero
B) \( \frac{1}{4 \pi \epsilon_0} \frac{2q}{(2)^2} \)
C) \( \frac{1}{4 \pi \epsilon_0} \frac{q}{4} \)
D) \( \frac{1}{4 \pi \epsilon_0} \frac{q}{2} \)
Q3: Four charges \( q_1 = +1 \mu C, q_2 = -1 \mu C, q_3 = +2 \mu C, q_4 = -2 \mu C \) are placed at equal distances 1 m from point P. Net electric field at P is:
A) Zero
B) \( 9 \times 10^3 \text{ N/C} \)
C) \( 18 \times 10^3 \text{ N/C} \)
D) \( 36 \times 10^3 \text{ N/C} \)

Answer Key

Q1: A
Calculation: \( E = 9 \times 10^9 \times \frac{3 \times 10^{-6}}{2^2} = 6.75 \times 10^3 \text{ N/C} \)
Q2: A
Fields due to equal and opposite charges at midpoint cancel out.
Q3: A
Equal positive and negative charges symmetrically placed produce zero net field.

Memory Aids

Electric field formula: \( E = k \frac{q}{r^2} \), where \( k = 9 \times 10^9 \text{ Nm}^2/\text{C}^2 \).
Electric field vectors add like forces (vector sum).
Field direction: away from positive, toward negative charges.

Electric Field Lines

Concept and Representation

Electric field lines map the electric field around charges.
Tangent to a field line at any point shows the direction of the net electric field.
Field lines start from positive charges and end at negative charges or infinity.
Field lines never intersect; intersection would imply multiple directions of the field.
Field lines do not form closed loops due to the conservative nature of electrostatic fields.
Density of field lines indicates field strength; closer lines mean stronger field.
Field lines are normal (perpendicular) to conductor surfaces where they start or end.

Diagram 12 – NEET Physics Electric Charges and Fields

Properties of Electric Field Lines

Continuous curves without breaks in charge-free regions.
Start from positive charges, end at negative charges or infinity.
Tangent at any point gives direction of electric field intensity.
Two field lines never cross.
Do not form closed loops.
Field strength proportional to line density.
Always perpendicular to conductor surfaces.

Dependence on Distance

Electric field strength decreases with distance from the charge.
Field lines spread out as distance increases, reducing density.
Field strength at a point is inversely proportional to the square of the distance from the charge: E ∝ 1/r².

Diagram 13 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1

Find the relative strength of the electric field at points P₁ and P₂ located at distances r₁ and r₂ from a point charge q.
Use formula: E = k|q|/r², where k = 9 × 10⁹ Nm²/C².
Calculate ratio: E₁/E₂ = r₂²/r₁².

Practice Set

Q1

What is the direction of electric field lines around a positive point charge?

A. Toward the charge
B. Away from the charge
C. Circular around the charge
D. Random directions

Q2

Which of the following is true about electric field lines?

A. They can intersect
B. They form closed loops
C. They start from positive and end at negative charges
D. They are parallel everywhere

Q3

At what distance from a point charge q is the electric field strength E reduced to one-fourth of its value at distance r?

A. 2r
B. r/2
C. 4r
D. r/4

Answer Key

Q1: B - Field lines radiate outward from positive charges.
Q2: C - Field lines start from positive and end at negative charges; they never intersect or form loops.
Q3: A - Electric field E ∝ 1/r², so doubling distance reduces field to one-fourth.

Memory Aids

"Positive push out, negative pull in" for field line directions.
Field strength E ∝ line density.
Field lines never cross = unique field direction.
Electric field E ∝ 1/r² for point charges.

Properties of Electric Field Lines

Continuous curves without breaks in charge-free regions.
Start from positive charges, end at negative charges or infinity.
Tangent at any point gives direction of electric field intensity.
Two field lines never cross.
Do not form closed loops.
Field strength proportional to line density.
Always perpendicular to conductor surfaces.

Dependence on Distance

Electric field strength decreases with distance from the charge.
Field lines spread out as distance increases, reducing density.
Field strength at a point is inversely proportional to the square of the distance from the charge: E ∝ 1/r².

Diagram 14 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1

Find the relative strength of the electric field at points P₁ and P₂ located at distances r₁ and r₂ from a point charge q.
Use formula: E = k|q|/r², where k = 9 × 10⁹ Nm²/C².
Calculate ratio: E₁/E₂ = r₂²/r₁².

Practice Set

Q1

What is the direction of electric field lines around a positive point charge?

A. Toward the charge
B. Away from the charge
C. Circular around the charge
D. Random directions

Q2

Which of the following is true about electric field lines?

A. They can intersect
B. They form closed loops
C. They start from positive and end at negative charges
D. They are parallel everywhere

Q3

At what distance from a point charge q is the electric field strength E reduced to one-fourth of its value at distance r?

A. 2r
B. r/2
C. 4r
D. r/4

Answer Key

Q1: B - Field lines radiate outward from positive charges.
Q2: C - Field lines start from positive and end at negative charges; they never intersect or form loops.
Q3: A - Electric field E ∝ 1/r², so doubling distance reduces field to one-fourth.

Memory Aids

"Positive push out, negative pull in" for field line directions.
Field strength E ∝ line density.
Field lines never cross = unique field direction.
Electric field E ∝ 1/r² for point charges.

Concept and Representation

Electric field lines map the electric field around charges.
Tangent to a field line at any point shows the direction of the net electric field.
Field lines start from positive charges and end at negative charges or infinity.
Field lines never intersect; intersection would imply multiple directions of the field.
Field lines do not form closed loops due to the conservative nature of electrostatic fields.
Density of field lines indicates field strength; closer lines mean stronger field.
Field lines are normal (perpendicular) to conductor surfaces where they start or end.

Diagram 15 – NEET Physics Electric Charges and Fields

Properties of Electric Field Lines

Continuous curves without breaks in charge-free regions.
Start from positive charges, end at negative charges or infinity.
Tangent at any point gives direction of electric field intensity.
Two field lines never cross.
Do not form closed loops.
Field strength proportional to line density.
Always perpendicular to conductor surfaces.

Dependence on Distance

Electric field strength decreases with distance from the charge.
Field lines spread out as distance increases, reducing density.
Field strength at a point is inversely proportional to the square of the distance from the charge: E ∝ 1/r².

Diagram 16 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1

Find the relative strength of the electric field at points P₁ and P₂ located at distances r₁ and r₂ from a point charge q.
Use formula: E = k|q|/r², where k = 9 × 10⁹ Nm²/C².
Calculate ratio: E₁/E₂ = r₂²/r₁².

Practice Set

Q1

What is the direction of electric field lines around a positive point charge?

A. Toward the charge
B. Away from the charge
C. Circular around the charge
D. Random directions

Q2

Which of the following is true about electric field lines?

A. They can intersect
B. They form closed loops
C. They start from positive and end at negative charges
D. They are parallel everywhere

Q3

At what distance from a point charge q is the electric field strength E reduced to one-fourth of its value at distance r?

A. 2r
B. r/2
C. 4r
D. r/4

Answer Key

Q1: B - Field lines radiate outward from positive charges.
Q2: C - Field lines start from positive and end at negative charges; they never intersect or form loops.
Q3: A - Electric field E ∝ 1/r², so doubling distance reduces field to one-fourth.

Memory Aids

"Positive push out, negative pull in" for field line directions.
Field strength E ∝ line density.
Field lines never cross = unique field direction.
Electric field E ∝ 1/r² for point charges.

Concept and Representation

Electric field lines map the electric field around charges.
Tangent to a field line at any point shows the direction of the net electric field.
Field lines start from positive charges and end at negative charges or infinity.
Field lines never intersect; intersection would imply multiple directions of the field.
Field lines do not form closed loops due to the conservative nature of electrostatic fields.
Density of field lines indicates field strength; closer lines mean stronger field.
Field lines are normal (perpendicular) to conductor surfaces where they start or end.

Diagram 17 – NEET Physics Electric Charges and Fields

Properties of Electric Field Lines

Continuous curves without breaks in charge-free regions.
Start from positive charges, end at negative charges or infinity.
Tangent at any point gives direction of electric field intensity.
Two field lines never cross.
Do not form closed loops.
Field strength proportional to line density.
Always perpendicular to conductor surfaces.

Dependence on Distance

Electric field strength decreases with distance from the charge.
Field lines spread out as distance increases, reducing density.
Field strength at a point is inversely proportional to the square of the distance from the charge: E ∝ 1/r².

Diagram 18 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1

Find the relative strength of the electric field at points P₁ and P₂ located at distances r₁ and r₂ from a point charge q.
Use formula: E = k|q|/r², where k = 9 × 10⁹ Nm²/C².
Calculate ratio: E₁/E₂ = r₂²/r₁².

Practice Set

Q1

What is the direction of electric field lines around a positive point charge?

A. Toward the charge
B. Away from the charge
C. Circular around the charge
D. Random directions

Q2

Which of the following is true about electric field lines?

A. They can intersect
B. They form closed loops
C. They start from positive and end at negative charges
D. They are parallel everywhere

Q3

At what distance from a point charge q is the electric field strength E reduced to one-fourth of its value at distance r?

A. 2r
B. r/2
C. 4r
D. r/4

Answer Key

Q1: B - Field lines radiate outward from positive charges.
Q2: C - Field lines start from positive and end at negative charges; they never intersect or form loops.
Q3: A - Electric field E ∝ 1/r², so doubling distance reduces field to one-fourth.

Memory Aids

"Positive push out, negative pull in" for field line directions.
Field strength E ∝ line density.
Field lines never cross = unique field direction.
Electric field E ∝ 1/r² for point charges.

Properties of Electric Field Lines

Continuous curves without breaks in charge-free regions.
Start from positive charges, end at negative charges or infinity.
Tangent at any point gives direction of electric field intensity.
Two field lines never cross.
Do not form closed loops.
Field strength proportional to line density.
Always perpendicular to conductor surfaces.

Dependence on Distance

Electric field strength decreases with distance from the charge.
Field lines spread out as distance increases, reducing density.
Field strength at a point is inversely proportional to the square of the distance from the charge: E ∝ 1/r².

Diagram 19 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1

Find the relative strength of the electric field at points P₁ and P₂ located at distances r₁ and r₂ from a point charge q.
Use formula: E = k|q|/r², where k = 9 × 10⁹ Nm²/C².
Calculate ratio: E₁/E₂ = r₂²/r₁².

Practice Set

Q1

What is the direction of electric field lines around a positive point charge?

A. Toward the charge
B. Away from the charge
C. Circular around the charge
D. Random directions

Q2

Which of the following is true about electric field lines?

A. They can intersect
B. They form closed loops
C. They start from positive and end at negative charges
D. They are parallel everywhere

Q3

At what distance from a point charge q is the electric field strength E reduced to one-fourth of its value at distance r?

A. 2r
B. r/2
C. 4r
D. r/4

Answer Key

Q1: B - Field lines radiate outward from positive charges.
Q2: C - Field lines start from positive and end at negative charges; they never intersect or form loops.
Q3: A - Electric field E ∝ 1/r², so doubling distance reduces field to one-fourth.

Memory Aids

"Positive push out, negative pull in" for field line directions.
Field strength E ∝ line density.
Field lines never cross = unique field direction.
Electric field E ∝ 1/r² for point charges.

Concept and Representation

Electric field lines map the electric field around charges.
Tangent to a field line at any point shows the direction of the net electric field.
Field lines start from positive charges and end at negative charges or infinity.
Field lines never intersect; intersection would imply multiple directions of the field.
Field lines do not form closed loops due to the conservative nature of electrostatic fields.
Density of field lines indicates field strength; closer lines mean stronger field.
Field lines are normal (perpendicular) to conductor surfaces where they start or end.

Diagram 20 – NEET Physics Electric Charges and Fields

Properties of Electric Field Lines

Continuous curves without breaks in charge-free regions.
Start from positive charges, end at negative charges or infinity.
Tangent at any point gives direction of electric field intensity.
Two field lines never cross.
Do not form closed loops.
Field strength proportional to line density.
Always perpendicular to conductor surfaces.

Dependence on Distance

Electric field strength decreases with distance from the charge.
Field lines spread out as distance increases, reducing density.
Field strength at a point is inversely proportional to the square of the distance from the charge: E ∝ 1/r².

Diagram 21 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1

Find the relative strength of the electric field at points P₁ and P₂ located at distances r₁ and r₂ from a point charge q.
Use formula: E = k|q|/r², where k = 9 × 10⁹ Nm²/C².
Calculate ratio: E₁/E₂ = r₂²/r₁².

Practice Set

Q1

What is the direction of electric field lines around a positive point charge?

A. Toward the charge
B. Away from the charge
C. Circular around the charge
D. Random directions

Q2

Which of the following is true about electric field lines?

A. They can intersect
B. They form closed loops
C. They start from positive and end at negative charges
D. They are parallel everywhere

Q3

At what distance from a point charge q is the electric field strength E reduced to one-fourth of its value at distance r?

A. 2r
B. r/2
C. 4r
D. r/4

Answer Key

Q1: B - Field lines radiate outward from positive charges.
Q2: C - Field lines start from positive and end at negative charges; they never intersect or form loops.
Q3: A - Electric field E ∝ 1/r², so doubling distance reduces field to one-fourth.

Memory Aids

"Positive push out, negative pull in" for field line directions.
Field strength E ∝ line density.
Field lines never cross = unique field direction.
Electric field E ∝ 1/r² for point charges.

Concept and Representation

Electric field lines map the electric field around charges.
Tangent to a field line at any point shows the direction of the net electric field.
Field lines start from positive charges and end at negative charges or infinity.
Field lines never intersect; intersection would imply multiple directions of the field.
Field lines do not form closed loops due to the conservative nature of electrostatic fields.
Density of field lines indicates field strength; closer lines mean stronger field.
Field lines are normal (perpendicular) to conductor surfaces where they start or end.

Diagram 22 – NEET Physics Electric Charges and Fields

Properties of Electric Field Lines

Continuous curves without breaks in charge-free regions.
Start from positive charges, end at negative charges or infinity.
Tangent at any point gives direction of electric field intensity.
Two field lines never cross.
Do not form closed loops.
Field strength proportional to line density.
Always perpendicular to conductor surfaces.

Dependence on Distance

Electric field strength decreases with distance from the charge.
Field lines spread out as distance increases, reducing density.
Field strength at a point is inversely proportional to the square of the distance from the charge: E ∝ 1/r².

Diagram 23 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1

Find the relative strength of the electric field at points P₁ and P₂ located at distances r₁ and r₂ from a point charge q.
Use formula: E = k|q|/r², where k = 9 × 10⁹ Nm²/C².
Calculate ratio: E₁/E₂ = r₂²/r₁².

Practice Set

Q1

What is the direction of electric field lines around a positive point charge?

A. Toward the charge
B. Away from the charge
C. Circular around the charge
D. Random directions

Q2

Which of the following is true about electric field lines?

A. They can intersect
B. They form closed loops
C. They start from positive and end at negative charges
D. They are parallel everywhere

Q3

At what distance from a point charge q is the electric field strength E reduced to one-fourth of its value at distance r?

A. 2r
B. r/2
C. 4r
D. r/4

Answer Key

Q1: B - Field lines radiate outward from positive charges.
Q2: C - Field lines start from positive and end at negative charges; they never intersect or form loops.
Q3: A - Electric field E ∝ 1/r², so doubling distance reduces field to one-fourth.

Memory Aids

"Positive push out, negative pull in" for field line directions.
Field strength E ∝ line density.
Field lines never cross = unique field direction.
Electric field E ∝ 1/r² for point charges.

Properties of Electric Field Lines

Continuous curves without breaks in charge-free regions.
Start from positive charges, end at negative charges or infinity.
Tangent at any point gives direction of electric field intensity.
Two field lines never cross.
Do not form closed loops.
Field strength proportional to line density.
Always perpendicular to conductor surfaces.

Dependence on Distance

Electric field strength decreases with distance from the charge.
Field lines spread out as distance increases, reducing density.
Field strength at a point is inversely proportional to the square of the distance from the charge: E ∝ 1/r².

Diagram 24 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1

Find the relative strength of the electric field at points P₁ and P₂ located at distances r₁ and r₂ from a point charge q.
Use formula: E = k|q|/r², where k = 9 × 10⁹ Nm²/C².
Calculate ratio: E₁/E₂ = r₂²/r₁².

Practice Set

Q1

What is the direction of electric field lines around a positive point charge?

A. Toward the charge
B. Away from the charge
C. Circular around the charge
D. Random directions

Q2

Which of the following is true about electric field lines?

A. They can intersect
B. They form closed loops
C. They start from positive and end at negative charges
D. They are parallel everywhere

Q3

At what distance from a point charge q is the electric field strength E reduced to one-fourth of its value at distance r?

A. 2r
B. r/2
C. 4r
D. r/4

Answer Key

Q1: B - Field lines radiate outward from positive charges.
Q2: C - Field lines start from positive and end at negative charges; they never intersect or form loops.
Q3: A - Electric field E ∝ 1/r², so doubling distance reduces field to one-fourth.

Memory Aids

"Positive push out, negative pull in" for field line directions.
Field strength E ∝ line density.
Field lines never cross = unique field direction.
Electric field E ∝ 1/r² for point charges.

Concept and Representation

Electric field lines map the electric field around charges.
Tangent to a field line at any point shows the direction of the net electric field.
Field lines start from positive charges and end at negative charges or infinity.
Field lines never intersect; intersection would imply multiple directions of the field.
Field lines do not form closed loops due to the conservative nature of electrostatic fields.
Density of field lines indicates field strength; closer lines mean stronger field.
Field lines are normal (perpendicular) to conductor surfaces where they start or end.

Diagram 25 – NEET Physics Electric Charges and Fields

Properties of Electric Field Lines

Continuous curves without breaks in charge-free regions.
Start from positive charges, end at negative charges or infinity.
Tangent at any point gives direction of electric field intensity.
Two field lines never cross.
Do not form closed loops.
Field strength proportional to line density.
Always perpendicular to conductor surfaces.

Dependence on Distance

Electric field strength decreases with distance from the charge.
Field lines spread out as distance increases, reducing density.
Field strength at a point is inversely proportional to the square of the distance from the charge: E ∝ 1/r².

Diagram 26 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1

Find the relative strength of the electric field at points P₁ and P₂ located at distances r₁ and r₂ from a point charge q.
Use formula: E = k|q|/r², where k = 9 × 10⁹ Nm²/C².
Calculate ratio: E₁/E₂ = r₂²/r₁².

Practice Set

Q1

What is the direction of electric field lines around a positive point charge?

A. Toward the charge
B. Away from the charge
C. Circular around the charge
D. Random directions

Q2

Which of the following is true about electric field lines?

A. They can intersect
B. They form closed loops
C. They start from positive and end at negative charges
D. They are parallel everywhere

Q3

At what distance from a point charge q is the electric field strength E reduced to one-fourth of its value at distance r?

A. 2r
B. r/2
C. 4r
D. r/4

Answer Key

Q1: B - Field lines radiate outward from positive charges.
Q2: C - Field lines start from positive and end at negative charges; they never intersect or form loops.
Q3: A - Electric field E ∝ 1/r², so doubling distance reduces field to one-fourth.

Memory Aids

"Positive push out, negative pull in" for field line directions.
Field strength E ∝ line density.
Field lines never cross = unique field direction.
Electric field E ∝ 1/r² for point charges.

Concept and Representation

Electric field lines map the electric field around charges.
Tangent to a field line at any point shows the direction of the net electric field.
Field lines start from positive charges and end at negative charges or infinity.
Field lines never intersect; intersection would imply multiple directions of the field.
Field lines do not form closed loops due to the conservative nature of electrostatic fields.
Density of field lines indicates field strength; closer lines mean stronger field.
Field lines are normal (perpendicular) to conductor surfaces where they start or end.

Diagram 27 – NEET Physics Electric Charges and Fields

Properties of Electric Field Lines

Continuous curves without breaks in charge-free regions.
Start from positive charges, end at negative charges or infinity.
Tangent at any point gives direction of electric field intensity.
Two field lines never cross.
Do not form closed loops.
Field strength proportional to line density.
Always perpendicular to conductor surfaces.

Dependence on Distance

Electric field strength decreases with distance from the charge.
Field lines spread out as distance increases, reducing density.
Field strength at a point is inversely proportional to the square of the distance from the charge: \( E \propto \frac{1}{r^2} \).

Diagram 28 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1

Find the relative strength of the electric field at points \(P_1\) and \(P_2\) located at distances \(r_1\) and \(r_2\) from a point charge \(q\).
Use formula: \( E = \frac{k|q|}{r^2} \), where \(k = 9 \times 10^9 \ \mathrm{Nm^2/C^2} \).
Calculate ratio: \( \frac{E_1}{E_2} = \frac{r_2^2}{r_1^2} \).

Practice Set

Q1

What is the direction of electric field lines around a positive point charge?

A. Toward the charge
B. Away from the charge
C. Circular around the charge
D. Random directions

Q2

Which of the following is true about electric field lines?

A. They can intersect
B. They form closed loops
C. They start from positive and end at negative charges
D. They are parallel everywhere

Q3

At what distance from a point charge \(q\) is the electric field strength \(E\) reduced to one-fourth of its value at distance \(r\)?

A. \(2r\)
B. \(\frac{r}{2}\)
C. \(4r\)
D. \(\frac{r}{4}\)

Answer Key

Q1: B - Field lines radiate outward from positive charges.
Q2: C - Field lines start from positive and end at negative charges; they never intersect or form loops.
Q3: A - Electric field \(E \propto \frac{1}{r^2}\), so doubling distance reduces field to one-fourth.

Memory Aids

"Positive push out, negative pull in" for field line directions.
Field strength \(E \propto \) line density.
Field lines never cross = unique field direction.
Electric field \(E \propto \frac{1}{r^2}\) for point charges.

Properties of Electric Field Lines

Continuous curves without breaks in charge-free regions.
Start from positive charges, end at negative charges or infinity.
Tangent at any point gives direction of electric field intensity.
Two field lines never cross.
Do not form closed loops.
Field strength proportional to line density.
Always perpendicular to conductor surfaces.

Dependence on Distance

Electric field strength decreases with distance from the charge.
Field lines spread out as distance increases, reducing density.
Field strength at a point is inversely proportional to the square of the distance from the charge: E ∝ 1/r².

Diagram 29 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1

Find the relative strength of the electric field at points P₁ and P₂ located at distances r₁ and r₂ from a point charge q.
Use formula: E = k|q|/r², where k = 9 × 10⁹ Nm²/C².
Calculate ratio: E₁/E₂ = r₂²/r₁².

Practice Set

Q1

What is the direction of electric field lines around a positive point charge?

A. Toward the charge
B. Away from the charge
C. Circular around the charge
D. Random directions

Q2

Which of the following is true about electric field lines?

A. They can intersect
B. They form closed loops
C. They start from positive and end at negative charges
D. They are parallel everywhere

Q3

At what distance from a point charge q is the electric field strength E reduced to one-fourth of its value at distance r?

A. 2r
B. r/2
C. 4r
D. r/4

Answer Key

Q1: B - Field lines radiate outward from positive charges.
Q2: C - Field lines start from positive and end at negative charges; they never intersect or form loops.
Q3: A - Electric field E ∝ 1/r², so doubling distance reduces field to one-fourth.

Memory Aids

"Positive push out, negative pull in" for field line directions.
Field strength E ∝ line density.
Field lines never cross = unique field direction.
Electric field E ∝ 1/r² for point charges.

Properties of Electric Field Lines

Continuous curves without breaks in charge-free regions.
Start from positive charges, end at negative charges or infinity.
Tangent at any point gives direction of electric field intensity.
Two field lines never cross.
Do not form closed loops.
Field strength proportional to line density.
Always perpendicular to conductor surfaces.

Dependence on Distance

Electric field strength decreases with distance from the charge.
Field lines spread out as distance increases, reducing density.
Field strength at a point is inversely proportional to the square of the distance from the charge: E ∝ 1/r².

Diagram 30 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1

Find the relative strength of the electric field at points P₁ and P₂ located at distances r₁ and r₂ from a point charge q.
Use formula: E = k|q|/r², where k = 9 × 10⁹ Nm²/C².
Calculate ratio: E₁/E₂ = r₂²/r₁².

Practice Set

Q1

What is the direction of electric field lines around a positive point charge?

A. Toward the charge
B. Away from the charge
C. Circular around the charge
D. Random directions

Q2

Which of the following is true about electric field lines?

A. They can intersect
B. They form closed loops
C. They start from positive and end at negative charges
D. They are parallel everywhere

Q3

At what distance from a point charge q is the electric field strength E reduced to one-fourth of its value at distance r?

A. 2r
B. r/2
C. 4r
D. r/4

Answer Key

Q1: B - Field lines radiate outward from positive charges.
Q2: C - Field lines start from positive and end at negative charges; they never intersect or form loops.
Q3: A - Electric field E ∝ 1/r², so doubling distance reduces field to one-fourth.

Memory Aids

"Positive push out, negative pull in" for field line directions.
Field strength E ∝ line density.
Field lines never cross = unique field direction.
Electric field E ∝ 1/r² for point charges.

Concept and Representation

Electric field lines map the electric field around charges.
Tangent to a field line at any point shows the direction of the net electric field.
Field lines start from positive charges and end at negative charges or infinity.
Field lines never intersect; intersection would imply multiple directions of the field.
Field lines do not form closed loops due to the conservative nature of electrostatic fields.
Density of field lines indicates field strength; closer lines mean stronger field.
Field lines are normal (perpendicular) to conductor surfaces where they start or end.

Diagram 31 – NEET Physics Electric Charges and Fields

Properties of Electric Field Lines

Continuous curves without breaks in charge-free regions.
Start from positive charges, end at negative charges or infinity.
Tangent at any point gives direction of electric field intensity.
Two field lines never cross.
Do not form closed loops.
Field strength proportional to line density.
Always perpendicular to conductor surfaces.

Dependence on Distance

Electric field strength decreases with distance from the charge.
Field lines spread out as distance increases, reducing density.
Field strength at a point is inversely proportional to the square of the distance from the charge: \( E \propto \frac{1}{r^2} \).

Diagram 32 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1

Find the relative strength of the electric field at points \(P_1\) and \(P_2\) located at distances \(r_1\) and \(r_2\) from a point charge \(q\).
Use formula: \( E = \frac{k|q|}{r^2} \), where \(k = 9 \times 10^9 \ \mathrm{Nm^2/C^2} \).
Calculate ratio: \( \frac{E_1}{E_2} = \frac{r_2^2}{r_1^2} \).

Practice Set

Q1

What is the direction of electric field lines around a positive point charge?

A. Toward the charge
B. Away from the charge
C. Circular around the charge
D. Random directions

Q2

Which of the following is true about electric field lines?

A. They can intersect
B. They form closed loops
C. They start from positive and end at negative charges
D. They are parallel everywhere

Q3

At what distance from a point charge \(q\) is the electric field strength \(E\) reduced to one-fourth of its value at distance \(r\)?

A. \(2r\)
B. \(\frac{r}{2}\)
C. \(4r\)
D. \(\frac{r}{4}\)

Answer Key

Q1: B - Field lines radiate outward from positive charges.
Q2: C - Field lines start from positive and end at negative charges; they never intersect or form loops.
Q3: A - Electric field \(E \propto \frac{1}{r^2}\), so doubling distance reduces field to one-fourth.

Memory Aids

"Positive push out, negative pull in" for field line directions.
Field strength \(E \propto \) line density.
Field lines never cross = unique field direction.
Electric field \(E \propto \frac{1}{r^2}\) for point charges.

Electric Flux

Concept and Formula

Electric flux (Φ) is the number of electric field lines passing through a surface.
Mathematically, Φ = ∫_S E · dA.
For uniform electric field, Φ = E · A = EA cos θ, where θ is the angle between E and the area vector A.
If θ = 90°, flux Φ = 0 (field lines parallel to surface).
If θ = 0°, flux is maximum Φ_max = EA.
Electric flux is a scalar quantity measured in volt-meter (V m) or newton meter squared per coulomb (N m² C^-1).
Electric field can be non-zero even if flux is zero, but zero flux implies zero net field through the surface.

Diagram 33 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric flux through a flat surface of area 2 m² with uniform electric field 5 N/C at 30° to the normal.

Use Φ = EA cos θ = 5 × 2 × cos 30° = 10 × 0.866 = 8.66 Nm²/C.

Practice Set

Q1: Electric flux through a surface is zero when the angle between E and A is:

A) 0°
B) 45°
C) 90°
D) 180°

Q2: A uniform electric field of 10 N/C is perpendicular to a surface of area 3 m². What is the electric flux?

A) 0 Nm²/C
B) 10 Nm²/C
C) 30 Nm²/C
D) 300 Nm²/C

Q3: Electric flux through a surface of area 4 m² with electric field 6 N/C at an angle 60° is:

A) 12 Nm²/C
B) 24 Nm²/C
C) 6 Nm²/C
D) 48 Nm²/C

Answer Key

Q1: C) 90° - Flux is zero when field lines are parallel to surface.

Q2: C) 30 Nm²/C - Φ = EA cos 0° = 10 × 3 = 30.

Q3: A) 12 Nm²/C - Φ = 6 × 4 × cos 60° = 24 × 0.5 = 12.

Memory Aids

Flux Φ = EA cos θ.
θ = 0° → max flux, θ = 90° → zero flux.
Flux is scalar, unit: V m or N m²/C.

Diagram 34 – NEET Physics Electric Charges and Fields

Diagram 35 – NEET Physics Electric Charges and Fields

Diagram 36 – NEET Physics Electric Charges and Fields

Concept and Formula

Electric flux (Φ) is the number of electric field lines passing through a surface.
Mathematically, Φ = ∫_S E · dA.
For uniform electric field, Φ = E · A = EA cos θ, where θ is the angle between E and the area vector A.
If θ = 90°, flux Φ = 0 (field lines parallel to surface).
If θ = 0°, flux is maximum Φ_max = EA.
Electric flux is a scalar quantity measured in volt-meter (V m) or newton meter squared per coulomb (N m² C^-1).
Electric field can be non-zero even if flux is zero, but zero flux implies zero net field through the surface.

Diagram 37 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric flux through a flat surface of area 2 m² with uniform electric field 5 N/C at 30° to the normal.

Use Φ = EA cos θ = 5 × 2 × cos 30° = 10 × 0.866 = 8.66 Nm²/C.

Practice Set

Q1: Electric flux through a surface is zero when the angle between E and A is:

A) 0°
B) 45°
C) 90°
D) 180°

Q2: A uniform electric field of 10 N/C is perpendicular to a surface of area 3 m². What is the electric flux?

A) 0 Nm²/C
B) 10 Nm²/C
C) 30 Nm²/C
D) 300 Nm²/C

Q3: Electric flux through a surface of area 4 m² with electric field 6 N/C at an angle 60° is:

A) 12 Nm²/C
B) 24 Nm²/C
C) 6 Nm²/C
D) 48 Nm²/C

Answer Key

Q1: C) 90° - Flux is zero when field lines are parallel to surface.

Q2: C) 30 Nm²/C - Φ = EA cos 0° = 10 × 3 = 30.

Q3: A) 12 Nm²/C - Φ = 6 × 4 × cos 60° = 24 × 0.5 = 12.

Memory Aids

Flux Φ = EA cos θ.
θ = 0° → max flux, θ = 90° → zero flux.
Flux is scalar, unit: V m or N m²/C.

Diagram 38 – NEET Physics Electric Charges and Fields

Diagram 39 – NEET Physics Electric Charges and Fields

Diagram 40 – NEET Physics Electric Charges and Fields

Concept and Formula

Electric flux (Φ) is the number of electric field lines passing through a surface.
Mathematically, Φ = ∫_S E · dA.
For uniform electric field, Φ = E · A = EA cos θ, where θ is the angle between E and the area vector A.
If θ = 90°, flux Φ = 0 (field lines parallel to surface).
If θ = 0°, flux is maximum Φ_max = EA.
Electric flux is a scalar quantity measured in volt-meter (V m) or newton meter squared per coulomb (N m² C^-1).
Electric field can be non-zero even if flux is zero, but zero flux implies zero net field through the surface.

Diagram 41 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric flux through a flat surface of area 2 m² with uniform electric field 5 N/C at 30° to the normal.

Use Φ = EA cos θ = 5 × 2 × cos 30° = 10 × 0.866 = 8.66 Nm²/C.

Practice Set

Q1: Electric flux through a surface is zero when the angle between E and A is:

A) 0°
B) 45°
C) 90°
D) 180°

Q2: A uniform electric field of 10 N/C is perpendicular to a surface of area 3 m². What is the electric flux?

A) 0 Nm²/C
B) 10 Nm²/C
C) 30 Nm²/C
D) 300 Nm²/C

Q3: Electric flux through a surface of area 4 m² with electric field 6 N/C at an angle 60° is:

A) 12 Nm²/C
B) 24 Nm²/C
C) 6 Nm²/C
D) 48 Nm²/C

Answer Key

Q1: C) 90° - Flux is zero when field lines are parallel to surface.

Q2: C) 30 Nm²/C - Φ = EA cos 0° = 10 × 3 = 30.

Q3: A) 12 Nm²/C - Φ = 6 × 4 × cos 60° = 24 × 0.5 = 12.

Memory Aids

Flux Φ = EA cos θ.
θ = 0° → max flux, θ = 90° → zero flux.
Flux is scalar, unit: V m or N m²/C.

Diagram 42 – NEET Physics Electric Charges and Fields

Diagram 43 – NEET Physics Electric Charges and Fields

Diagram 44 – NEET Physics Electric Charges and Fields

Concept and Formula

Electric flux (Φ) is the number of electric field lines passing through a surface.
Mathematically, Φ = ∫_S E · dA.
For uniform electric field, Φ = E · A = EA cos θ, where θ is the angle between E and the area vector A.
If θ = 90°, flux Φ = 0 (field lines parallel to surface).
If θ = 0°, flux is maximum Φ_max = EA.
Electric flux is a scalar quantity measured in volt-meter (V m) or newton meter squared per coulomb (N m² C^-1).
Electric field can be non-zero even if flux is zero, but zero flux implies zero net field through the surface.

Diagram 45 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric flux through a flat surface of area 2 m² with uniform electric field 5 N/C at 30° to the normal.

Use Φ = EA cos θ = 5 × 2 × cos 30° = 10 × 0.866 = 8.66 Nm²/C.

Practice Set

Q1: Electric flux through a surface is zero when the angle between E and A is:

A) 0°
B) 45°
C) 90°
D) 180°

Q2: A uniform electric field of 10 N/C is perpendicular to a surface of area 3 m². What is the electric flux?

A) 0 Nm²/C
B) 10 Nm²/C
C) 30 Nm²/C
D) 300 Nm²/C

Q3: Electric flux through a surface of area 4 m² with electric field 6 N/C at an angle 60° is:

A) 12 Nm²/C
B) 24 Nm²/C
C) 6 Nm²/C
D) 48 Nm²/C

Answer Key

Q1: C) 90° - Flux is zero when field lines are parallel to surface.

Q2: C) 30 Nm²/C - Φ = EA cos 0° = 10 × 3 = 30.

Q3: A) 12 Nm²/C - Φ = 6 × 4 × cos 60° = 24 × 0.5 = 12.

Memory Aids

Flux Φ = EA cos θ.
θ = 0° → max flux, θ = 90° → zero flux.
Flux is scalar, unit: V m or N m²/C.

Diagram 46 – NEET Physics Electric Charges and Fields

Diagram 47 – NEET Physics Electric Charges and Fields

Diagram 48 – NEET Physics Electric Charges and Fields

Concept and Formula

Electric flux (Φ) is the number of electric field lines passing through a surface.
Mathematically, Φ = ∫_S E · dA.
For uniform electric field, Φ = E · A = EA cos θ, where θ is the angle between E and the area vector A.
If θ = 90°, flux Φ = 0 (field lines parallel to surface).
If θ = 0°, flux is maximum Φ_max = EA.
Electric flux is a scalar quantity measured in volt-meter (V m) or newton meter squared per coulomb (N m² C^-1).
Electric field can be non-zero even if flux is zero, but zero flux implies zero net field through the surface.

Diagram 49 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric flux through a flat surface of area 2 m² with uniform electric field 5 N/C at 30° to the normal.

Use Φ = EA cos θ = 5 × 2 × cos 30° = 10 × 0.866 = 8.66 Nm²/C.

Practice Set

Q1: Electric flux through a surface is zero when the angle between E and A is:

A) 0°
B) 45°
C) 90°
D) 180°

Q2: A uniform electric field of 10 N/C is perpendicular to a surface of area 3 m². What is the electric flux?

A) 0 Nm²/C
B) 10 Nm²/C
C) 30 Nm²/C
D) 300 Nm²/C

Q3: Electric flux through a surface of area 4 m² with electric field 6 N/C at an angle 60° is:

A) 12 Nm²/C
B) 24 Nm²/C
C) 6 Nm²/C
D) 48 Nm²/C

Answer Key

Q1: C) 90° - Flux is zero when field lines are parallel to surface.

Q2: C) 30 Nm²/C - Φ = EA cos 0° = 10 × 3 = 30.

Q3: A) 12 Nm²/C - Φ = 6 × 4 × cos 60° = 24 × 0.5 = 12.

Memory Aids

Flux Φ = EA cos θ.
θ = 0° → max flux, θ = 90° → zero flux.
Flux is scalar, unit: V m or N m²/C.

Diagram 50 – NEET Physics Electric Charges and Fields

Diagram 51 – NEET Physics Electric Charges and Fields

Diagram 52 – NEET Physics Electric Charges and Fields

Concept and Formula

Electric flux (Φ) is the number of electric field lines passing through a surface.
Mathematically, Φ = ∫_S E · dA.
For uniform electric field, Φ = E · A = EA cos θ, where θ is the angle between E and the area vector A.
If θ = 90°, flux Φ = 0 (field lines parallel to surface).
If θ = 0°, flux is maximum Φ_max = EA.
Electric flux is a scalar quantity measured in volt-meter (V m) or newton meter squared per coulomb (N m² C^-1).
Electric field can be non-zero even if flux is zero, but zero flux implies zero net field through the surface.

Diagram 53 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric flux through a flat surface of area 2 m² with uniform electric field 5 N/C at 30° to the normal.

Use Φ = EA cos θ = 5 × 2 × cos 30° = 10 × 0.866 = 8.66 Nm²/C.

Practice Set

Q1: Electric flux through a surface is zero when the angle between E and A is:

A) 0°
B) 45°
C) 90°
D) 180°

Q2: A uniform electric field of 10 N/C is perpendicular to a surface of area 3 m². What is the electric flux?

A) 0 Nm²/C
B) 10 Nm²/C
C) 30 Nm²/C
D) 300 Nm²/C

Q3: Electric flux through a surface of area 4 m² with electric field 6 N/C at an angle 60° is:

A) 12 Nm²/C
B) 24 Nm²/C
C) 6 Nm²/C
D) 48 Nm²/C

Answer Key

Q1: C) 90° - Flux is zero when field lines are parallel to surface.

Q2: C) 30 Nm²/C - Φ = EA cos 0° = 10 × 3 = 30.

Q3: A) 12 Nm²/C - Φ = 6 × 4 × cos 60° = 24 × 0.5 = 12.

Memory Aids

Flux Φ = EA cos θ.
θ = 0° → max flux, θ = 90° → zero flux.
Flux is scalar, unit: V m or N m²/C.

Diagram 54 – NEET Physics Electric Charges and Fields

Diagram 55 – NEET Physics Electric Charges and Fields

Diagram 56 – NEET Physics Electric Charges and Fields

Concept and Formula

Electric flux (Φ) is the number of electric field lines passing through a surface.
Mathematically, Φ = ∫_S E · dA.
For uniform electric field, Φ = E · A = EA cos θ, where θ is the angle between E and the area vector A.
If θ = 90°, flux Φ = 0 (field lines parallel to surface).
If θ = 0°, flux is maximum Φ_max = EA.
Electric flux is a scalar quantity measured in volt-meter (V m) or newton meter squared per coulomb (N m² C^-1).
Electric field can be non-zero even if flux is zero, but zero flux implies zero net field through the surface.

Diagram 57 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric flux through a flat surface of area 2 m² with uniform electric field 5 N/C at 30° to the normal.

Use Φ = EA cos θ = 5 × 2 × cos 30° = 10 × 0.866 = 8.66 Nm²/C.

Practice Set

Q1: Electric flux through a surface is zero when the angle between E and A is:

A) 0°
B) 45°
C) 90°
D) 180°

Q2: A uniform electric field of 10 N/C is perpendicular to a surface of area 3 m². What is the electric flux?

A) 0 Nm²/C
B) 10 Nm²/C
C) 30 Nm²/C
D) 300 Nm²/C

Q3: Electric flux through a surface of area 4 m² with electric field 6 N/C at an angle 60° is:

A) 12 Nm²/C
B) 24 Nm²/C
C) 6 Nm²/C
D) 48 Nm²/C

Answer Key

Q1: C) 90° - Flux is zero when field lines are parallel to surface.

Q2: C) 30 Nm²/C - Φ = EA cos 0° = 10 × 3 = 30.

Q3: A) 12 Nm²/C - Φ = 6 × 4 × cos 60° = 24 × 0.5 = 12.

Memory Aids

Flux Φ = EA cos θ.
θ = 0° → max flux, θ = 90° → zero flux.
Flux is scalar, unit: V m or N m²/C.

Diagram 58 – NEET Physics Electric Charges and Fields

Diagram 59 – NEET Physics Electric Charges and Fields

Diagram 60 – NEET Physics Electric Charges and Fields

Concept and Formula

Electric flux (Φ) is the number of electric field lines passing through a surface.
Mathematically, Φ = ∫_S E · dA.
For uniform electric field, Φ = E · A = EA cos θ, where θ is the angle between E and the area vector A.
If θ = 90°, flux Φ = 0 (field lines parallel to surface).
If θ = 0°, flux is maximum Φ_max = EA.
Electric flux is a scalar quantity measured in volt-meter (V m) or newton meter squared per coulomb (N m² C^-1).
Electric field can be non-zero even if flux is zero, but zero flux implies zero net field through the surface.

Diagram 61 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric flux through a flat surface of area 2 m² with uniform electric field 5 N/C at 30° to the normal.

Use Φ = EA cos θ = 5 × 2 × cos 30° = 10 × 0.866 = 8.66 Nm²/C.

Practice Set

Q1: Electric flux through a surface is zero when the angle between E and A is:

A) 0°
B) 45°
C) 90°
D) 180°

Q2: A uniform electric field of 10 N/C is perpendicular to a surface of area 3 m². What is the electric flux?

A) 0 Nm²/C
B) 10 Nm²/C
C) 30 Nm²/C
D) 300 Nm²/C

Q3: Electric flux through a surface of area 4 m² with electric field 6 N/C at an angle 60° is:

A) 12 Nm²/C
B) 24 Nm²/C
C) 6 Nm²/C
D) 48 Nm²/C

Answer Key

Q1: C) 90° - Flux is zero when field lines are parallel to surface.

Q2: C) 30 Nm²/C - Φ = EA cos 0° = 10 × 3 = 30.

Q3: A) 12 Nm²/C - Φ = 6 × 4 × cos 60° = 24 × 0.5 = 12.

Memory Aids

Flux Φ = EA cos θ.
θ = 0° → max flux, θ = 90° → zero flux.
Flux is scalar, unit: V m or N m²/C.

Diagram 62 – NEET Physics Electric Charges and Fields

Diagram 63 – NEET Physics Electric Charges and Fields

Diagram 64 – NEET Physics Electric Charges and Fields

Concept and Formula

Electric flux (Φ) is the number of electric field lines passing through a surface.
Mathematically, Φ = ∫_S E · dA.
For uniform electric field, Φ = E · A = EA cos θ, where θ is the angle between E and the area vector A.
If θ = 90°, flux Φ = 0 (field lines parallel to surface).
If θ = 0°, flux is maximum Φ_max = EA.
Electric flux is a scalar quantity measured in volt-meter (V m) or newton meter squared per coulomb (N m² C^-1).
Electric field can be non-zero even if flux is zero, but zero flux implies zero net field through the surface.

Diagram 65 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric flux through a flat surface of area 2 m² with uniform electric field 5 N/C at 30° to the normal.

Use Φ = EA cos θ = 5 × 2 × cos 30° = 10 × 0.866 = 8.66 Nm²/C.

Practice Set

Q1: Electric flux through a surface is zero when the angle between E and A is:

A) 0°
B) 45°
C) 90°
D) 180°

Q2: A uniform electric field of 10 N/C is perpendicular to a surface of area 3 m². What is the electric flux?

A) 0 Nm²/C
B) 10 Nm²/C
C) 30 Nm²/C
D) 300 Nm²/C

Q3: Electric flux through a surface of area 4 m² with electric field 6 N/C at an angle 60° is:

A) 12 Nm²/C
B) 24 Nm²/C
C) 6 Nm²/C
D) 48 Nm²/C

Answer Key

Q1: C) 90° - Flux is zero when field lines are parallel to surface.

Q2: C) 30 Nm²/C - Φ = EA cos 0° = 10 × 3 = 30.

Q3: A) 12 Nm²/C - Φ = 6 × 4 × cos 60° = 24 × 0.5 = 12.

Memory Aids

Flux Φ = EA cos θ.
θ = 0° → max flux, θ = 90° → zero flux.
Flux is scalar, unit: V m or N m²/C.

Diagram 66 – NEET Physics Electric Charges and Fields

Diagram 67 – NEET Physics Electric Charges and Fields

Diagram 68 – NEET Physics Electric Charges and Fields

Concept and Formula

Electric flux (Φ) is the number of electric field lines passing through a surface.
Mathematically, Φ = ∫_S E · dA.
For uniform electric field, Φ = E · A = EA cos θ, where θ is the angle between E and the area vector A.
If θ = 90°, flux Φ = 0 (field lines parallel to surface).
If θ = 0°, flux is maximum Φ_max = EA.
Electric flux is a scalar quantity measured in volt-meter (V m) or newton meter squared per coulomb (N m² C^-1).
Electric field can be non-zero even if flux is zero, but zero flux implies zero net field through the surface.

Diagram 69 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric flux through a flat surface of area 2 m² with uniform electric field 5 N/C at 30° to the normal.

Use Φ = EA cos θ = 5 × 2 × cos 30° = 10 × 0.866 = 8.66 Nm²/C.

Practice Set

Q1: Electric flux through a surface is zero when the angle between E and A is:

A) 0°
B) 45°
C) 90°
D) 180°

Q2: A uniform electric field of 10 N/C is perpendicular to a surface of area 3 m². What is the electric flux?

A) 0 Nm²/C
B) 10 Nm²/C
C) 30 Nm²/C
D) 300 Nm²/C

Q3: Electric flux through a surface of area 4 m² with electric field 6 N/C at an angle 60° is:

A) 12 Nm²/C
B) 24 Nm²/C
C) 6 Nm²/C
D) 48 Nm²/C

Answer Key

Q1: C) 90° - Flux is zero when field lines are parallel to surface.

Q2: C) 30 Nm²/C - Φ = EA cos 0° = 10 × 3 = 30.

Q3: A) 12 Nm²/C - Φ = 6 × 4 × cos 60° = 24 × 0.5 = 12.

Memory Aids

Flux Φ = EA cos θ.
θ = 0° → max flux, θ = 90° → zero flux.
Flux is scalar, unit: V m or N m²/C.

Diagram 70 – NEET Physics Electric Charges and Fields

Diagram 71 – NEET Physics Electric Charges and Fields

Diagram 72 – NEET Physics Electric Charges and Fields

Concept and Formula

Electric flux (Φ) is the number of electric field lines passing through a surface.
Mathematically, Φ = ∫_S E · dA.
For uniform electric field, Φ = E · A = EA cos θ, where θ is the angle between E and the area vector A.
If θ = 90°, flux Φ = 0 (field lines parallel to surface).
If θ = 0°, flux is maximum Φ_max = EA.
Electric flux is a scalar quantity measured in volt-meter (V m) or newton meter squared per coulomb (N m² C^-1).
Electric field can be non-zero even if flux is zero, but zero flux implies zero net field through the surface.

Diagram 73 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric flux through a flat surface of area 2 m² with uniform electric field 5 N/C at 30° to the normal.

Use Φ = EA cos θ = 5 × 2 × cos 30° = 10 × 0.866 = 8.66 Nm²/C.

Practice Set

Q1: Electric flux through a surface is zero when the angle between E and A is:

A) 0°
B) 45°
C) 90°
D) 180°

Q2: A uniform electric field of 10 N/C is perpendicular to a surface of area 3 m². What is the electric flux?

A) 0 Nm²/C
B) 10 Nm²/C
C) 30 Nm²/C
D) 300 Nm²/C

Q3: Electric flux through a surface of area 4 m² with electric field 6 N/C at an angle 60° is:

A) 12 Nm²/C
B) 24 Nm²/C
C) 6 Nm²/C
D) 48 Nm²/C

Answer Key

Q1: C) 90° - Flux is zero when field lines are parallel to surface.

Q2: C) 30 Nm²/C - Φ = EA cos 0° = 10 × 3 = 30.

Q3: A) 12 Nm²/C - Φ = 6 × 4 × cos 60° = 24 × 0.5 = 12.

Memory Aids

Flux Φ = EA cos θ.
θ = 0° → max flux, θ = 90° → zero flux.
Flux is scalar, unit: V m or N m²/C.

Diagram 74 – NEET Physics Electric Charges and Fields

Diagram 75 – NEET Physics Electric Charges and Fields

Diagram 76 – NEET Physics Electric Charges and Fields

Concept and Formula

Electric flux (Φ) is the number of electric field lines passing through a surface.
Mathematically, Φ = ∫_S E · dA.
For uniform electric field, Φ = E · A = EA cos θ, where θ is the angle between E and the area vector A.
If θ = 90°, flux Φ = 0 (field lines parallel to surface).
If θ = 0°, flux is maximum Φ_max = EA.
Electric flux is a scalar quantity measured in volt-meter (V m) or newton meter squared per coulomb (N m² C^-1).
Electric field can be non-zero even if flux is zero, but zero flux implies zero net field through the surface.

Diagram 77 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric flux through a flat surface of area 2 m² with uniform electric field 5 N/C at 30° to the normal.

Use Φ = EA cos θ = 5 × 2 × cos 30° = 10 × 0.866 = 8.66 Nm²/C.

Practice Set

Q1: Electric flux through a surface is zero when the angle between E and A is:

A) 0°
B) 45°
C) 90°
D) 180°

Q2: A uniform electric field of 10 N/C is perpendicular to a surface of area 3 m². What is the electric flux?

A) 0 Nm²/C
B) 10 Nm²/C
C) 30 Nm²/C
D) 300 Nm²/C

Q3: Electric flux through a surface of area 4 m² with electric field 6 N/C at an angle 60° is:

A) 12 Nm²/C
B) 24 Nm²/C
C) 6 Nm²/C
D) 48 Nm²/C

Answer Key

Q1: C) 90° - Flux is zero when field lines are parallel to surface.

Q2: C) 30 Nm²/C - Φ = EA cos 0° = 10 × 3 = 30.

Q3: A) 12 Nm²/C - Φ = 6 × 4 × cos 60° = 24 × 0.5 = 12.

Memory Aids

Flux Φ = EA cos θ.
θ = 0° → max flux, θ = 90° → zero flux.
Flux is scalar, unit: V m or N m²/C.

Diagram 78 – NEET Physics Electric Charges and Fields

Diagram 79 – NEET Physics Electric Charges and Fields

Diagram 80 – NEET Physics Electric Charges and Fields

Concept and Formula

Electric flux (Φ) is the number of electric field lines passing through a surface.
Mathematically, Φ = ∫_S E · dA.
For uniform electric field, Φ = E · A = EA cos θ, where θ is the angle between E and the area vector A.
If θ = 90°, flux Φ = 0 (field lines parallel to surface).
If θ = 0°, flux is maximum Φ_max = EA.
Electric flux is a scalar quantity measured in volt-meter (V m) or newton meter squared per coulomb (N m² C^-1).
Electric field can be non-zero even if flux is zero, but zero flux implies zero net field through the surface.

Diagram 81 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric flux through a flat surface of area 2 m² with uniform electric field 5 N/C at 30° to the normal.

Use Φ = EA cos θ = 5 × 2 × cos 30° = 10 × 0.866 = 8.66 Nm²/C.

Practice Set

Q1: Electric flux through a surface is zero when the angle between E and A is:

A) 0°
B) 45°
C) 90°
D) 180°

Q2: A uniform electric field of 10 N/C is perpendicular to a surface of area 3 m². What is the electric flux?

A) 0 Nm²/C
B) 10 Nm²/C
C) 30 Nm²/C
D) 300 Nm²/C

Q3: Electric flux through a surface of area 4 m² with electric field 6 N/C at an angle 60° is:

A) 12 Nm²/C
B) 24 Nm²/C
C) 6 Nm²/C
D) 48 Nm²/C

Answer Key

Q1: C) 90° - Flux is zero when field lines are parallel to surface.

Q2: C) 30 Nm²/C - Φ = EA cos 0° = 10 × 3 = 30.

Q3: A) 12 Nm²/C - Φ = 6 × 4 × cos 60° = 24 × 0.5 = 12.

Memory Aids

Flux Φ = EA cos θ.
θ = 0° → max flux, θ = 90° → zero flux.
Flux is scalar, unit: V m or N m²/C.

Diagram 82 – NEET Physics Electric Charges and Fields

Diagram 83 – NEET Physics Electric Charges and Fields

Diagram 84 – NEET Physics Electric Charges and Fields

Concept and Formula

Electric flux (\(\Phi\)) is the number of electric field lines passing through a surface.
Mathematically, \(\Phi = \int_{S} \vec{E} \cdot d\vec{A}\).
For uniform electric field, \(\Phi = \vec{E} \cdot \vec{A} = EA \cos \theta\), where \(\theta\) is the angle between \(\vec{E}\) and the area vector \(\vec{A}\).
If \(\theta = 90^\circ\), flux \(\Phi = 0\) (field lines parallel to surface).
If \(\theta = 0^\circ\), flux is maximum \(\Phi_{max} = EA\).
Electric flux is a scalar quantity measured in volt-meter (V m) or newton meter squared per coulomb (N m\(^2\) C\(^{-1}\)).
Electric field can be non-zero even if flux is zero, but zero flux implies zero net field through the surface.

Diagram 85 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric flux through a flat surface of area 2 m\(^2\) with uniform electric field 5 N/C at \(30^\circ\) to the normal.

Use \(\Phi = EA \cos \theta = 5 \times 2 \times \cos 30^\circ = 10 \times 0.866 = 8.66 \text{ Nm}^2/\text{C}\).

Practice Set

Q1: Electric flux through a surface is zero when the angle between \(\vec{E}\) and \(\vec{A}\) is:

A) 0\(^\circ\)
B) 45\(^\circ\)
C) 90\(^\circ\)
D) 180\(^\circ\)

Q2: A uniform electric field of 10 N/C is perpendicular to a surface of area 3 m\(^2\). What is the electric flux?

A) 0 Nm\(^2\)/C
B) 10 Nm\(^2\)/C
C) 30 Nm\(^2\)/C
D) 300 Nm\(^2\)/C

Q3: Electric flux through a surface of area 4 m\(^2\) with electric field 6 N/C at an angle 60\(^\circ\) is:

A) 12 Nm\(^2\)/C
B) 24 Nm\(^2\)/C
C) 6 Nm\(^2\)/C
D) 48 Nm\(^2\)/C

Answer Key

Q1: C) 90\(^\circ\) - Flux is zero when field lines are parallel to surface.

Q2: C) 30 Nm\(^2\)/C - \(\Phi = EA \cos 0^\circ = 10 \times 3 = 30\).

Q3: A) 12 Nm\(^2\)/C - \(\Phi = 6 \times 4 \times \cos 60^\circ = 24 \times 0.5 = 12\).

Memory Aids

Flux \(\Phi = EA \cos \theta\).
\(\theta = 0^\circ\) → max flux, \(\theta = 90^\circ\) → zero flux.
Flux is scalar, unit: V m or N m\(^2\)/C.

Diagram 86 – NEET Physics Electric Charges and Fields

Diagram 87 – NEET Physics Electric Charges and Fields

Diagram 88 – NEET Physics Electric Charges and Fields

Electric Dipole

Definition and Dipole Moment

Electric dipole: two equal and opposite charges separated by a small distance.
Dipole moment p = q (2a), points from negative to positive charge.
Distance between charges = 2a.

Diagram 89 – NEET Physics Electric Charges and Fields

Electric Field on Axial Line (End-on Position)

Electric field at point on axis: E = E_+q + E_-q = q p̂/(4 π ε₀) [ 1/(r - a)² - 1/(r + a)² ]
Also, E = 2 p r/(4 π ε₀ (r² - a²)²)
At large distance (r ≫ a), E = 1/(4 π ε₀) 2p/r³

Diagram 90 – NEET Physics Electric Charges and Fields

Electric Field on Equatorial Line (Broadside-on Position)

Electric field at equatorial point: E = - (E_+q + E_-q) cos θ p̂ = 1/(4 π ε₀) -p/(r² + a²)^3/2
At large distance (r ≫ a), E = - p/(4 π ε₀ r³)

Diagram 91 – NEET Physics Electric Charges and Fields

Physical Significance

Nonpolar molecules: centers of positive and negative charges coincide (e.g., CO₂, CH₄) → no permanent dipole moment.
Polar molecules: centers do not coincide (e.g., H₂O, NH₃) → permanent dipole moment.
External electric field can induce dipole moment in nonpolar molecules.

Solved Examples

Ex 1

Calculate electric field at axial point r = 10a for dipole moment p.
Use formula: E = 1/(4 π ε₀) 2p/r³.
Substitute r = 10a, calculate E.

Practice Set

Q1: Dipole moment p points from (A) positive to negative charge (B) negative to positive charge (C) both directions (D) none
Q2: Electric field at large distance on axial line is proportional to (A) 1/r² (B) 1/r³ (C) 1/r (D) r²
Q3: For a dipole with p = 4 × 10^-29 C·m, distance r = 0.1 m, calculate approximate axial electric field E using ε₀ = 8.85 × 10^-12 F/m. (A) 7.2 × 10³ N/C (B) 1.8 × 10⁴ N/C (C) 3.6 × 10³ N/C (D) 9.0 × 10² N/C

Answer Key

Q1: B - Dipole moment vector points from negative to positive charge.
Q2: B - Electric field on axial line at large distance varies as 1/r³.
Q3: A - Using E = 1/(4 π ε₀) 2p/r³, calculate E ≈ 7.2 × 10³ N/C.

Memory Aids

Dipole moment p = q × 2a, points negative to positive.
Axial field ∝ 2p/r³, equatorial field ∝ -p/r³.
At large distances, axial field is twice equatorial field.

Definition and Dipole Moment

Electric dipole: two equal and opposite charges separated by a small distance.
Dipole moment (vector) = q (2a), points from negative to positive charge.
Distance between charges = 2a.

Diagram 92 – NEET Physics Electric Charges and Fields

Electric Field on Axial Line (End-on Position)

Electric field at point on axis: E = E_+q + E_-q = q p̂/(4 π ε₀) [ 1/(r - a)² - 1/(r + a)² ]
Also, E = 2 p r/(4 π ε₀ (r² - a²)²)
At large distance (r ≫ a), E = 1/(4 π ε₀) 2p/r³

Diagram 93 – NEET Physics Electric Charges and Fields

Electric Field on Equatorial Line (Broadside-on Position)

Electric field at equatorial point: E = - (E_+q + E_-q) cos θ p̂ = 1/(4 π ε₀) -p/(r² + a²)^3/2
At large distance (r ≫ a), E = - p/(4 π ε₀ r³)

Diagram 94 – NEET Physics Electric Charges and Fields

Physical Significance

Nonpolar molecules: centers of positive and negative charges coincide (e.g., CO₂, CH₄) → no permanent dipole moment.
Polar molecules: centers do not coincide (e.g., H₂O, NH₃) → permanent dipole moment.
External electric field can induce dipole moment in nonpolar molecules.

Solved Examples

Ex 1

Calculate electric field at axial point r = 10a for dipole moment p.
Use formula: E = 1/(4 π ε₀) 2p/r³.
Substitute r = 10a, calculate E.

Practice Set

Q1: Dipole moment p points from (A) positive to negative charge (B) negative to positive charge (C) both directions (D) none
Q2: Electric field at large distance on axial line is proportional to (A) 1/r² (B) 1/r³ (C) 1/r (D) r²
Q3: For a dipole with p = 4 × 10^-29 C·m, distance r = 0.1 m, calculate approximate axial electric field E using ε₀ = 8.85 × 10^-12 F/m. (A) 7.2 × 10³ N/C (B) 1.8 × 10⁴ N/C (C) 3.6 × 10³ N/C (D) 9.0 × 10² N/C

Answer Key

Q1: B - Dipole moment vector points from negative to positive charge.
Q2: B - Electric field on axial line at large distance varies as 1/r³.
Q3: A - Using E = 1/(4 π ε₀) 2p/r³, calculate E ≈ 7.2 × 10³ N/C.

Memory Aids

Dipole moment p = q × 2a, points negative to positive.
Axial field ∝ 2p/r³, equatorial field ∝ -p/r³.
At large distances, axial field is twice equatorial field.

Definition and Dipole Moment

Electric dipole: two equal and opposite charges separated by a small distance.
Dipole moment p = q (2a), points from negative to positive charge.
Distance between charges = 2a.

Diagram 95 – NEET Physics Electric Charges and Fields

Electric Field on Axial Line (End-on Position)

Electric field at point on axis: E = E_+q + E_-q = q p̂/(4 π ε₀) [ 1/(r - a)² - 1/(r + a)² ]
Also, E = 2 p r/(4 π ε₀ (r² - a²)²)
At large distance (r ≫ a), E = 1/(4 π ε₀) 2p/r³

Diagram 96 – NEET Physics Electric Charges and Fields

Electric Field on Equatorial Line (Broadside-on Position)

Electric field at equatorial point: E = - (E_+q + E_-q) cos θ p̂ = 1/(4 π ε₀) -p/(r² + a²)^3/2
At large distance (r ≫ a), E = - p/(4 π ε₀ r³)

Diagram 97 – NEET Physics Electric Charges and Fields

Physical Significance

Nonpolar molecules: centers of positive and negative charges coincide (e.g., CO₂, CH₄) → no permanent dipole moment.
Polar molecules: centers do not coincide (e.g., H₂O, NH₃) → permanent dipole moment.
External electric field can induce dipole moment in nonpolar molecules.

Solved Examples

Ex 1

Calculate electric field at axial point r = 10a for dipole moment p.
Use formula: E = 1/(4 π ε₀) 2p/r³.
Substitute r = 10a, calculate E.

Practice Set

Q1: Dipole moment p points from (A) positive to negative charge (B) negative to positive charge (C) both directions (D) none
Q2: Electric field at large distance on axial line is proportional to (A) 1/r² (B) 1/r³ (C) 1/r (D) r²
Q3: For a dipole with p = 4 × 10^-29 C·m, distance r = 0.1 m, calculate approximate axial electric field E using ε₀ = 8.85 × 10^-12 F/m. (A) 7.2 × 10³ N/C (B) 1.8 × 10⁴ N/C (C) 3.6 × 10³ N/C (D) 9.0 × 10² N/C

Answer Key

Q1: B - Dipole moment vector points from negative to positive charge.
Q2: B - Electric field on axial line at large distance varies as 1/r³.
Q3: A - Using E = 1/(4 π ε₀) 2p/r³, calculate E ≈ 7.2 × 10³ N/C.

Memory Aids

Dipole moment p = q × 2a, points negative to positive.
Axial field ∝ 2p/r³, equatorial field ∝ -p/r³.
At large distances, axial field is twice equatorial field.

Definition and Dipole Moment

Electric dipole: two equal and opposite charges separated by a small distance.
Dipole moment [?]vec{p} = q (2 [?]vec{a}) , points from negative to positive charge.
Distance between charges = 2a.

Diagram 98 – NEET Physics Electric Charges and Fields

Electric Field on Axial Line (End-on Position)

Electric field at point on axis: [?]vec{E} = [?]vec{E}_{+q} + [?]vec{E}_{-q} = \frac{q \hat{p}}{4 \pi \epsilon_0} \left[ \frac{1}{(r - a)^2} - \frac{1}{(r + a)^2} \right]
Also, [?]vec{E} = \frac{2 \u001b[?]vec{p} r}{4 \pi \epsilon_0 (r^2 - a^2)^2}
At large distance [?] (r \gg a) [?] , [?]vec{E} = \frac{1}{4 \pi \epsilon_0} \frac{2 \u001b[?]vec{p}}{r^3}

Diagram 99 – NEET Physics Electric Charges and Fields

Electric Field on Equatorial Line (Broadside-on Position)

Electric field at equatorial point: [?]vec{E} = - ([?]vec{E}_{+q} + [?]vec{E}_{-q}) cos θ [?]hat{p} = \frac{1}{4 \pi \epsilon_0} \frac{- \u001b[?]vec{p}}{(r^2 + a^2)^{3/2}}
At large distance [?] (r \gg a) [?] , [?]vec{E} = - \frac{\u001b[?]vec{p}}{4 \pi \epsilon_0 r^3}

Diagram 100 – NEET Physics Electric Charges and Fields

Physical Significance

Nonpolar molecules: centers of positive and negative charges coincide (e.g., CO(_2), CH(_4)) → no permanent dipole moment.
Polar molecules: centers do not coincide (e.g., H(_2)O, NH(_3)) → permanent dipole moment.
External electric field can induce dipole moment in nonpolar molecules.

Solved Examples

Ex 1

Calculate electric field at axial point \(r = 10a\) for dipole moment \(p\).
Use formula: \( E = \frac{1}{4 \pi \epsilon_0} \frac{2p}{r^3} \).
Substitute \(r = 10a\), calculate \(E\).

Practice Set

Q1: Dipole moment \( [?]vec{p} \) points from (A) positive to negative charge (B) negative to positive charge (C) both directions (D) none
Q2: Electric field at large distance on axial line is proportional to (A) \(1/r^2\) (B) \(1/r^3\) (C) \(1/r\) (D) \(r^2\)
Q3: For a dipole with \(p = 4 \times 10^{-29} \) C·m, distance \(r = 0.1\) m, calculate approximate axial electric field \(E\) using \( \epsilon_0 = 8.85 \times 10^{-12} \) F/m. (A) \(7.2 \times 10^3\) N/C (B) \(1.8 \times 10^4\) N/C (C) \(3.6 \times 10^3\) N/C (D) \(9.0 \times 10^2\) N/C

Answer Key

Q1: B - Dipole moment vector points from negative to positive charge.
Q2: B - Electric field on axial line at large distance varies as \(1/r^3\).
Q3: A - Using \(E = \frac{1}{4 \pi \epsilon_0} \frac{2p}{r^3}\), calculate \(E \approx 7.2 \times 10^3\) N/C.

Memory Aids

Dipole moment \(\vec{p} = q \times 2a\), points negative to positive.
Axial field \(\propto 2p/r^3\), equatorial field \(\propto -p/r^3\).
At large distances, axial field is twice equatorial field.

Definition and Dipole Moment

Electric dipole: two equal and opposite charges separated by a small distance.
Dipole moment p = q (2a), points from negative to positive charge.
Distance between charges = 2a.

Diagram 101 – NEET Physics Electric Charges and Fields

Electric Field on Axial Line (End-on Position)

Electric field at point on axis: E = E_+q + E_-q = q p̂/(4 π ε₀) [ 1/(r - a)² - 1/(r + a)² ]
Also, E = 2 p r/(4 π ε₀ (r² - a²)²)
At large distance (r ≫ a), E = 1/(4 π ε₀) 2p/r³

Diagram 102 – NEET Physics Electric Charges and Fields

Electric Field on Equatorial Line (Broadside-on Position)

Electric field at equatorial point: E = - (E_+q + E_-q) cos θ p̂ = 1/(4 π ε₀) -p/(r² + a²)^3/2
At large distance (r ≫ a), E = - p/(4 π ε₀ r³)

Diagram 103 – NEET Physics Electric Charges and Fields

Physical Significance

Nonpolar molecules: centers of positive and negative charges coincide (e.g., CO₂, CH₄) → no permanent dipole moment.
Polar molecules: centers do not coincide (e.g., H₂O, NH₃) → permanent dipole moment.
External electric field can induce dipole moment in nonpolar molecules.

Solved Examples

Ex 1

Calculate electric field at axial point r = 10a for dipole moment p.
Use formula: E = 1/(4 π ε₀) 2p/r³.
Substitute r = 10a, calculate E.

Practice Set

Q1: Dipole moment p points from (A) positive to negative charge (B) negative to positive charge (C) both directions (D) none
Q2: Electric field at large distance on axial line is proportional to (A) 1/r² (B) 1/r³ (C) 1/r (D) r²
Q3: For a dipole with p = 4 × 10^-29 C·m, distance r = 0.1 m, calculate approximate axial electric field E using ε₀ = 8.85 × 10^-12 F/m. (A) 7.2 × 10³ N/C (B) 1.8 × 10⁴ N/C (C) 3.6 × 10³ N/C (D) 9.0 × 10² N/C

Answer Key

Q1: B - Dipole moment vector points from negative to positive charge.
Q2: B - Electric field on axial line at large distance varies as 1/r³.
Q3: A - Using E = 1/(4 π ε₀) 2p/r³, calculate E ≈ 7.2 × 10³ N/C.

Memory Aids

Dipole moment p = q × 2a, points negative to positive.
Axial field ∝ 2p/r³, equatorial field ∝ -p/r³.
At large distances, axial field is twice equatorial field.

Definition and Dipole Moment

Electric dipole: two equal and opposite charges separated by a small distance.
Dipole moment p = q (2a), points from negative to positive charge.
Distance between charges = 2a.

Diagram 104 – NEET Physics Electric Charges and Fields

Electric Field on Axial Line (End-on Position)

Electric field at point on axis: E = E_+q + E_-q = q p̂/(4 π ε₀) [ 1/(r - a)² - 1/(r + a)² ]
Also, E = 2 p r/(4 π ε₀ (r² - a²)²)
At large distance (r ≫ a), E = 1/(4 π ε₀) 2p/r³

Diagram 105 – NEET Physics Electric Charges and Fields

Electric Field on Equatorial Line (Broadside-on Position)

Electric field at equatorial point: E = - (E_+q + E_-q) cos θ p̂ = 1/(4 π ε₀) -p/(r² + a²)^3/2
At large distance (r ≫ a), E = - p/(4 π ε₀ r³)

Diagram 106 – NEET Physics Electric Charges and Fields

Physical Significance

Nonpolar molecules: centers of positive and negative charges coincide (e.g., CO₂, CH₄) → no permanent dipole moment.
Polar molecules: centers do not coincide (e.g., H₂O, NH₃) → permanent dipole moment.
External electric field can induce dipole moment in nonpolar molecules.

Solved Examples

Ex 1

Calculate electric field at axial point r = 10a for dipole moment p.
Use formula: E = 1/(4 π ε₀) 2p/r³.
Substitute r = 10a, calculate E.

Practice Set

Q1: Dipole moment p points from (A) positive to negative charge (B) negative to positive charge (C) both directions (D) none
Q2: Electric field at large distance on axial line is proportional to (A) 1/r² (B) 1/r³ (C) 1/r (D) r²
Q3: For a dipole with p = 4 × 10^-29 C·m, distance r = 0.1 m, calculate approximate axial electric field E using ε₀ = 8.85 × 10^-12 F/m. (A) 7.2 × 10³ N/C (B) 1.8 × 10⁴ N/C (C) 3.6 × 10³ N/C (D) 9.0 × 10² N/C

Answer Key

Q1: B - Dipole moment vector points from negative to positive charge.
Q2: B - Electric field on axial line at large distance varies as 1/r³.
Q3: A - Using E = 1/(4 π ε₀) 2p/r³, calculate E ≈ 7.2 × 10³ N/C.

Memory Aids

Dipole moment p = q × 2a, points negative to positive.
Axial field ∝ 2p/r³, equatorial field ∝ -p/r³.
At large distances, axial field is twice equatorial field.

Definition and Dipole Moment

Electric dipole: two equal and opposite charges separated by a small distance.
Dipole moment (vector) = q (2a), points from negative to positive charge.
Distance between charges = 2a.

Diagram 107 – NEET Physics Electric Charges and Fields

Electric Field on Axial Line (End-on Position)

Electric field at point on axis: E = E_+q + E_-q = q p̂/(4 π ε₀) [ 1/(r - a)² - 1/(r + a)² ]
Also, E = 2 p r/(4 π ε₀ (r² - a²)²)
At large distance (r ≫ a), E = 1/(4 π ε₀) 2p/r³

Diagram 108 – NEET Physics Electric Charges and Fields

Electric Field on Equatorial Line (Broadside-on Position)

Electric field at equatorial point: E = - (E_+q + E_-q) cos θ p̂ = 1/(4 π ε₀) -p/(r² + a²)^3/2
At large distance (r ≫ a), E = - p/(4 π ε₀ r³)

Diagram 109 – NEET Physics Electric Charges and Fields

Physical Significance

Nonpolar molecules: centers of positive and negative charges coincide (e.g., CO₂, CH₄) → no permanent dipole moment.
Polar molecules: centers do not coincide (e.g., H₂O, NH₃) → permanent dipole moment.
External electric field can induce dipole moment in nonpolar molecules.

Solved Examples

Ex 1

Calculate electric field at axial point r = 10a for dipole moment p.
Use formula: E = 1/(4 π ε₀) 2p/r³.
Substitute r = 10a, calculate E.

Practice Set

Q1: Dipole moment p points from (A) positive to negative charge (B) negative to positive charge (C) both directions (D) none
Q2: Electric field at large distance on axial line is proportional to (A) 1/r² (B) 1/r³ (C) 1/r (D) r²
Q3: For a dipole with p = 4 × 10^-29 C·m, distance r = 0.1 m, calculate approximate axial electric field E using ε₀ = 8.85 × 10^-12 F/m. (A) 7.2 × 10³ N/C (B) 1.8 × 10⁴ N/C (C) 3.6 × 10³ N/C (D) 9.0 × 10² N/C

Answer Key

Q1: B - Dipole moment vector points from negative to positive charge.
Q2: B - Electric field on axial line at large distance varies as 1/r³.
Q3: A - Using E = 1/(4 π ε₀) 2p/r³, calculate E ≈ 7.2 × 10³ N/C.

Memory Aids

Dipole moment p = q × 2a, points negative to positive.
Axial field ∝ 2p/r³, equatorial field ∝ -p/r³.
At large distances, axial field is twice equatorial field.

Definition and Dipole Moment

Electric dipole: two equal and opposite charges separated by a small distance.
Dipole moment p = q (2a), points from negative to positive charge.
Distance between charges = 2a.

Diagram 110 – NEET Physics Electric Charges and Fields

Electric Field on Axial Line (End-on Position)

Electric field at point on axis: E = E_+q + E_-q = q p̂/(4 π ε₀) [ 1/(r - a)² - 1/(r + a)² ]
Also, E = 2 p r/(4 π ε₀ (r² - a²)²)
At large distance (r ≫ a), E = 1/(4 π ε₀) 2p/r³

Diagram 111 – NEET Physics Electric Charges and Fields

Electric Field on Equatorial Line (Broadside-on Position)

Electric field at equatorial point: E = - (E_+q + E_-q) cos θ p̂ = 1/(4 π ε₀) -p/(r² + a²)^3/2
At large distance (r ≫ a), E = - p/(4 π ε₀ r³)

Diagram 112 – NEET Physics Electric Charges and Fields

Physical Significance

Nonpolar molecules: centers of positive and negative charges coincide (e.g., CO₂, CH₄) → no permanent dipole moment.
Polar molecules: centers do not coincide (e.g., H₂O, NH₃) → permanent dipole moment.
External electric field can induce dipole moment in nonpolar molecules.

Solved Examples

Ex 1

Calculate electric field at axial point r = 10a for dipole moment p.
Use formula: E = 1/(4 π ε₀) 2p/r³.
Substitute r = 10a, calculate E.

Practice Set

Q1: Dipole moment p points from (A) positive to negative charge (B) negative to positive charge (C) both directions (D) none
Q2: Electric field at large distance on axial line is proportional to (A) 1/r² (B) 1/r³ (C) 1/r (D) r²
Q3: For a dipole with p = 4 × 10^-29 C·m, distance r = 0.1 m, calculate approximate axial electric field E using ε₀ = 8.85 × 10^-12 F/m. (A) 7.2 × 10³ N/C (B) 1.8 × 10⁴ N/C (C) 3.6 × 10³ N/C (D) 9.0 × 10² N/C

Answer Key

Q1: B - Dipole moment vector points from negative to positive charge.
Q2: B - Electric field on axial line at large distance varies as 1/r³.
Q3: A - Using E = 1/(4 π ε₀) 2p/r³, calculate E ≈ 7.2 × 10³ N/C.

Memory Aids

Dipole moment p = q × 2a, points negative to positive.
Axial field ∝ 2p/r³, equatorial field ∝ -p/r³.
At large distances, axial field is twice equatorial field.

Definition and Dipole Moment

Electric dipole: two equal and opposite charges separated by a small distance.
Dipole moment \( \vec{p} = q (2 \vec{a}) \), points from negative to positive charge.
Distance between charges = \(2a\).

Diagram 113 – NEET Physics Electric Charges and Fields

Electric Field on Axial Line (End-on Position)

Electric field at point on axis: \( \vec{E} = \vec{E}_{+q} + \vec{E}_{-q} = \frac{q \hat{p}}{4 \pi \epsilon_0} \left[ \frac{1}{(r - a)^2} - \frac{1}{(r + a)^2} \right] \)
Also, \( \vec{E} = \frac{2 \vec{p} r}{4 \pi \epsilon_0 (r^2 - a^2)^2} \)
At large distance \( (r \gg a) \), \( \vec{E} = \frac{1}{4 \pi \epsilon_0} \frac{2 \vec{p}}{r^3} \)

Diagram 114 – NEET Physics Electric Charges and Fields

Electric Field on Equatorial Line (Broadside-on Position)

Electric field at equatorial point: \( \vec{E} = - (\vec{E}_{+q} + \vec{E}_{-q}) \cos \theta \hat{p} = \frac{1}{4 \pi \epsilon_0} \frac{- \vec{p}}{(r^2 + a^2)^{3/2}} \)
At large distance \( (r \gg a) \), \( \vec{E} = - \frac{\vec{p}}{4 \pi \epsilon_0 r^3} \)

Diagram 115 – NEET Physics Electric Charges and Fields

Physical Significance

Nonpolar molecules: centers of positive and negative charges coincide (e.g., CO\(_2\), CH\(_4\)) → no permanent dipole moment.
Polar molecules: centers do not coincide (e.g., H\(_2\)O, NH\(_3\)) → permanent dipole moment.
External electric field can induce dipole moment in nonpolar molecules.

Solved Examples

Ex 1

Calculate electric field at axial point \(r = 10a\) for dipole moment \(p\).
Use formula: \( E = \frac{1}{4 \pi \epsilon_0} \frac{2p}{r^3} \).
Substitute \(r = 10a\), calculate \(E\).

Practice Set

Q1: Dipole moment \( \vec{p} \) points from (A) positive to negative charge (B) negative to positive charge (C) both directions (D) none
Q2: Electric field at large distance on axial line is proportional to (A) \(1/r^2\) (B) \(1/r^3\) (C) \(1/r\) (D) \(r^2\)
Q3: For a dipole with \(p = 4 \times 10^{-29} \) C·m, distance \(r = 0.1\) m, calculate approximate axial electric field \(E\) using \( \epsilon_0 = 8.85 \times 10^{-12} \) F/m. (A) \(7.2 \times 10^3\) N/C (B) \(1.8 \times 10^4\) N/C (C) \(3.6 \times 10^3\) N/C (D) \(9.0 \times 10^2\) N/C

Answer Key

Q1: B - Dipole moment vector points from negative to positive charge.
Q2: B - Electric field on axial line at large distance varies as \(1/r^3\).
Q3: A - Using \(E = \frac{1}{4 \pi \epsilon_0} \frac{2p}{r^3}\), calculate \(E \approx 7.2 \times 10^3\) N/C.

Memory Aids

Dipole moment \(\vec{p} = q \times 2a\), points negative to positive.
Axial field \(\propto 2p/r^3\), equatorial field \(\propto -p/r^3\).
At large distances, axial field is twice equatorial field.

Electric Field at General Point

Field Components and Resultant

Resolve dipole moment p into components: p cos θ (axial), p sin θ (equatorial).
Electric field due to axial component: E₁ = 2k p cos θ/r³.
Electric field due to equatorial component: E₂ = k p sin θ/r³.
Resultant field magnitude: E = k p √(3 cos² θ + 1)/r³.
Angle α between E and r: tan α = tan θ/2.

Diagram 116 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1

Find resultant electric field at point P where θ = 60°, r = 0.2 m, dipole moment p = 5 × 10^-29 C·m.
Calculate E₁, E₂ using formulas.
Calculate resultant E and angle α.

Practice Set

Q1: The angle α between resultant field and r is given by (A) tan^-1(tan θ) (B) tan^-1(tan θ / 2) (C) tan^-1(2 tan θ) (D) tan^-1(sin θ)
Q2: Resultant electric field magnitude varies as (A) 1/r² (B) 1/r³ (C) 1/r (D) r³
Q3: For θ = 45°, p = 3 × 10^-29 C·m, r = 0.1 m, calculate E₁ and E₂ (use k = 9 × 10⁹ Nm²/C²). (A) E₁ = 3.8 × 10⁴, E₂ = 1.9 × 10⁴ (B) E₁ = 2.7 × 10⁴, E₂ = 2.7 × 10⁴ (C) E₁ = 1.9 × 10⁴, E₂ = 3.8 × 10⁴ (D) E₁ = 4.0 × 10⁴, E₂ = 2.0 × 10⁴

Answer Key

Q1: B - α = tan^-1(tan θ / 2) from formula.
Q2: B - Electric field varies as inverse cube of distance 1/r³.
Q3: A - Calculate using E₁ = 2kp cos θ / r³, E₂ = kp sin θ / r³.

Memory Aids

Resolve dipole moment into axial and equatorial components.
Resultant field magnitude involves √(3 cos² θ + 1).
Angle α is half the tangent of θ in tangent form.

Field Components and Resultant

Resolve dipole moment p into components: p cos θ (axial), p sin θ (equatorial).
Electric field due to axial component: E₁ = 2k p cos θ/r³.
Electric field due to equatorial component: E₂ = k p sin θ/r³.
Resultant field magnitude: E = k p √(3 cos² θ + 1)/r³.
Angle α between E and r: tan α = tan θ/2.

Diagram 117 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1

Find resultant electric field at point P where θ = 60°, r = 0.2 m, dipole moment p = 5 × 10^-29 C·m.
Calculate E₁, E₂ using formulas.
Calculate resultant E and angle α.

Practice Set

Q1: The angle α between resultant field and r is given by (A) tan^-1(tan θ) (B) tan^-1(tan θ / 2) (C) tan^-1(2 tan θ) (D) tan^-1(sin θ)
Q2: Resultant electric field magnitude varies as (A) 1/r² (B) 1/r³ (C) 1/r (D) r³
Q3: For θ = 45°, p = 3 × 10^-29 C·m, r = 0.1 m, calculate E₁ and E₂ (use k = 9 × 10⁹ Nm²/C²). (A) E₁ = 3.8 × 10⁴, E₂ = 1.9 × 10⁴ (B) E₁ = 2.7 × 10⁴, E₂ = 2.7 × 10⁴ (C) E₁ = 1.9 × 10⁴, E₂ = 3.8 × 10⁴ (D) E₁ = 4.0 × 10⁴, E₂ = 2.0 × 10⁴

Answer Key

Q1: B - α = tan^-1(tan θ / 2) from formula.
Q2: B - Electric field varies as inverse cube of distance 1/r³.
Q3: A - Calculate using E₁ = 2kp cos θ / r³, E₂ = kp sin θ / r³.

Memory Aids

Resolve dipole moment into axial and equatorial components.
Resultant field magnitude involves √(3 cos² θ + 1).
Angle α is half the tangent of θ in tangent form.

Field Components and Resultant

Resolve dipole moment (vector) into components: p cos θ (axial), p sin θ (equatorial).
Electric field due to axial component: E₁ = 2k p cos θ/r³.
Electric field due to equatorial component: E₂ = k p sin θ/r³.
Resultant field magnitude: E = k p √(3 cos² θ + 1)/r³.
Angle α between E and r: tan α = tan θ/2.

Diagram 118 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1

Find resultant electric field at point P where θ = 60°, r = 0.2 m, dipole moment p = 5 × 10^-29 C·m.
Calculate E₁, E₂ using formulas.
Calculate resultant E and angle α.

Practice Set

Q1: The angle α between resultant field and r is given by (A) tan^-1(tan θ) (B) tan^-1(tan θ / 2) (C) tan^-1(2 tan θ) (D) tan^-1(sin θ)
Q2: Resultant electric field magnitude varies as (A) 1/r² (B) 1/r³ (C) 1/r (D) r³
Q3: For θ = 45°, p = 3 × 10^-29 C·m, r = 0.1 m, calculate E₁ and E₂ (use k = 9 × 10⁹ Nm²/C²). (A) E₁ = 3.8 × 10⁴, E₂ = 1.9 × 10⁴ (B) E₁ = 2.7 × 10⁴, E₂ = 2.7 × 10⁴ (C) E₁ = 1.9 × 10⁴, E₂ = 3.8 × 10⁴ (D) E₁ = 4.0 × 10⁴, E₂ = 2.0 × 10⁴

Answer Key

Q1: B - α = tan^-1(tan θ / 2) from formula.
Q2: B - Electric field varies as inverse cube of distance 1/r³.
Q3: A - Calculate using E₁ = 2kp cos θ / r³, E₂ = kp sin θ / r³.

Memory Aids

Resolve dipole moment into axial and equatorial components.
Resultant field magnitude involves √(3 cos² θ + 1).
Angle α is half the tangent of θ in tangent form.

Field Components and Resultant

Resolve dipole moment p into components: p cos θ (axial), p sin θ (equatorial).
Electric field due to axial component: E₁ = 2k p cos θ/r³.
Electric field due to equatorial component: E₂ = k p sin θ/r³.
Resultant field magnitude: E = k p √(3 cos² θ + 1)/r³.
Angle α between E and r: tan α = tan θ/2.

Diagram 119 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1

Find resultant electric field at point P where θ = 60°, r = 0.2 m, dipole moment p = 5 × 10^-29 C·m.
Calculate E₁, E₂ using formulas.
Calculate resultant E and angle α.

Practice Set

Q1: The angle α between resultant field and r is given by (A) tan^-1(tan θ) (B) tan^-1(tan θ / 2) (C) tan^-1(2 tan θ) (D) tan^-1(sin θ)
Q2: Resultant electric field magnitude varies as (A) 1/r² (B) 1/r³ (C) 1/r (D) r³
Q3: For θ = 45°, p = 3 × 10^-29 C·m, r = 0.1 m, calculate E₁ and E₂ (use k = 9 × 10⁹ Nm²/C²). (A) E₁ = 3.8 × 10⁴, E₂ = 1.9 × 10⁴ (B) E₁ = 2.7 × 10⁴, E₂ = 2.7 × 10⁴ (C) E₁ = 1.9 × 10⁴, E₂ = 3.8 × 10⁴ (D) E₁ = 4.0 × 10⁴, E₂ = 2.0 × 10⁴

Answer Key

Q1: B - α = tan^-1(tan θ / 2) from formula.
Q2: B - Electric field varies as inverse cube of distance 1/r³.
Q3: A - Calculate using E₁ = 2kp cos θ / r³, E₂ = kp sin θ / r³.

Memory Aids

Resolve dipole moment into axial and equatorial components.
Resultant field magnitude involves √(3 cos² θ + 1).
Angle α is half the tangent of θ in tangent form.

Field Components and Resultant

Resolve dipole moment \( \vec{p} \) into components: \( p \cos \theta \) (axial), \( p \sin \theta \) (equatorial).
Electric field due to axial component: \( E_1 = \frac{2k p \cos \theta}{r^3} \).
Electric field due to equatorial component: \( E_2 = \frac{k p \sin \theta}{r^3} \).
Resultant field magnitude: \( E = k p \frac{\sqrt{3 \cos^2 \theta + 1}}{r^3} \).
Angle \( \alpha \) between \( \vec{E} \) and \( \vec{r} \): \( \tan \alpha = \frac{\tan \theta}{2} \).

Diagram 120 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1

Find resultant electric field at point \(P\) where \( \theta = 60^\circ \), \(r = 0.2\) m, dipole moment \(p = 5 \times 10^{-29} \) C·m.
Calculate \(E_1, E_2\) using formulas.
Calculate resultant \(E\) and angle \(\alpha\).

Practice Set

Q1: The angle \(\alpha\) between resultant field and \(\vec{r}\) is given by (A) \(\tan^{-1}(\tan \theta)\) (B) \(\tan^{-1}(\tan \theta / 2)\) (C) \(\tan^{-1}(2 \tan \theta)\) (D) \(\tan^{-1}(\sin \theta)\)
Q2: Resultant electric field magnitude varies as (A) \(1/r^2\) (B) \(1/r^3\) (C) \(1/r\) (D) \(r^3\)
Q3: For \( \theta = 45^\circ \), \(p = 3 \times 10^{-29} \) C·m, \(r = 0.1\) m, calculate \(E_1\) and \(E_2\) (use \(k = 9 \times 10^9\) Nm²/C²). (A) \(E_1 = 3.8 \times 10^4, E_2 = 1.9 \times 10^4\) (B) \(E_1 = 2.7 \times 10^4, E_2 = 2.7 \times 10^4\) (C) \(E_1 = 1.9 \times 10^4, E_2 = 3.8 \times 10^4\) (D) \(E_1 = 4.0 \times 10^4, E_2 = 2.0 \times 10^4\)

Answer Key

Q1: B - \( \alpha = \tan^{-1}(\tan \theta / 2) \) from formula.
Q2: B - Electric field varies as inverse cube of distance \(1/r^3\).
Q3: A - Calculate using \(E_1 = 2kp \cos \theta / r^3\), \(E_2 = kp \sin \theta / r^3\).

Memory Aids

Resolve dipole moment into axial and equatorial components.
Resultant field magnitude involves \( \sqrt{3 \cos^2 \theta + 1} \).
Angle \( \alpha \) is half the tangent of \( \theta \) in tangent form.

Dipole in a Uniform External Field

Concepts and Formulas

An electric dipole consists of charges +q and -q separated by distance 2a.
Placed in uniform electric field E, forces on charges are equal and opposite, net force = 0.
Forces are not collinear, causing torque τ = F r_⊥ = qE × 2a sin θ = pE sin θ, where p = q × 2a.
Vector form: τ = p × E, θ is angle between p and E.
Torque extremes: τ_min = 0 at θ = 0°, 180°; τ_max = pE at θ = 90°.
Torque aligns dipole along the electric field direction.

Diagram 121 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate torque on dipole with p = 4 × 10^-29 C m, E = 5 × 10⁴ N/C, θ = 60°.
- Use τ = pE sin θ = 4 × 10^-29 × 5 × 10⁴ × sin 60°.
- τ = 1.73 × 10^-24 N m.

Practice Set

Q1: What is the net force on an electric dipole in a uniform electric field?
- A) Zero
- B) qE
- C) 2qE
- D) Depends on θ
Q2: Torque on a dipole is maximum when θ is:
- A) 0°
- B) 45°
- C) 90°
- D) 180°
Q3: A dipole with moment 2 × 10^-29 C m is in a field of 10⁵ N/C at 30°. Calculate torque.
- A) 1 × 10^-24 N m
- B) 5 × 10^-25 N m
- C) 1 × 10^-25 N m
- D) 2 × 10^-24 N m

Answer Key

Q1: A - Forces on +q and -q cancel, net force is zero in uniform field.
Q2: C - Torque τ = pE sin θ is max at 90°.
Q3: B - τ = pE sin 30° = 2 × 10^-29 × 10⁵ × 0.5 = 5 × 10^-25 N m.

Memory Aids

Torque τ = pE sin θ aligns dipole with field.
Net force zero in uniform field.
Max torque at 90°, zero at 0°, 180°.

Concepts and Formulas

An electric dipole consists of charges +q and -q separated by distance 2a.
Placed in uniform electric field E, forces on charges are equal and opposite, net force = 0.
Forces are not collinear, causing torque τ = F r_⊥ = qE × 2a sin θ = pE sin θ, where p = q × 2a.
Vector form: τ = p × E, θ is angle between p and E.
Torque extremes: τ_min = 0 at θ = 0°, 180°; τ_max = pE at θ = 90°.
Torque aligns dipole along the electric field direction.

Diagram 122 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate torque on dipole with p = 4 × 10^-29 C m, E = 5 × 10⁴ N/C, θ = 60°.
- Use τ = pE sin θ = 4 × 10^-29 × 5 × 10⁴ × sin 60°.
- τ = 1.73 × 10^-24 N m.

Practice Set

Q1: What is the net force on an electric dipole in a uniform electric field?
- A) Zero
- B) qE
- C) 2qE
- D) Depends on θ
Q2: Torque on a dipole is maximum when θ is:
- A) 0°
- B) 45°
- C) 90°
- D) 180°
Q3: A dipole with moment 2 × 10^-29 C m is in a field of 10⁵ N/C at 30°. Calculate torque.
- A) 1 × 10^-24 N m
- B) 5 × 10^-25 N m
- C) 1 × 10^-25 N m
- D) 2 × 10^-24 N m

Answer Key

Q1: A - Forces on +q and -q cancel, net force is zero in uniform field.
Q2: C - Torque τ = pE sin θ is max at 90°.
Q3: B - τ = pE sin 30° = 2 × 10^-29 × 10⁵ × 0.5 = 5 × 10^-25 N m.

Memory Aids

Torque τ = pE sin θ aligns dipole with field.
Net force zero in uniform field.
Max torque at 90°, zero at 0°, 180°.

Concepts and Formulas

An electric dipole consists of charges +q and -q separated by distance 2a.
Placed in uniform electric field vec{E}, forces on charges are equal and opposite, net force = 0.
Forces are not collinear, causing torque tau = F r_perp = qE × 2a sin θ = pE sin θ, where p = q × 2a.
Vector form: vec{tau} = vec{p} × vec{E}, θ is angle between vec{p} and vec{E}.
Torque extremes: tau_min = 0 at θ = 0°, 180°; tau_max = pE at θ = 90°.
Torque aligns dipole along the electric field direction.

Diagram 123 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate torque on dipole with p = 4 × 10^-29 text{ C m}, E = 5 × 10⁴ text{ N/C}, theta = 60°.
- Use tau = pE sin theta = 4 × 10^-29 × 5 × 10⁴ × sin 60°.
- tau = 1.73 × 10^-24 text{ N m}.

Practice Set

Q1: What is the net force on an electric dipole in a uniform electric field?
- A) Zero
- B) qE
- C) 2qE
- D) Depends on theta
Q2: Torque on a dipole is maximum when theta is:
- A) 0°
- B) 45°
- C) 90°
- D) 180°
Q3: A dipole with moment 2 × 10^-29 text{ C m} is in a field of 10⁵ text{ N/C} at 30°. Calculate torque.
- A) 1 × 10^-24 text{ N m}
- B) 5 × 10^-25 text{ N m}
- C) 1 × 10^-25 text{ N m}
- D) 2 × 10^-24 text{ N m}

Answer Key

Q1: A - Forces on +q and -q cancel, net force is zero in uniform field.
Q2: C - Torque tau = pE sin theta is max at 90°.
Q3: B - tau = pE sin 30° = 2 × 10^-29 × 10⁵ × 0.5 = 5 × 10^-25 text{ N m}.

Memory Aids

Torque tau = pE sin theta aligns dipole with field.
Net force zero in uniform field.
Max torque at 90°, zero at 0°, 180°.

Concepts and Formulas

An electric dipole consists of charges +q and -q separated by distance 2a.
Placed in uniform electric field E, forces on charges are equal and opposite, net force = 0.
Forces are not collinear, causing torque τ = F r_⊥ = qE × 2a sin θ = pE sin θ, where p = q × 2a.
Vector form: τ = p × E, θ is angle between p and E.
Torque extremes: τ_min = 0 at θ = 0°, 180°; τ_max = pE at θ = 90°.
Torque aligns dipole along the electric field direction.

Diagram 124 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate torque on dipole with p = 4 × 10^-29 C m, E = 5 × 10⁴ N/C, θ = 60°.
- Use τ = pE sin θ = 4 × 10^-29 × 5 × 10⁴ × sin 60°.
- τ = 1.73 × 10^-24 N m.

Practice Set

Q1: What is the net force on an electric dipole in a uniform electric field?
- A) Zero
- B) qE
- C) 2qE
- D) Depends on θ
Q2: Torque on a dipole is maximum when θ is:
- A) 0°
- B) 45°
- C) 90°
- D) 180°
Q3: A dipole with moment 2 × 10^-29 C m is in a field of 10⁵ N/C at 30°. Calculate torque.
- A) 1 × 10^-24 N m
- B) 5 × 10^-25 N m
- C) 1 × 10^-25 N m
- D) 2 × 10^-24 N m

Answer Key

Q1: A - Forces on +q and -q cancel, net force is zero in uniform field.
Q2: C - Torque τ = pE sin θ is max at 90°.
Q3: B - τ = pE sin 30° = 2 × 10^-29 × 10⁵ × 0.5 = 5 × 10^-25 N m.

Memory Aids

Torque τ = pE sin θ aligns dipole with field.
Net force zero in uniform field.
Max torque at 90°, zero at 0°, 180°.

Concepts and Formulas

An electric dipole consists of charges +q and -q separated by distance 2a.
Placed in uniform electric field \(\vec{E}\), forces on charges are equal and opposite, net force = 0.
Forces are not collinear, causing torque \(\tau = F r_\perp = qE \times 2a \sin \theta = pE \sin \theta\), where \(p = q \times 2a\).
Vector form: \(\vec{\tau} = \vec{p} \times \vec{E}\), \(\theta\) is angle between \(\vec{p}\) and \(\vec{E}\).
Torque extremes: \(\tau_{min} = 0\) at \(\theta = 0^\circ, 180^\circ\); \(\tau_{max} = pE\) at \(\theta = 90^\circ\).
Torque aligns dipole along the electric field direction.

Diagram 125 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate torque on dipole with \(p = 4 \times 10^{-29} \text{ C m}\), \(E = 5 \times 10^{4} \text{ N/C}\), \(\theta = 60^\circ\).
- Use \(\tau = pE \sin \theta = 4 \times 10^{-29} \times 5 \times 10^{4} \times \sin 60^\circ\).
- \(\tau = 1.73 \times 10^{-24} \text{ N m}\).

Practice Set

Q1: What is the net force on an electric dipole in a uniform electric field?
- A) Zero
- B) \(qE\)
- C) \(2qE\)
- D) Depends on \(\theta\)
Q2: Torque on a dipole is maximum when \(\theta\) is:
- A) 0°
- B) 45°
- C) 90°
- D) 180°
Q3: A dipole with moment \(2 \times 10^{-29} \text{ C m}\) is in a field of \(10^{5} \text{ N/C}\) at \(30^\circ\). Calculate torque.
- A) \(1 \times 10^{-24} \text{ N m}\)
- B) \(5 \times 10^{-25} \text{ N m}\)
- C) \(1 \times 10^{-25} \text{ N m}\)
- D) \(2 \times 10^{-24} \text{ N m}\)

Answer Key

Q1: A - Forces on +q and -q cancel, net force is zero in uniform field.
Q2: C - Torque \(\tau = pE \sin \theta\) is max at \(90^\circ\).
Q3: B - \(\tau = pE \sin 30^\circ = 2 \times 10^{-29} \times 10^{5} \times 0.5 = 5 \times 10^{-25} \text{ N m}\).

Memory Aids

Torque \(\tau = pE \sin \theta\) aligns dipole with field.
Net force zero in uniform field.
Max torque at \(90^\circ\), zero at \(0^\circ, 180^\circ\).

Dipole in a Non-Uniform Electric Field

Concepts and Formulas

In non-uniform field, forces on +q and -q differ, causing net force and torque on dipole.
Net force direction depends on dipole orientation relative to field gradient.
If E not parallel or antiparallel to p, dipole experiences both net force and torque.
Time period of oscillating dipole in uniform field: T = 2 π √(I/pE), where I is moment of inertia.

Diagram 126 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Dipole in increasing field experiences net force. If force on +q is 2 × 10^-6 N right and on -q is 1 × 10^-6 N left, net force?
- Net force = 2 × 10^-6 - 1 × 10^-6 = 1 × 10^-6 N right.

Practice Set

Q1: What happens to dipole in non-uniform field?
- A) Only torque
- B) Only net force
- C) Both net force and torque
- D) No force or torque
Q2: If E is not parallel to p, dipole experiences:
- A) No force
- B) Only torque
- C) Only net force
- D) Net force and torque
Q3: Time period of oscillating dipole depends on:
- A) √(I/pE)
- B) √(pE/I)
- C) I/pE
- D) pE/I

Answer Key

Q1: C - Non-uniform field causes both net force and torque.
Q2: D - Non-parallel E and p cause net force and torque.
Q3: A - T = 2 π √(I/pE) time period formula.

Memory Aids

Non-uniform field: net force + torque.
Oscillation period T = 2 π √(I/pE).
Force direction depends on dipole orientation.

Concepts and Formulas

In non-uniform field, forces on +q and -q differ, causing net force and torque on dipole.
Net force direction depends on dipole orientation relative to field gradient.
If E not parallel or antiparallel to p, dipole experiences both net force and torque.
Time period of oscillating dipole in uniform field: T = 2 π √(I/pE), where I is moment of inertia.

Diagram 127 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Dipole in increasing field experiences net force. If force on +q is 2 × 10^-6 N right and on -q is 1 × 10^-6 N left, net force?
- Net force = 2 × 10^-6 - 1 × 10^-6 = 1 × 10^-6 N right.

Practice Set

Q1: What happens to dipole in non-uniform field?
- A) Only torque
- B) Only net force
- C) Both net force and torque
- D) No force or torque
Q2: If E is not parallel to p, dipole experiences:
- A) No force
- B) Only torque
- C) Only net force
- D) Net force and torque
Q3: Time period of oscillating dipole depends on:
- A) √(I/pE)
- B) √(pE/I)
- C) I/pE
- D) pE/I

Answer Key

Q1: C - Non-uniform field causes both net force and torque.
Q2: D - Non-parallel E and p cause net force and torque.
Q3: A - T = 2 π √(I/pE) time period formula.

Memory Aids

Non-uniform field: net force + torque.
Oscillation period T = 2 π √(I/pE).
Force direction depends on dipole orientation.

Concepts and Formulas

In non-uniform field, forces on +q and -q differ, causing net force and torque on dipole.
Net force direction depends on dipole orientation relative to field gradient.
If vec{E} not parallel or antiparallel to vec{p}, dipole experiences both net force and torque.
Time period of oscillating dipole in uniform field: T = 2 π √(I/pE), where I is moment of inertia.

Diagram 128 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Dipole in increasing field experiences net force. If force on +q is 2 × 10^-6 N right and on -q is 1 × 10^-6 N left, net force?
- Net force = 2 × 10^-6 - 1 × 10^-6 = 1 × 10^-6 N right.

Practice Set

Q1: What happens to dipole in non-uniform field?
- A) Only torque
- B) Only net force
- C) Both net force and torque
- D) No force or torque
Q2: If vec{E} is not parallel to vec{p}, dipole experiences:
- A) No force
- B) Only torque
- C) Only net force
- D) Net force and torque
Q3: Time period of oscillating dipole depends on:
- A) √(I/pE)
- B) √(pE/I)
- C) I/pE
- D) pE/I

Answer Key

Q1: C - Non-uniform field causes both net force and torque.
Q2: D - Non-parallel vec{E} and vec{p} cause net force and torque.
Q3: A - T = 2 π √(I/pE) time period formula.

Memory Aids

Non-uniform field: net force + torque.
Oscillation period T = 2 π √(I/pE).
Force direction depends on dipole orientation.

Concepts and Formulas

In non-uniform field, forces on +q and -q differ, causing net force and torque on dipole.
Net force direction depends on dipole orientation relative to field gradient.
If E not parallel or antiparallel to p, dipole experiences both net force and torque.
Time period of oscillating dipole in uniform field: T = 2 π √(I/pE), where I is moment of inertia.

Diagram 129 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Dipole in increasing field experiences net force. If force on +q is 2 × 10^-6 N right and on -q is 1 × 10^-6 N left, net force?
- Net force = 2 × 10^-6 - 1 × 10^-6 = 1 × 10^-6 N right.

Practice Set

Q1: What happens to dipole in non-uniform field?
- A) Only torque
- B) Only net force
- C) Both net force and torque
- D) No force or torque
Q2: If E is not parallel to p, dipole experiences:
- A) No force
- B) Only torque
- C) Only net force
- D) Net force and torque
Q3: Time period of oscillating dipole depends on:
- A) √(I/pE)
- B) √(pE/I)
- C) I/pE
- D) pE/I

Answer Key

Q1: C - Non-uniform field causes both net force and torque.
Q2: D - Non-parallel E and p cause net force and torque.
Q3: A - T = 2 π √(I/pE) time period formula.

Memory Aids

Non-uniform field: net force + torque.
Oscillation period T = 2 π √(I/pE).
Force direction depends on dipole orientation.

Concepts and Formulas

In non-uniform field, forces on +q and -q differ, causing net force and torque on dipole.
Net force direction depends on dipole orientation relative to field gradient.
If \(\vec{E}\) not parallel or antiparallel to \(\vec{p}\), dipole experiences both net force and torque.
Time period of oscillating dipole in uniform field: \(T = 2 \pi \sqrt{\frac{I}{pE}}\), where \(I\) is moment of inertia.

Diagram 130 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Dipole in increasing field experiences net force. If force on +q is \(2 \times 10^{-6} \text{ N}\) right and on -q is \(1 \times 10^{-6} \text{ N}\) left, net force?
- Net force = \(2 \times 10^{-6} - 1 \times 10^{-6} = 1 \times 10^{-6} \text{ N}\) right.

Practice Set

Q1: What happens to dipole in non-uniform field?
- A) Only torque
- B) Only net force
- C) Both net force and torque
- D) No force or torque
Q2: If \(\vec{E}\) is not parallel to \(\vec{p}\), dipole experiences:
- A) No force
- B) Only torque
- C) Only net force
- D) Net force and torque
Q3: Time period of oscillating dipole depends on:
- A) \(\sqrt{\frac{I}{pE}}\)
- B) \(\sqrt{\frac{pE}{I}}\)
- C) \(\frac{I}{pE}\)
- D) \(\frac{pE}{I}\)

Answer Key

Q1: C - Non-uniform field causes both net force and torque.
Q2: D - Non-parallel \(\vec{E}\) and \(\vec{p}\) cause net force and torque.
Q3: A - \(T = 2 \pi \sqrt{\frac{I}{pE}}\) time period formula.

Memory Aids

Non-uniform field: net force + torque.
Oscillation period \(T = 2 \pi \sqrt{I/pE}\).
Force direction depends on dipole orientation.

Continuous Charge Distribution

Charge Densities and Distribution

Electric charge is treated as continuous at macroscopic scale due to dense electron concentration.
Charge distributions:
- Line charge: Charge spread along a line segment. Charge ΔQ = λΔl, where λ is linear charge density (C/m).
- Surface charge: Charge spread over a surface area. Charge ΔQ = σΔS, where σ is surface charge density (C/m²).
- Volume charge: Charge distributed in volume. Charge ΔQ = ρΔV, where ρ is volume charge density (C/m³).
Charge densities:
- Linear charge density: λ = ΔQ/Δl (C/m)
- Surface charge density: σ = ΔQ/ΔS (C/m²)
- Volume charge density: ρ = ΔQ/ΔV (C/m³)
Electric field from continuous charge distributions:
- E_λ = 1/(4 π ε₀) ∑ λΔl/r'² r̂' (line charge)
- E_σ = 1/(4 π ε₀) ∑ σΔS/r'² r̂' (surface charge)
- E_ρ = 1/(4 π ε₀) ∑ ρΔV/r'² r̂' (volume charge)

Diagram 131 – NEET Physics Electric Charges and Fields

Diagram 132 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate charge on 2 m length of wire with λ = 3 × 10^-6 C/m.
- Use Q = λl = 3 × 10^-6 × 2 = 6 × 10^-6 C.
Ex 2: Find surface charge on 0.5 m² with σ = 4 × 10^-6 C/m².
- Use Q = σS = 4 × 10^-6 × 0.5 = 2 × 10^-6 C.

Practice Set

Q1: Linear charge density λ units are:
- A) C/m
- B) C/m²
- C) C/m³
- D) C
Q2: Surface charge density σ is charge per unit:
- A) Length
- B) Area
- C) Volume
- D) Mass
Q3: Volume charge density ρ units are:
- A) C/m
- B) C/m²
- C) C/m³
- D) C

Answer Key

Q1: A) C/m — Linear charge density is charge per unit length.
Q2: B) Area — Surface charge density is charge per unit area.
Q3: C) C/m³ — Volume charge density is charge per unit volume.

Memory Aids

λ for line (length), units C/m.
σ for surface (area), units C/m².
ρ for volume, units C/m³.
Electric field sums contributions from small charge elements.

Charge Densities and Distribution

Electric charge is treated as continuous at macroscopic scale due to dense electron concentration.
Charge distributions:
- Line charge: Charge spread along a line segment. Charge ΔQ = λΔl, where λ is linear charge density (C/m).
- Surface charge: Charge spread over a surface area. Charge ΔQ = σΔS, where σ is surface charge density (C/m²).
- Volume charge: Charge distributed in volume. Charge ΔQ = ρΔV, where ρ is volume charge density (C/m³).
Charge densities:
- Linear charge density: λ = ΔQ/Δl (C/m)
- Surface charge density: σ = ΔQ/ΔS (C/m²)
- Volume charge density: ρ = ΔQ/ΔV (C/m³)
Electric field from continuous charge distributions:
- E_λ = 1/(4 π ε₀) ∑ λΔl/r'² r̂' (line charge)
- E_σ = 1/(4 π ε₀) ∑ σΔS/r'² r̂' (surface charge)
- E_ρ = 1/(4 π ε₀) ∑ ρΔV/r'² r̂' (volume charge)

Diagram 133 – NEET Physics Electric Charges and Fields

Diagram 134 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate charge on 2 m length of wire with λ = 3 × 10^-6 C/m.
- Use Q = λl = 3 × 10^-6 × 2 = 6 × 10^-6 C.
Ex 2: Find surface charge on 0.5 m² with σ = 4 × 10^-6 C/m².
- Use Q = σS = 4 × 10^-6 × 0.5 = 2 × 10^-6 C.

Practice Set

Q1: Linear charge density λ units are:
- A) C/m
- B) C/m²
- C) C/m³
- D) C
Q2: Surface charge density σ is charge per unit:
- A) Length
- B) Area
- C) Volume
- D) Mass
Q3: Volume charge density ρ units are:
- A) C/m
- B) C/m²
- C) C/m³
- D) C

Answer Key

Q1: A) C/m — Linear charge density is charge per unit length.
Q2: B) Area — Surface charge density is charge per unit area.
Q3: C) C/m³ — Volume charge density is charge per unit volume.

Memory Aids

λ for line (length), units C/m.
σ for surface (area), units C/m².
ρ for volume, units C/m³.
Electric field sums contributions from small charge elements.

Charge Densities and Distribution

Electric charge is treated as continuous at macroscopic scale due to dense electron concentration.
Charge distributions:
- Line charge: Charge spread along a line segment. Charge ΔQ = λΔl, where λ is linear charge density (C/m).
- Surface charge: Charge spread over a surface area. Charge ΔQ = σΔS, where σ is surface charge density (C/m²).
- Volume charge: Charge distributed in volume. Charge ΔQ = ρΔV, where ρ is volume charge density (C/m³).
Charge densities:
- Linear charge density: λ = ΔQ/Δl (C/m)
- Surface charge density: σ = ΔQ/ΔS (C/m²)
- Volume charge density: ρ = ΔQ/ΔV (C/m³)
Electric field from continuous charge distributions:
- E_λ = 1/(4 π ε₀) ∑ λΔl/r'² r̂' (line charge)
- E_σ = 1/(4 π ε₀) ∑ σΔS/r'² r̂' (surface charge)
- E_ρ = 1/(4 π ε₀) ∑ ρΔV/r'² r̂' (volume charge)

Diagram 135 – NEET Physics Electric Charges and Fields

Diagram 136 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate charge on 2 m length of wire with λ = 3 × 10^-6 C/m.
- Use Q = λl = 3 × 10^-6 × 2 = 6 × 10^-6 C.
Ex 2: Find surface charge on 0.5 m² with σ = 4 × 10^-6 C/m².
- Use Q = σS = 4 × 10^-6 × 0.5 = 2 × 10^-6 C.

Practice Set

Q1: Linear charge density λ units are:
- A) C/m
- B) C/m²
- C) C/m³
- D) C
Q2: Surface charge density σ is charge per unit:
- A) Length
- B) Area
- C) Volume
- D) Mass
Q3: Volume charge density ρ units are:
- A) C/m
- B) C/m²
- C) C/m³
- D) C

Answer Key

Q1: A) C/m — Linear charge density is charge per unit length.
Q2: B) Area — Surface charge density is charge per unit area.
Q3: C) C/m³ — Volume charge density is charge per unit volume.

Memory Aids

λ for line (length), units C/m.
σ for surface (area), units C/m².
ρ for volume, units C/m³.
Electric field sums contributions from small charge elements.

Charge Densities and Distribution

Electric charge is treated as continuous at macroscopic scale due to dense electron concentration.
Charge distributions:
- Line charge: Charge spread along a line segment. Charge ΔQ = λΔl, where λ is linear charge density (C/m).
- Surface charge: Charge spread over a surface area. Charge ΔQ = σΔS, where σ is surface charge density (C/m²).
- Volume charge: Charge distributed in volume. Charge ΔQ = ρΔV, where ρ is volume charge density (C/m³).
Charge densities:
- Linear charge density: λ = ΔQ/Δl (C/m)
- Surface charge density: σ = ΔQ/ΔS (C/m²)
- Volume charge density: ρ = ΔQ/ΔV (C/m³)
Electric field from continuous charge distributions:
- E_λ = 1/(4 π ε₀) ∑ λΔl/r'² r̂' (line charge)
- E_σ = 1/(4 π ε₀) ∑ σΔS/r'² r̂' (surface charge)
- E_ρ = 1/(4 π ε₀) ∑ ρΔV/r'² r̂' (volume charge)

Diagram 137 – NEET Physics Electric Charges and Fields

Diagram 138 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate charge on 2 m length of wire with λ = 3 × 10^-6 C/m.
- Use Q = λl = 3 × 10^-6 × 2 = 6 × 10^-6 C.
Ex 2: Find surface charge on 0.5 m² with σ = 4 × 10^-6 C/m².
- Use Q = σS = 4 × 10^-6 × 0.5 = 2 × 10^-6 C.

Practice Set

Q1: Linear charge density λ units are:
- A) C/m
- B) C/m²
- C) C/m³
- D) C
Q2: Surface charge density σ is charge per unit:
- A) Length
- B) Area
- C) Volume
- D) Mass
Q3: Volume charge density ρ units are:
- A) C/m
- B) C/m²
- C) C/m³
- D) C

Answer Key

Q1: A) C/m — Linear charge density is charge per unit length.
Q2: B) Area — Surface charge density is charge per unit area.
Q3: C) C/m³ — Volume charge density is charge per unit volume.

Memory Aids

λ for line (length), units C/m.
σ for surface (area), units C/m².
ρ for volume, units C/m³.
Electric field sums contributions from small charge elements.

Charge Densities and Distribution

Electric charge is treated as continuous at macroscopic scale due to dense electron concentration.
Charge distributions:
- Line charge: Charge spread along a line segment. Charge ΔQ = λΔl, where λ is linear charge density (C/m).
- Surface charge: Charge spread over a surface area. Charge ΔQ = σΔS, where σ is surface charge density (C/m²).
- Volume charge: Charge distributed in volume. Charge ΔQ = ρΔV, where ρ is volume charge density (C/m³).
Charge densities:
- Linear charge density: λ = ΔQ/Δl (C/m)
- Surface charge density: σ = ΔQ/ΔS (C/m²)
- Volume charge density: ρ = ΔQ/ΔV (C/m³)
Electric field from continuous charge distributions:
- E_λ = 1/(4 π ε₀) ∑ λΔl/r'² r̂' (line charge)
- E_σ = 1/(4 π ε₀) ∑ σΔS/r'² r̂' (surface charge)
- E_ρ = 1/(4 π ε₀) ∑ ρΔV/r'² r̂' (volume charge)

Diagram 139 – NEET Physics Electric Charges and Fields

Diagram 140 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate charge on 2 m length of wire with λ = 3 × 10^-6 C/m.
- Use Q = λl = 3 × 10^-6 × 2 = 6 × 10^-6 C.
Ex 2: Find surface charge on 0.5 m² with σ = 4 × 10^-6 C/m².
- Use Q = σS = 4 × 10^-6 × 0.5 = 2 × 10^-6 C.

Practice Set

Q1: Linear charge density λ units are:
- A) C/m
- B) C/m²
- C) C/m³
- D) C
Q2: Surface charge density σ is charge per unit:
- A) Length
- B) Area
- C) Volume
- D) Mass
Q3: Volume charge density ρ units are:
- A) C/m
- B) C/m²
- C) C/m³
- D) C

Answer Key

Q1: A) C/m — Linear charge density is charge per unit length.
Q2: B) Area — Surface charge density is charge per unit area.
Q3: C) C/m³ — Volume charge density is charge per unit volume.

Memory Aids

λ for line (length), units C/m.
σ for surface (area), units C/m².
ρ for volume, units C/m³.
Electric field sums contributions from small charge elements.

Charge Densities and Distribution

Electric charge is treated as continuous at macroscopic scale due to dense electron concentration.
Charge distributions:
- Line charge: Charge spread along a line segment. Charge \(\Delta Q = \lambda \Delta l\), where \(\lambda\) is linear charge density (C/m).
- Surface charge: Charge spread over a surface area. Charge \(\Delta Q = \sigma \Delta S\), where \(\sigma\) is surface charge density (C/m²).
- Volume charge: Charge distributed in volume. Charge \(\Delta Q = \rho \Delta V\), where \(\rho\) is volume charge density (C/m³).
Charge densities:
- Linear charge density: \(\lambda = \frac{\Delta Q}{\Delta l}\) (C/m)
- Surface charge density: \(\sigma = \frac{\Delta Q}{\Delta S}\) (C/m²)
- Volume charge density: \(\rho = \frac{\Delta Q}{\Delta V}\) (C/m³)
Electric field from continuous charge distributions:
- \(\vec{E}_{\lambda} = \frac{1}{4 \pi \varepsilon_0} \sum \frac{\lambda \Delta l}{r'^2} \hat{r}'\) (line charge)
- \(\vec{E}_{\sigma} = \frac{1}{4 \pi \varepsilon_0} \sum \frac{\sigma \Delta S}{r'^2} \hat{r}'\) (surface charge)
- \(\vec{E}_{\rho} = \frac{1}{4 \pi \varepsilon_0} \sum \frac{\rho \Delta V}{r'^2} \hat{r}'\) (volume charge)

Diagram 141 – NEET Physics Electric Charges and Fields

Diagram 142 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate charge on 2 m length of wire with \(\lambda = 3 \times 10^{-6} \text{ C/m} \).
- Use \(Q = \lambda l = 3 \times 10^{-6} \times 2 = 6 \times 10^{-6} \text{ C}\).
Ex 2: Find surface charge on 0.5 m² with \(\sigma = 4 \times 10^{-6} \text{ C/m}^2\).
- Use \(Q = \sigma S = 4 \times 10^{-6} \times 0.5 = 2 \times 10^{-6} \text{ C}\).

Practice Set

Q1: Linear charge density \(\lambda\) units are:
- A) C/m
- B) C/m²
- C) C/m³
- D) C
Q2: Surface charge density \(\sigma\) is charge per unit:
- A) Length
- B) Area
- C) Volume
- D) Mass
Q3: Volume charge density \(\rho\) units are:
- A) C/m
- B) C/m²
- C) C/m³
- D) C

Answer Key

Q1: A) C/m — Linear charge density is charge per unit length.
Q2: B) Area — Surface charge density is charge per unit area.
Q3: C) C/m³ — Volume charge density is charge per unit volume.

Memory Aids

\(\lambda\) for line (length), units C/m.
\(\sigma\) for surface (area), units C/m².
\(\rho\) for volume, units C/m³.
Electric field sums contributions from small charge elements.

Electric Field Linear Charge

Electric Field from Linear Charges

Finite wire: E_P = kλ/r sin α, depends on wire angle α and distance r.
Infinite wire: E_P = 2kλ/r, depends on λ and r only.
Charged arc: E_P = kλ/r 2 sin α, depends on arc angle α, λ, and r.
k = 1/(4 π ε₀) is Coulomb's constant.

Diagram 143 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Find electric field at P due to infinite wire with λ = 5 × 10^-6 C/m at r = 0.1 m.
- Use E_P = 2kλ/r = (2 × 9 × 10⁹ × 5 × 10^-6)/0.1 = 9 × 10⁵ N/C.
Ex 2: Calculate E_P for charged arc with λ = 4 × 10^-6 C/m, α = 30°, r = 0.2 m.
- Use E_P = kλ/r 2 sin α = (9 × 10⁹ × 4 × 10^-6)/0.2 × 2 × 0.5 = 1.8 × 10⁵ N/C.

Practice Set

Q1: Electric field due to infinite wire varies as:
- A) 1/r²
- B) 1/r
- C) r
- D) Constant
Q2: For a finite wire, electric field depends on:
- A) Wire length only
- B) Angle subtended α and distance r
- C) Charge only
- D) None
Q3: Electric field at P due to charged arc is proportional to:
- A) sin α
- B) cos α
- C) tan α
- D) α²

Answer Key

Q1: B) 1/r — Infinite wire field decreases inversely with distance.
Q2: B) Angle α and distance r — Finite wire field depends on geometry.
Q3: A) sin α — Arc field proportional to sine of arc angle.

Memory Aids

Infinite wire field ∝ 1/r.
Finite wire field depends on angle subtended.
Arc field involves 2 sin α factor.

Electric Field from Linear Charges

Finite wire: E_P = kλ/r sin α, depends on wire angle α and distance r.
Infinite wire: E_P = 2kλ/r, depends on λ and r only.
Charged arc: E_P = kλ/r 2 sin α, depends on arc angle α, λ, and r.
k = 1/(4 π ε₀) is Coulomb's constant.

Diagram 144 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Find electric field at P due to infinite wire with λ = 5 × 10^-6 C/m at r = 0.1 m.
- Use E_P = 2kλ/r = (2 × 9 × 10⁹ × 5 × 10^-6)/0.1 = 9 × 10⁵ N/C.
Ex 2: Calculate E_P for charged arc with λ = 4 × 10^-6 C/m, α = 30°, r = 0.2 m.
- Use E_P = kλ/r 2 sin α = (9 × 10⁹ × 4 × 10^-6)/0.2 × 2 × 0.5 = 1.8 × 10⁵ N/C.

Practice Set

Q1: Electric field due to infinite wire varies as:
- A) 1/r²
- B) 1/r
- C) r
- D) Constant
Q2: For a finite wire, electric field depends on:
- A) Wire length only
- B) Angle subtended α and distance r
- C) Charge only
- D) None
Q3: Electric field at P due to charged arc is proportional to:
- A) sin α
- B) cos α
- C) tan α
- D) α²

Answer Key

Q1: B) 1/r — Infinite wire field decreases inversely with distance.
Q2: B) Angle α and distance r — Finite wire field depends on geometry.
Q3: A) sin α — Arc field proportional to sine of arc angle.

Memory Aids

Infinite wire field ∝ 1/r.
Finite wire field depends on angle subtended.
Arc field involves 2 sin α factor.

Electric Field from Linear Charges

Finite wire: E_P = kλ/r sin α, depends on wire angle α and distance r.
Infinite wire: E_P = 2kλ/r, depends on λ and r only.
Charged arc: E_P = kλ/r 2 sin α, depends on arc angle α, λ, and r.
k = 1/(4 π ε₀) is Coulomb's constant.

Diagram 145 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Find electric field at P due to infinite wire with λ = 5 × 10^-6 C/m at r = 0.1 m.
- Use E_P = 2kλ/r = (2 × 9 × 10⁹ × 5 × 10^-6)/0.1 = 9 × 10⁵ N/C.
Ex 2: Calculate E_P for charged arc with λ = 4 × 10^-6 C/m, α = 30°, r = 0.2 m.
- Use E_P = kλ/r 2 sin α = (9 × 10⁹ × 4 × 10^-6)/0.2 × 2 × 0.5 = 1.8 × 10⁵ N/C.

Practice Set

Q1: Electric field due to infinite wire varies as:
- A) 1/r²
- B) 1/r
- C) r
- D) Constant
Q2: For a finite wire, electric field depends on:
- A) Wire length only
- B) Angle subtended α and distance r
- C) Charge only
- D) None
Q3: Electric field at P due to charged arc is proportional to:
- A) sin α
- B) cos α
- C) tan α
- D) α²

Answer Key

Q1: B) 1/r — Infinite wire field decreases inversely with distance.
Q2: B) Angle α and distance r — Finite wire field depends on geometry.
Q3: A) sin α — Arc field proportional to sine of arc angle.

Memory Aids

Infinite wire field ∝ 1/r.
Finite wire field depends on angle subtended.
Arc field involves 2 sin α factor.

Electric Field from Linear Charges

Finite wire: E_P = kλ/r sin α, depends on wire angle α and distance r.
Infinite wire: E_P = 2kλ/r, depends on λ and r only.
Charged arc: E_P = kλ/r 2 sin α, depends on arc angle α, λ, and r.
k = 1/(4 π ε₀) is Coulomb's constant.

Diagram 146 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Find electric field at P due to infinite wire with λ = 5 × 10^-6 C/m at r = 0.1 m.
- Use E_P = 2kλ/r = (2 × 9 × 10⁹ × 5 × 10^-6)/0.1 = 9 × 10⁵ N/C.
Ex 2: Calculate E_P for charged arc with λ = 4 × 10^-6 C/m, α = 30°, r = 0.2 m.
- Use E_P = kλ/r 2 sin α = (9 × 10⁹ × 4 × 10^-6)/0.2 × 2 × 0.5 = 1.8 × 10⁵ N/C.

Practice Set

Q1: Electric field due to infinite wire varies as:
- A) 1/r²
- B) 1/r
- C) r
- D) Constant
Q2: For a finite wire, electric field depends on:
- A) Wire length only
- B) Angle subtended α and distance r
- C) Charge only
- D) None
Q3: Electric field at P due to charged arc is proportional to:
- A) sin α
- B) cos α
- C) tan α
- D) α²

Answer Key

Q1: B) 1/r — Infinite wire field decreases inversely with distance.
Q2: B) Angle α and distance r — Finite wire field depends on geometry.
Q3: A) sin α — Arc field proportional to sine of arc angle.

Memory Aids

Infinite wire field ∝ 1/r.
Finite wire field depends on angle subtended.
Arc field involves 2 sin α factor.

Electric Field from Linear Charges

Finite wire: \(E_P = \frac{k \lambda}{r} \sin \alpha\), depends on wire angle \(\alpha\) and distance \(r\).
Infinite wire: \(E_P = \frac{2k \lambda}{r}\), depends on \(\lambda\) and \(r\) only.
Charged arc: \(E_P = \frac{k \lambda}{r} 2 \sin \alpha\), depends on arc angle \(\alpha\), \(\lambda\), and \(r\).
\(k = \frac{1}{4 \pi \varepsilon_0}\) is Coulomb's constant.

Diagram 147 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Find electric field at P due to infinite wire with \(\lambda = 5 \times 10^{-6} \text{ C/m} \) at \(r = 0.1 \text{ m}\).
- Use \(E_P = \frac{2k \lambda}{r} = \frac{2 \times 9 \times 10^9 \times 5 \times 10^{-6}}{0.1} = 9 \times 10^5 \text{ N/C}\).
Ex 2: Calculate \(E_P\) for charged arc with \(\lambda = 4 \times 10^{-6} \text{ C/m} \), \(\alpha = 30^\circ\), \(r = 0.2 \text{ m}\).
- Use \(E_P = \frac{k \lambda}{r} 2 \sin \alpha = \frac{9 \times 10^9 \times 4 \times 10^{-6}}{0.2} \times 2 \times 0.5 = 1.8 \times 10^5 \text{ N/C}\).

Practice Set

Q1: Electric field due to infinite wire varies as:
- A) \(1/r^2\)
- B) \(1/r\)
- C) \(r\)
- D) Constant
Q2: For a finite wire, electric field depends on:
- A) Wire length only
- B) Angle subtended \(\alpha\) and distance \(r\)
- C) Charge only
- D) None
Q3: Electric field at P due to charged arc is proportional to:
- A) \(\sin \alpha\)
- B) \(\cos \alpha\)
- C) \(\tan \alpha\)
- D) \(\alpha^2\)

Answer Key

Q1: B) \(1/r\) — Infinite wire field decreases inversely with distance.
Q2: B) Angle \(\alpha\) and distance \(r\) — Finite wire field depends on geometry.
Q3: A) \(\sin \alpha\) — Arc field proportional to sine of arc angle.

Memory Aids

Infinite wire field \(\propto 1/r\).
Finite wire field depends on angle subtended.
Arc field involves \(2 \sin \alpha\) factor.

Electric Field Charged Ring

Electric Field on Ring Axis

Consider ring radius a, total charge Q, point P on axis at distance x.
Small charge element dQ produces field dE = dQ/(4 π ε₀ (AP)²) at P.
Field component along axis: dE cos θ = x dQ/(4 π ε₀ (a² + x²)^3/2).
Net field at P: [ E = x Q/(4 π ε₀ (a² + x²)^3/2) ]
At ring center x=0, E=0 due to symmetry.
Field points along ring axis; maximum at specific x.

Diagram 148 – NEET Physics Electric Charges and Fields

Diagram 149 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric field at x=0.1 m on axis of ring with Q=5 × 10^-6 C, a=0.05 m.
- Use E = kQx/(a² + x²)^3/2 = (9 × 10⁹ × 5 × 10^-6 × 0.1)/((0.05² + 0.1²)^3/2).
- Calculate denominator and evaluate E.

Practice Set

Q1: Electric field at center of charged ring is:
- A) Maximum
- B) Zero
- C) Depends on charge
- D) Infinite
Q2: Electric field on axis of ring depends on:
- A) Distance from center x
- B) Ring radius a
- C) Total charge Q
- D) All above
Q3: Formula for electric field on axis of ring is:
- A) kQ/x²
- B) kQx/(a² + x²)^3/2
- C) kQa/(a² + x²)^3/2
- D) kQx/(a² + x²)²

Answer Key

Q1: B) Zero — Field cancels at ring center by symmetry.
Q2: D) All above — Field depends on x, a, and Q.
Q3: B) kQx/(a² + x²)^3/2 — Standard formula for ring axis field.

Memory Aids

Field zero at ring center.
Field max at some distance on axis.
Formula involves x and a in denominator power 3/2.

Electric Field on Ring Axis

Consider ring radius a, total charge Q, point P on axis at distance x.
Small charge element dQ produces field dE = dQ/(4 π ε₀ (AP)²) at P.
Field component along axis: dE cos θ = x dQ/(4 π ε₀ (a² + x²)^3/2).
Net field at P: [ E = x Q/(4 π ε₀ (a² + x²)^3/2) ]
At ring center x=0, E=0 due to symmetry.
Field points along ring axis; maximum at specific x.

Diagram 150 – NEET Physics Electric Charges and Fields

Diagram 151 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric field at x=0.1 m on axis of ring with Q=5 × 10^-6 C, a=0.05 m.
- Use E = kQx/(a² + x²)^3/2 = (9 × 10⁹ × 5 × 10^-6 × 0.1)/((0.05² + 0.1²)^3/2).
- Calculate denominator and evaluate E.

Practice Set

Q1: Electric field at center of charged ring is:
- A) Maximum
- B) Zero
- C) Depends on charge
- D) Infinite
Q2: Electric field on axis of ring depends on:
- A) Distance from center x
- B) Ring radius a
- C) Total charge Q
- D) All above
Q3: Formula for electric field on axis of ring is:
- A) kQ/x²
- B) kQx/(a² + x²)^3/2
- C) kQa/(a² + x²)^3/2
- D) kQx/(a² + x²)²

Answer Key

Q1: B) Zero — Field cancels at ring center by symmetry.
Q2: D) All above — Field depends on x, a, and Q.
Q3: B) kQx/(a² + x²)^3/2 — Standard formula for ring axis field.

Memory Aids

Field zero at ring center.
Field max at some distance on axis.
Formula involves x and a in denominator power 3/2.

Electric Field on Ring Axis

Consider ring radius a, total charge Q, point P on axis at distance x.
Small charge element dQ produces field dE = dQ/(4 π ε₀ (AP)²) at P.
Field component along axis: dE cos θ = x dQ/(4 π ε₀ (a² + x²)^3/2).
Net field at P: [ E = x Q/(4 π ε₀ (a² + x²)^3/2) ]
At ring center x=0, E=0 due to symmetry.
Field points along ring axis; maximum at specific x.

Diagram 152 – NEET Physics Electric Charges and Fields

Diagram 153 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric field at x=0.1 m on axis of ring with Q=5 × 10^-6 C, a=0.05 m.
- Use E = kQx/(a² + x²)^3/2 = (9 × 10⁹ × 5 × 10^-6 × 0.1)/((0.05² + 0.1²)^3/2).
- Calculate denominator and evaluate E.

Practice Set

Q1: Electric field at center of charged ring is:
- A) Maximum
- B) Zero
- C) Depends on charge
- D) Infinite
Q2: Electric field on axis of ring depends on:
- A) Distance from center x
- B) Ring radius a
- C) Total charge Q
- D) All above
Q3: Formula for electric field on axis of ring is:
- A) kQ/x²
- B) kQx/(a² + x²)^3/2
- C) kQa/(a² + x²)^3/2
- D) kQx/(a² + x²)²

Answer Key

Q1: B) Zero — Field cancels at ring center by symmetry.
Q2: D) All above — Field depends on x, a, and Q.
Q3: B) kQx/(a² + x²)^3/2 — Standard formula for ring axis field.

Memory Aids

Field zero at ring center.
Field max at some distance on axis.
Formula involves x and a in denominator power 3/2.

Electric Field on Ring Axis

Consider ring radius a, total charge Q, point P on axis at distance x.
Small charge element dQ produces field dE = dQ/(4 π ε₀ (AP)²) at P.
Field component along axis: dE cos θ = x dQ/(4 π ε₀ (a² + x²)^3/2).
Net field at P: [ E = x Q/(4 π ε₀ (a² + x²)^3/2) ]
At ring center x=0, E=0 due to symmetry.
Field points along ring axis; maximum at specific x.

Diagram 154 – NEET Physics Electric Charges and Fields

Diagram 155 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric field at x=0.1 m on axis of ring with Q=5 × 10^-6 C, a=0.05 m.
- Use E = kQx/(a² + x²)^3/2 = (9 × 10⁹ × 5 × 10^-6 × 0.1)/((0.05² + 0.1²)^3/2).
- Calculate denominator and evaluate E.

Practice Set

Q1: Electric field at center of charged ring is:
- A) Maximum
- B) Zero
- C) Depends on charge
- D) Infinite
Q2: Electric field on axis of ring depends on:
- A) Distance from center x
- B) Ring radius a
- C) Total charge Q
- D) All above
Q3: Formula for electric field on axis of ring is:
- A) kQ/x²
- B) kQx/(a² + x²)^3/2
- C) kQa/(a² + x²)^3/2
- D) kQx/(a² + x²)²

Answer Key

Q1: B) Zero — Field cancels at ring center by symmetry.
Q2: D) All above — Field depends on x, a, and Q.
Q3: B) kQx/(a² + x²)^3/2 — Standard formula for ring axis field.

Memory Aids

Field zero at ring center.
Field max at some distance on axis.
Formula involves x and a in denominator power 3/2.

Electric Field on Ring Axis

Consider ring radius a, total charge Q, point P on axis at distance x.
Small charge element dQ produces field dE = dQ/(4 π ε₀ (AP)²) at P.
Field component along axis: dE cos θ = x dQ/(4 π ε₀ (a² + x²)^3/2).
Net field at P: [ E = x Q/(4 π ε₀ (a² + x²)^3/2) ]
At ring center x=0, E=0 due to symmetry.
Field points along ring axis; maximum at specific x.

Diagram 156 – NEET Physics Electric Charges and Fields

Diagram 157 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric field at x=0.1 m on axis of ring with Q=5 × 10^-6 C, a=0.05 m.
- Use E = kQx/(a² + x²)^3/2 = (9 × 10⁹ × 5 × 10^-6 × 0.1)/((0.05² + 0.1²)^3/2).
- Calculate denominator and evaluate E.

Practice Set

Q1: Electric field at center of charged ring is:
- A) Maximum
- B) Zero
- C) Depends on charge
- D) Infinite
Q2: Electric field on axis of ring depends on:
- A) Distance from center x
- B) Ring radius a
- C) Total charge Q
- D) All above
Q3: Formula for electric field on axis of ring is:
- A) kQ/x²
- B) kQx/(a² + x²)^3/2
- C) kQa/(a² + x²)^3/2
- D) kQx/(a² + x²)²

Answer Key

Q1: B) Zero — Field cancels at ring center by symmetry.
Q2: D) All above — Field depends on x, a, and Q.
Q3: B) kQx/(a² + x²)^3/2 — Standard formula for ring axis field.

Memory Aids

Field zero at ring center.
Field max at some distance on axis.
Formula involves x and a in denominator power 3/2.

Electric Field on Ring Axis

Consider ring radius a, total charge Q, point P on axis at distance x.
Small charge element dQ produces field dE = dQ/(4 π ε₀ (AP)²) at P.
Field component along axis: dE cos θ = x dQ/(4 π ε₀ (a² + x²)^3/2).
Net field at P: [ E = x Q/(4 π ε₀ (a² + x²)^3/2) ]
At ring center x=0, E=0 due to symmetry.
Field points along ring axis; maximum at specific x.

Diagram 158 – NEET Physics Electric Charges and Fields

Diagram 159 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric field at x=0.1 m on axis of ring with Q=5 × 10^-6 C, a=0.05 m.
- Use E = kQx/(a² + x²)^3/2 = (9 × 10⁹ × 5 × 10^-6 × 0.1)/((0.05² + 0.1²)^3/2).
- Calculate denominator and evaluate E.

Practice Set

Q1: Electric field at center of charged ring is:
- A) Maximum
- B) Zero
- C) Depends on charge
- D) Infinite
Q2: Electric field on axis of ring depends on:
- A) Distance from center x
- B) Ring radius a
- C) Total charge Q
- D) All above
Q3: Formula for electric field on axis of ring is:
- A) kQ/x²
- B) kQx/(a² + x²)^3/2
- C) kQa/(a² + x²)^3/2
- D) kQx/(a² + x²)²

Answer Key

Q1: B) Zero — Field cancels at ring center by symmetry.
Q2: D) All above — Field depends on x, a, and Q.
Q3: B) kQx/(a² + x²)^3/2 — Standard formula for ring axis field.

Memory Aids

Field zero at ring center.
Field max at some distance on axis.
Formula involves x and a in denominator power 3/2.

Electric Field on Ring Axis

Consider ring radius a, total charge Q, point P on axis at distance x.
Small charge element dQ produces field dE = dQ/(4 π ε₀ (AP)²) at P.
Field component along axis: dE cos θ = x dQ/(4 π ε₀ (a² + x²)^3/2).
Net field at P: [ E = x Q/(4 π ε₀ (a² + x²)^3/2) ]
At ring center x=0, E=0 due to symmetry.
Field points along ring axis; maximum at specific x.

Diagram 160 – NEET Physics Electric Charges and Fields

Diagram 161 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric field at x=0.1 m on axis of ring with Q=5 × 10^-6 C, a=0.05 m.
- Use E = kQx/(a² + x²)^3/2 = (9 × 10⁹ × 5 × 10^-6 × 0.1)/((0.05² + 0.1²)^3/2).
- Calculate denominator and evaluate E.

Practice Set

Q1: Electric field at center of charged ring is:
- A) Maximum
- B) Zero
- C) Depends on charge
- D) Infinite
Q2: Electric field on axis of ring depends on:
- A) Distance from center x
- B) Ring radius a
- C) Total charge Q
- D) All above
Q3: Formula for electric field on axis of ring is:
- A) kQ/x²
- B) kQx/(a² + x²)^3/2
- C) kQa/(a² + x²)^3/2
- D) kQx/(a² + x²)²

Answer Key

Q1: B) Zero — Field cancels at ring center by symmetry.
Q2: D) All above — Field depends on x, a, and Q.
Q3: B) kQx/(a² + x²)^3/2 — Standard formula for ring axis field.

Memory Aids

Field zero at ring center.
Field max at some distance on axis.
Formula involves x and a in denominator power 3/2.

Electric Field on Ring Axis

Consider ring radius \(a\), total charge \(Q\), point \(P\) on axis at distance \(x\).
Small charge element \(dQ\) produces field \(dE = \frac{dQ}{4 \pi \varepsilon_0 (AP)^2}\) at \(P\).
Field component along axis: \(dE \cos \theta = \frac{x dQ}{4 \pi \varepsilon_0 (a^2 + x^2)^{3/2}}\).
Net field at \(P\): \[ E = \frac{x Q}{4 \pi \varepsilon_0 (a^2 + x^2)^{3/2}} \]
At ring center \(x=0\), \(E=0\) due to symmetry.
Field points along ring axis; maximum at specific \(x\).

Diagram 162 – NEET Physics Electric Charges and Fields

Diagram 163 – NEET Physics Electric Charges and Fields

Solved Examples

Ex 1: Calculate electric field at \(x=0.1 \text{ m}\) on axis of ring with \(Q=5 \times 10^{-6} \text{ C}\), \(a=0.05 \text{ m}\).
- Use \(E = \frac{k Q x}{(a^2 + x^2)^{3/2}} = \frac{9 \times 10^9 \times 5 \times 10^{-6} \times 0.1}{(0.05^2 + 0.1^2)^{3/2}}\).
- Calculate denominator and evaluate \(E\).

Practice Set

Q1: Electric field at center of charged ring is:
- A) Maximum
- B) Zero
- C) Depends on charge
- D) Infinite
Q2: Electric field on axis of ring depends on:
- A) Distance from center \(x\)
- B) Ring radius \(a\)
- C) Total charge \(Q\)
- D) All above
Q3: Formula for electric field on axis of ring is:
- A) \(\frac{k Q}{x^2}\)
- B) \(\frac{k Q x}{(a^2 + x^2)^{3/2}}\)
- C) \(\frac{k Q a}{(a^2 + x^2)^{3/2}}\)
- D) \(\frac{k Q x}{(a^2 + x^2)^2}\)

Answer Key

Q1: B) Zero — Field cancels at ring center by symmetry.
Q2: D) All above — Field depends on \(x\), \(a\), and \(Q\).
Q3: B) \(\frac{k Q x}{(a^2 + x^2)^{3/2}}\) — Standard formula for ring axis field.

Memory Aids

Field zero at ring center.
Field max at some distance on axis.
Formula involves \(x\) and \(a\) in denominator power 3/2.

Gauss's Law

Explanation

Gauss's law states total electric flux Φ over a closed surface S in vacuum is 1/ε₀ times the total charge enclosed q_enclosed:
Φ = ∮_S E · dS = q_enclosed/ε₀
Electric field E due to point charge Q at distance R is:
E = 1/(4 π ε₀) Q/R²
For a spherical Gaussian surface of radius R, flux:
Φ = E × 4 π R² = Q/ε₀
Gauss's law holds for any closed surface, regardless of shape or size.
q_enclosed is the sum of all charges inside the surface.
Electric field includes contributions from charges inside and outside, but Gauss's law relates only to enclosed charges.
Gaussian surface must not pass through discrete charges but can pass through continuous charge distributions.
Useful for calculating electrostatic fields in symmetric charge distributions.
Based on inverse square law; violation indicates deviation from Coulomb's law.

Solved Examples

Ex 1: Calculate electric flux through a spherical surface of radius 0.5 m enclosing charge 2 μC.
Step 1: Use Φ = q/ε₀
Step 2: q = 2 × 10^-6 C, ε₀ = 8.85 × 10^-12 C²/Nm²
Step 3: Φ = (2 × 10^-6)/(8.85 × 10^-12) = 2.26 × 10⁵ Nm²/C
Ex 2: Find electric field on surface of sphere radius 0.1 m with enclosed charge 5 μC.
Step 1: Use E = 1/(4 π ε₀) Q/R²
Step 2: Q = 5 × 10^-6 C, R = 0.1 m
Step 3: E = (9 × 10⁹ × 5 × 10^-6)/(0.1)² = 4.5 × 10⁶ N/C

Practice Set

Q1: Gauss's law relates electric flux to:
A) Total charge outside surface
B) Total charge inside surface
C) Electric field only
D) Surface area only
Q2: Electric flux through a spherical surface of radius 0.2 m enclosing charge 3 μC is:
A) 3.39 × 10⁵ Nm²/C
B) 2.5 × 10⁵ Nm²/C
C) 1.5 × 10⁵ Nm²/C
D) 4.0 × 10⁵ Nm²/C
Q3: Electric field at 0.3 m from a point charge 6 μC is:
A) 2 × 10⁶ N/C
B) 6 × 10⁶ N/C
C) 1.8 × 10⁶ N/C
D) 3 × 10⁶ N/C

Answer Key

Q1: B - Gauss's law relates flux to total charge enclosed inside the surface.
Q2: A - Φ = q/ε₀ = (3 × 10^-6)/(8.85 × 10^-12) = 3.39 × 10⁵ Nm²/C
Q3: C - E = (9 × 10⁹ × 6 × 10^-6)/(0.3)² = 1.8 × 10⁶ N/C

Memory Aids

Flux Φ = q_enclosed/ε₀
Electric field E = 1/(4 π ε₀) Q/R²
Gaussian surface: closed surface enclosing charge
Gauss's law confirms inverse square law

Explanation

Gauss's law states total electric flux Φ over a closed surface S in vacuum is 1/ε₀ times the total charge enclosed q_enclosed:
Φ = ∮_S E · dS = q_enclosed/ε₀
Electric field E due to point charge Q at distance R is:
E = 1/(4 π ε₀) Q/R²
For a spherical Gaussian surface of radius R, flux:
Φ = E × 4 π R² = Q/ε₀
Gauss's law holds for any closed surface, regardless of shape or size.
q_enclosed is the sum of all charges inside the surface.
Electric field includes contributions from charges inside and outside, but Gauss's law relates only to enclosed charges.
Gaussian surface must not pass through discrete charges but can pass through continuous charge distributions.
Useful for calculating electrostatic fields in symmetric charge distributions.
Based on inverse square law; violation indicates deviation from Coulomb's law.

Solved Examples

Ex 1: Calculate electric flux through a spherical surface of radius 0.5 m enclosing charge 2 μC.
Step 1: Use Φ = q/ε₀
Step 2: q = 2 × 10^-6 C, ε₀ = 8.85 × 10^-12 C²/Nm²
Step 3: Φ = (2 × 10^-6)/(8.85 × 10^-12) = 2.26 × 10⁵ Nm²/C
Ex 2: Find electric field on surface of sphere radius 0.1 m with enclosed charge 5 μC.
Step 1: Use E = 1/(4 π ε₀) Q/R²
Step 2: Q = 5 × 10^-6 C, R = 0.1 m
Step 3: E = (9 × 10⁹ × 5 × 10^-6)/(0.1)² = 4.5 × 10⁶ N/C

Practice Set

Q1: Gauss's law relates electric flux to:
A) Total charge outside surface
B) Total charge inside surface
C) Electric field only
D) Surface area only
Q2: Electric flux through a spherical surface of radius 0.2 m enclosing charge 3 μC is:
A) 3.39 × 10⁵ Nm²/C
B) 2.5 × 10⁵ Nm²/C
C) 1.5 × 10⁵ Nm²/C
D) 4.0 × 10⁵ Nm²/C
Q3: Electric field at 0.3 m from a point charge 6 μC is:
A) 2 × 10⁶ N/C
B) 6 × 10⁶ N/C
C) 1.8 × 10⁶ N/C
D) 3 × 10⁶ N/C

Answer Key

Q1: B - Gauss's law relates flux to total charge enclosed inside the surface.
Q2: A - Φ = q/ε₀ = (3 × 10^-6)/(8.85 × 10^-12) = 3.39 × 10⁵ Nm²/C
Q3: C - E = (9 × 10⁹ × 6 × 10^-6)/(0.3)² = 1.8 × 10⁶ N/C

Memory Aids

Flux Φ = q_enclosed/ε₀
Electric field E = 1/(4 π ε₀) Q/R²
Gaussian surface: closed surface enclosing charge
Gauss's law confirms inverse square law

Explanation

Gauss's law states total electric flux Φ over a closed surface S in vacuum is 1/ε₀ times the total charge enclosed q_enclosed:
Φ = ∮_S E · dS = q_enclosed/ε₀
Electric field E due to point charge Q at distance R is:
E = 1/(4 π ε₀) Q/R²
For a spherical Gaussian surface of radius R, flux:
Φ = E × 4 π R² = Q/ε₀
Gauss's law holds for any closed surface, regardless of shape or size.
q_enclosed is the sum of all charges inside the surface.
Electric field includes contributions from charges inside and outside, but Gauss's law relates only to enclosed charges.
Gaussian surface must not pass through discrete charges but can pass through continuous charge distributions.
Useful for calculating electrostatic fields in symmetric charge distributions.
Based on inverse square law; violation indicates deviation from Coulomb's law.

Solved Examples

Ex 1: Calculate electric flux through a spherical surface of radius 0.5 m enclosing charge 2 μC.
Step 1: Use Φ = q/ε₀
Step 2: q = 2 × 10^-6 C, ε₀ = 8.85 × 10^-12 C²/Nm²
Step 3: Φ = (2 × 10^-6)/(8.85 × 10^-12) = 2.26 × 10⁵ Nm²/C
Ex 2: Find electric field on surface of sphere radius 0.1 m with enclosed charge 5 μC.
Step 1: Use E = 1/(4 π ε₀) Q/R²
Step 2: Q = 5 × 10^-6 C, R = 0.1 m
Step 3: E = (9 × 10⁹ × 5 × 10^-6)/(0.1)² = 4.5 × 10⁶ N/C

Practice Set

Q1: Gauss's law relates electric flux to:
A) Total charge outside surface
B) Total charge inside surface
C) Electric field only
D) Surface area only
Q2: Electric flux through a spherical surface of radius 0.2 m enclosing charge 3 μC is:
A) 3.39 × 10⁵ Nm²/C
B) 2.5 × 10⁵ Nm²/C
C) 1.5 × 10⁵ Nm²/C
D) 4.0 × 10⁵ Nm²/C
Q3: Electric field at 0.3 m from a point charge 6 μC is:
A) 2 × 10⁶ N/C
B) 6 × 10⁶ N/C
C) 1.8 × 10⁶ N/C
D) 3 × 10⁶ N/C

Answer Key

Q1: B - Gauss's law relates flux to total charge enclosed inside the surface.
Q2: A - Φ = q/ε₀ = (3 × 10^-6)/(8.85 × 10^-12) = 3.39 × 10⁵ Nm²/C
Q3: C - E = (9 × 10⁹ × 6 × 10^-6)/(0.3)² = 1.8 × 10⁶ N/C

Memory Aids

Flux Φ = q_enclosed/ε₀
Electric field E = 1/(4 π ε₀) Q/R²
Gaussian surface: closed surface enclosing charge
Gauss's law confirms inverse square law

Explanation

Gauss's law states total electric flux \( \Phi \) over a closed surface \( S \) in vacuum is \( \frac{1}{\varepsilon_0} \) times the total charge enclosed \( q_{\text{enclosed}} \):
\( \Phi = \oint_S \vec{E} \cdot d\vec{S} = \frac{q_{\text{enclosed}}}{\varepsilon_0} \)
Electric field \( E \) due to point charge \( Q \) at distance \( R \) is:
\( E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R^2} \)
For a spherical Gaussian surface of radius \( R \), flux:
\( \Phi = E \times 4 \pi R^2 = \frac{Q}{\varepsilon_0} \)
Gauss's law holds for any closed surface, regardless of shape or size.
\( q_{\text{enclosed}} \) is the sum of all charges inside the surface.
Electric field includes contributions from charges inside and outside, but Gauss's law relates only to enclosed charges.
Gaussian surface must not pass through discrete charges but can pass through continuous charge distributions.
Useful for calculating electrostatic fields in symmetric charge distributions.
Based on inverse square law; violation indicates deviation from Coulomb's law.

Solved Examples

Ex 1: Calculate electric flux through a spherical surface of radius 0.5 m enclosing charge 2 \( \mu C \).
Step 1: Use \( \Phi = \frac{q}{\varepsilon_0} \)
Step 2: \( q = 2 \times 10^{-6} C, \varepsilon_0 = 8.85 \times 10^{-12} C^2/Nm^2 \)
Step 3: \( \Phi = \frac{2 \times 10^{-6}}{8.85 \times 10^{-12}} = 2.26 \times 10^{5} \text{ Nm}^2/\text{C} \)
Ex 2: Find electric field on surface of sphere radius 0.1 m with enclosed charge 5 \( \mu C \).
Step 1: Use \( E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R^2} \)
Step 2: \( Q = 5 \times 10^{-6} C, R = 0.1 m \)
Step 3: \( E = \frac{9 \times 10^{9} \times 5 \times 10^{-6}}{(0.1)^2} = 4.5 \times 10^{6} \text{ N/C} \)

Practice Set

Q1: Gauss's law relates electric flux to:
A) Total charge outside surface
B) Total charge inside surface
C) Electric field only
D) Surface area only
Q2: Electric flux through a spherical surface of radius 0.2 m enclosing charge 3 \( \mu C \) is:
A) \( 3.39 \times 10^{5} \text{ Nm}^2/\text{C} \)
B) \( 2.5 \times 10^{5} \text{ Nm}^2/\text{C} \)
C) \( 1.5 \times 10^{5} \text{ Nm}^2/\text{C} \)
D) \( 4.0 \times 10^{5} \text{ Nm}^2/\text{C} \)
Q3: Electric field at 0.3 m from a point charge 6 \( \mu C \) is:
A) \( 2 \times 10^{6} \text{ N/C} \)
B) \( 6 \times 10^{6} \text{ N/C} \)
C) \( 1.8 \times 10^{6} \text{ N/C} \)
D) \( 3 \times 10^{6} \text{ N/C} \)

Answer Key

Q1: B - Gauss's law relates flux to total charge enclosed inside the surface.
Q2: A - \( \Phi = \frac{q}{\varepsilon_0} = \frac{3 \times 10^{-6}}{8.85 \times 10^{-12}} = 3.39 \times 10^{5} \text{ Nm}^2/\text{C} \)
Q3: C - \( E = \frac{9 \times 10^{9} \times 6 \times 10^{-6}}{(0.3)^2} = 1.8 \times 10^{6} \text{ N/C} \)

Memory Aids

Flux \( \Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0} \)
Electric field \( E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R^2} \)
Gaussian surface: closed surface enclosing charge
Gauss's law confirms inverse square law

Applications of Gauss's Law

Electric Fields from Charge Distributions

Gauss's law helps find electric fields vector E for symmetric charge setups.
Key charge distributions: line charge, thin sheet (non-conducting and conducting), spherical shell, solid sphere.
Electric field formulas depend on geometry and charge arrangement.

Diagram 164 – NEET Physics Electric Charges and Fields

Charge Distribution Details

Infinitely long line charge: E = λ/(2πε₀r), field ∝ 1/r, radial direction.
Infinitely large non-conducting thin sheet: E = σ/(2ε₀), uniform field, perpendicular to sheet.
Infinitely large charged conducting sheet: E = σ/ε₀, uniform field, perpendicular to surface.
Uniformly charged spherical shell or conducting sphere: Outside: E = kQ/r², Inside: E=0.
Uniformly charged solid non-conducting sphere: Outside: E = kQ/r², Inside: E = kQr/R³, linear increase with r.

Solved Examples

Ex 1: Find E at r=0.1 m from a line charge with λ=5×10^-6 C/m.
- Use E=λ/(2πε₀r).
- Calculate E=5×10^-6/(2π × 8.85×10^-12 × 0.1) ≈ 8.99×10⁴ N/C.
Ex 2: Electric field near a charged conducting sheet with surface charge density σ=1×10^-6 C/m².
- Use E=σ/ε₀ = 1×10^-6/8.85×10^-12 ≈ 1.13×10⁵ N/C.

Practice Set

Q1: Electric field due to an infinite line charge varies as?
- A) 1/r²
- B) 1/r
- C) Constant
- D) r
Q2: Electric field inside a uniformly charged spherical shell is?
- A) kQ/r²
- B) Zero
- C) kQr/R³
- D) Constant non-zero
Q3: For a uniformly charged solid sphere, electric field inside at distance r is?
- A) kQ/r²
- B) Zero
- C) kQr/R³
- D) σ/2ε₀

Answer Key

Q1: B) 1/r - Field from line charge decreases inversely with distance.
Q2: B) Zero - Inside spherical shell, electric field is zero by Gauss's law.
Q3: C) kQr/R³ - Inside solid sphere, field increases linearly with r.

Memory Aids

Line charge: E ∝ 1/r.
Sheet charge: uniform E, independent of r.
Spherical shell: zero inside, ∝ 1/r² outside.
Solid sphere: linear E inside, 1/r² outside.

Electric Fields from Charge Distributions

Gauss's law helps find electric fields vector E for symmetric charge setups.
Key charge distributions: line charge, thin sheet (non-conducting and conducting), spherical shell, solid sphere.
Electric field formulas depend on geometry and charge arrangement.

Diagram 165 – NEET Physics Electric Charges and Fields

Charge Distribution Details

Infinitely long line charge: E = λ/(2πε₀r), field ∝ 1/r, radial direction.
Infinitely large non-conducting thin sheet: E = σ/(2ε₀), uniform field, perpendicular to sheet.
Infinitely large charged conducting sheet: E = σ/ε₀, uniform field, perpendicular to surface.
Uniformly charged spherical shell or conducting sphere: Outside: E = kQ/r², Inside: E=0.
Uniformly charged solid non-conducting sphere: Outside: E = kQ/r², Inside: E = kQr/R³, linear increase with r.

Solved Examples

Ex 1: Find E at r=0.1 m from a line charge with λ=5×10^-6 C/m.
- Use E=λ/(2πε₀r).
- Calculate E=5×10^-6/(2π × 8.85×10^-12 × 0.1) ≈ 8.99×10⁴ N/C.
Ex 2: Electric field near a charged conducting sheet with surface charge density σ=1×10^-6 C/m².
- Use E=σ/ε₀ = 1×10^-6/8.85×10^-12 ≈ 1.13×10⁵ N/C.

Practice Set

Q1: Electric field due to an infinite line charge varies as?
- A) 1/r²
- B) 1/r
- C) Constant
- D) r
Q2: Electric field inside a uniformly charged spherical shell is?
- A) kQ/r²
- B) Zero
- C) kQr/R³
- D) Constant non-zero
Q3: For a uniformly charged solid sphere, electric field inside at distance r is?
- A) kQ/r²
- B) Zero
- C) kQr/R³
- D) σ/2ε₀

Answer Key

Q1: B) 1/r - Field from line charge decreases inversely with distance.
Q2: B) Zero - Inside spherical shell, electric field is zero by Gauss's law.
Q3: C) kQr/R³ - Inside solid sphere, field increases linearly with r.

Memory Aids

Line charge: E ∝ 1/r.
Sheet charge: uniform E, independent of r.
Spherical shell: zero inside, ∝ 1/r² outside.
Solid sphere: linear E inside, 1/r² outside.

Electric Fields from Charge Distributions

Gauss's law helps find electric fields vector E for symmetric charge setups.
Key charge distributions: line charge, thin sheet (non-conducting and conducting), spherical shell, solid sphere.
Electric field formulas depend on geometry and charge arrangement.

Diagram 166 – NEET Physics Electric Charges and Fields

Charge Distribution Details

Infinitely long line charge: E = λ/(2πε₀r), field ∝ 1/r, radial direction.
Infinitely large non-conducting thin sheet: E = σ/(2ε₀), uniform field, perpendicular to sheet.
Infinitely large charged conducting sheet: E = σ/ε₀, uniform field, perpendicular to surface.
Uniformly charged spherical shell or conducting sphere: Outside: E = kQ/r², Inside: E=0.
Uniformly charged solid non-conducting sphere: Outside: E = kQ/r², Inside: E = kQr/R³, linear increase with r.

Solved Examples

Ex 1: Find E at r=0.1 m from a line charge with λ=5×10^-6 C/m.
- Use E=λ/(2πε₀r).
- Calculate E=5×10^-6/(2π × 8.85×10^-12 × 0.1) ≈ 8.99×10⁴ N/C.
Ex 2: Electric field near a charged conducting sheet with surface charge density σ=1×10^-6 C/m².
- Use E=σ/ε₀ = 1×10^-6/8.85×10^-12 ≈ 1.13×10⁵ N/C.

Practice Set

Q1: Electric field due to an infinite line charge varies as?
- A) 1/r²
- B) 1/r
- C) Constant
- D) r
Q2: Electric field inside a uniformly charged spherical shell is?
- A) kQ/r²
- B) Zero
- C) kQr/R³
- D) Constant non-zero
Q3: For a uniformly charged solid sphere, electric field inside at distance r is?
- A) kQ/r²
- B) Zero
- C) kQr/R³
- D) σ/2ε₀

Answer Key

Q1: B) 1/r - Field from line charge decreases inversely with distance.
Q2: B) Zero - Inside spherical shell, electric field is zero by Gauss's law.
Q3: C) kQr/R³ - Inside solid sphere, field increases linearly with r.

Memory Aids

Line charge: E ∝ 1/r.
Sheet charge: uniform E, independent of r.
Spherical shell: zero inside, ∝ 1/r² outside.
Solid sphere: linear E inside, 1/r² outside.

Electric Fields from Charge Distributions

Gauss's law helps find electric fields \(\vec{E}\) for symmetric charge setups.
Key charge distributions: line charge, thin sheet (non-conducting and conducting), spherical shell, solid sphere.
Electric field formulas depend on geometry and charge arrangement.

Diagram 167 – NEET Physics Electric Charges and Fields

Charge Distribution Details

Infinitely long line charge: \(E = \frac{\lambda}{2\pi \epsilon_0 r}\), field \(\propto 1/r\), radial direction.
Infinitely large non-conducting thin sheet: \(E = \frac{\sigma}{2\epsilon_0}\), uniform field, perpendicular to sheet.
Infinitely large charged conducting sheet: \(E = \frac{\sigma}{\epsilon_0}\), uniform field, perpendicular to surface.
Uniformly charged spherical shell or conducting sphere: Outside: \(E = \frac{kQ}{r^2}\), Inside: \(E=0\).
Uniformly charged solid non-conducting sphere: Outside: \(E = \frac{kQ}{r^2}\), Inside: \(E = \frac{kQr}{R^3}\), linear increase with \(r\).

Solved Examples

Ex 1: Find \(E\) at \(r=0.1\,m\) from a line charge with \(\lambda=5\times10^{-6}\,C/m\).
- Use \(E=\frac{\lambda}{2\pi \epsilon_0 r}\).
- Calculate \(E=\frac{5\times10^{-6}}{2\pi \times 8.85\times10^{-12} \times 0.1} \approx 8.99\times10^{4}\,N/C\).
Ex 2: Electric field near a charged conducting sheet with surface charge density \(\sigma=1\times10^{-6}\,C/m^2\).
- Use \(E=\frac{\sigma}{\epsilon_0} = \frac{1\times10^{-6}}{8.85\times10^{-12}} \approx 1.13\times10^{5}\,N/C\).

Practice Set

Q1: Electric field due to an infinite line charge varies as?
- A) \(1/r^2\)
- B) \(1/r\)
- C) Constant
- D) \(r\)
Q2: Electric field inside a uniformly charged spherical shell is?
- A) \(kQ/r^2\)
- B) Zero
- C) \(kQr/R^3\)
- D) Constant non-zero
Q3: For a uniformly charged solid sphere, electric field inside at distance \(r\) is?
- A) \(kQ/r^2\)
- B) Zero
- C) \(kQr/R^3\)
- D) \(\sigma/2\epsilon_0\)

Answer Key

Q1: B) \(1/r\) - Field from line charge decreases inversely with distance.
Q2: B) Zero - Inside spherical shell, electric field is zero by Gauss's law.
Q3: C) \(kQr/R^3\) - Inside solid sphere, field increases linearly with \(r\).

Memory Aids

Line charge: \(E \propto 1/r\).
Sheet charge: uniform \(E\), independent of \(r\).
Spherical shell: zero inside, \(\propto 1/r^2\) outside.
Solid sphere: linear \(E\) inside, \(1/r^2\) outside.

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