Chapter 17: Electrostatic Potential and Capacitance

Introduction

Concepts and Formulas

• Test charge ₀ > 0 moved slowly from point R to P against repulsive force by fixed charge > 0 at origin.
• Work done by external force: W_RP = ∫_R^P vector F_ext · dvector r = - ∫_R^P vector F_E · dvector r
• Work done against electrostatic repulsion is stored as potential energy.
• Potential energy difference: ΔU = U_P - U_R = W_RP
• Work done by electrostatic field depends only on initial and final points (path independent).
• Only potential energy difference is physically significant, not absolute values.

Solved Examples

• Ex 1: Calculate work done to move q₀ = 2 μC from R to P against repulsive force of q = 3 μC at origin, distance r = 0.5 m.
  • Use W_RP = ΔU = k qq₀/r
  • Substitute k = 9 × 10⁹, q = 3 × 10^-6, q₀ = 2 × 10^-6, r = 0.5
  • Calculate W_RP = 9 × 10⁹ × (3 × 10^-6 × 2 × 10^-6)/0.5 = 0.108 J

Practice Set

• Q1: Work done by external force moving positive charge against repulsive force is
  • A) Negative
  • B) Zero
  • C) Positive
  • D) Infinite
• Q2: Potential energy difference ΔU between two points depends on
  • A) Path taken
  • B) Initial and final points only
  • C) Speed of charge
  • D) Charge magnitude only
• Q3: Work done to move 1 μC charge from 1 m to 0.5 m near 2 μC charge is
  • A) 18 J
  • B) 0.018 J
  • C) 0.18 J
  • D) 1.8 J

Answer Key

• Q1: C - Work done by external force against repulsion is positive.
• Q2: B - Work depends only on initial and final points (path independent).
• Q3: B - Use W = k qq₀/r, calculate difference for given distances.

Memory Aids

• Work done = Potential energy difference.
• Electrostatic work is path independent.
• Only potential energy difference matters, not absolute values.

Concepts and Formulas

• Test charge q₀ > 0 moved slowly from point R to P against repulsive force by fixed charge q > 0 at origin.
• Work done by external force: W_RP = ∫_R^P vector F_ext · dvector r = - ∫_R^P vector F_E · dvector r
• Work done against electrostatic repulsion is stored as potential energy.
• Potential energy difference: ΔU = U_P - U_R = W_RP
• Work done by electrostatic field depends only on initial and final points (path independent).
• Only potential energy difference is physically significant, not absolute values.

Solved Examples

• Ex 1: Calculate work done to move q₀ = 2 μC from R to P against repulsive force of q = 3 μC at origin, distance r = 0.5 m.
  • Use W_RP = ΔU = k qq₀/r
  • Substitute k = 9 × 10⁹, q = 3 × 10^-6, q₀ = 2 × 10^-6, r = 0.5
  • Calculate W_RP = 9 × 10⁹ × (3 × 10^-6 × 2 × 10^-6)/0.5 = 0.108 J

Practice Set

• Q1: Work done by external force moving positive charge against repulsive force is
  • A) Negative
  • B) Zero
  • C) Positive
  • D) Infinite
• Q2: Potential energy difference ΔU between two points depends on
  • A) Path taken
  • B) Initial and final points only
  • C) Speed of charge
  • D) Charge magnitude only
• Q3: Work done to move 1 μC charge from 1 m to 0.5 m near 2 μC charge is
  • A) 18 J
  • B) 0.018 J
  • C) 0.18 J
  • D) 1.8 J

Answer Key

• Q1: C - Work done by external force against repulsion is positive.
• Q2: B - Work depends only on initial and final points (path independent).
• Q3: B - Use W = k qq₀/r, calculate difference for given distances.

Memory Aids

• Work done = Potential energy difference.
• Electrostatic work is path independent.
• Only potential energy difference matters, not absolute values.

Concepts and Formulas

• Test charge ₀ > 0 moved slowly from point R to P against repulsive force by fixed charge > 0 at origin.
• Work done by external force: W_RP = ∫_R^P vector F_ext · dvector r = - ∫_R^P vector F_E · dvector r
• Work done against electrostatic repulsion is stored as potential energy.
• Potential energy difference: ΔU = U_P - U_R = W_RP
• Work done by electrostatic field depends only on initial and final points (path independent).
• Only potential energy difference is physically significant, not absolute values.

Solved Examples

• Ex 1: Calculate work done to move q₀ = 2 μC from R to P against repulsive force of q = 3 μC at origin, distance r = 0.5 m.
  • Use W_RP = ΔU = k qq₀/r
  • Substitute k = 9 × 10⁹, q = 3 × 10^-6, q₀ = 2 × 10^-6, r = 0.5
  • Calculate W_RP = 9 × 10⁹ × (3 × 10^-6 × 2 × 10^-6)/0.5 = 0.108 J

Practice Set

• Q1: Work done by external force moving positive charge against repulsive force is
  • A) Negative
  • B) Zero
  • C) Positive
  • D) Infinite
• Q2: Potential energy difference ΔU between two points depends on
  • A) Path taken
  • B) Initial and final points only
  • C) Speed of charge
  • D) Charge magnitude only
• Q3: Work done to move 1 μC charge from 1 m to 0.5 m near 2 μC charge is
  • A) 18 J
  • B) 0.018 J
  • C) 0.18 J
  • D) 1.8 J

Answer Key

• Q1: C - Work done by external force against repulsion is positive.
• Q2: B - Work depends only on initial and final points (path independent).
• Q3: B - Use W = k qq₀/r, calculate difference for given distances.

Memory Aids

• Work done = Potential energy difference.
• Electrostatic work is path independent.
• Only potential energy difference matters, not absolute values.

Concepts and Formulas

• Test charge q₀ > 0 moved slowly from point R to P against repulsive force by fixed charge q > 0 at origin.
• Work done by external force: W_RP = ∫_R^P vector F_ext · dvector r = - ∫_R^P vector F_E · dvector r
• Work done against electrostatic repulsion is stored as potential energy.
• Potential energy difference: ΔU = U_P - U_R = W_RP
• Work done by electrostatic field depends only on initial and final points (path independent).
• Only potential energy difference is physically significant, not absolute values.

Solved Examples

• Ex 1: Calculate work done to move q₀ = 2 μC from R to P against repulsive force of q = 3 μC at origin, distance r = 0.5 m.
  • Use W_RP = ΔU = k qq₀/r
  • Substitute k = 9 × 10⁹, q = 3 × 10^-6, q₀ = 2 × 10^-6, r = 0.5
  • Calculate W_RP = 9 × 10⁹ × (3 × 10^-6 × 2 × 10^-6)/0.5 = 0.108 J

Practice Set

• Q1: Work done by external force moving positive charge against repulsive force is
  • A) Negative
  • B) Zero
  • C) Positive
  • D) Infinite
• Q2: Potential energy difference ΔU between two points depends on
  • A) Path taken
  • B) Initial and final points only
  • C) Speed of charge
  • D) Charge magnitude only
• Q3: Work done to move 1 μC charge from 1 m to 0.5 m near 2 μC charge is
  • A) 18 J
  • B) 0.018 J
  • C) 0.18 J
  • D) 1.8 J

Answer Key

• Q1: C - Work done by external force against repulsion is positive.
• Q2: B - Work depends only on initial and final points (path independent).
• Q3: B - Use W = k qq₀/r, calculate difference for given distances.

Memory Aids

• Work done = Potential energy difference.
• Electrostatic work is path independent.
• Only potential energy difference matters, not absolute values.

Concepts and Formulas

• Test charge ₀ > 0 moved slowly from point R to P against repulsive force by fixed charge > 0 at origin.
• Work done by external force: W_RP = ∫_R^P vector F_ext · dvector r = - ∫_R^P vector F_E · dvector r
• Work done against electrostatic repulsion is stored as potential energy.
• Potential energy difference: ΔU = U_P - U_R = W_RP
• Work done by electrostatic field depends only on initial and final points (path independent).
• Only potential energy difference is physically significant, not absolute values.

Solved Examples

• Ex 1: Calculate work done to move q₀ = 2 μC from R to P against repulsive force of q = 3 μC at origin, distance r = 0.5 m.
  • Use W_RP = ΔU = k qq₀/r
  • Substitute k = 9 × 10⁹, q = 3 × 10^-6, q₀ = 2 × 10^-6, r = 0.5
  • Calculate W_RP = 9 × 10⁹ × (3 × 10^-6 × 2 × 10^-6)/0.5 = 0.108 J

Practice Set

• Q1: Work done by external force moving positive charge against repulsive force is
  • A) Negative
  • B) Zero
  • C) Positive
  • D) Infinite
• Q2: Potential energy difference ΔU between two points depends on
  • A) Path taken
  • B) Initial and final points only
  • C) Speed of charge
  • D) Charge magnitude only
• Q3: Work done to move 1 μC charge from 1 m to 0.5 m near 2 μC charge is
  • A) 18 J
  • B) 0.018 J
  • C) 0.18 J
  • D) 1.8 J

Answer Key

• Q1: C - Work done by external force against repulsion is positive.
• Q2: B - Work depends only on initial and final points (path independent).
• Q3: B - Use W = k qq₀/r, calculate difference for given distances.

Memory Aids

• Work done = Potential energy difference.
• Electrostatic work is path independent.
• Only potential energy difference matters, not absolute values.

Concepts and Formulas

• Test charge q₀ > 0 moved slowly from point R to P against repulsive force by fixed charge q > 0 at origin.
• Work done by external force: W_RP = ∫_R^P vector F_ext · dvector r = - ∫_R^P vector F_E · dvector r
• Work done against electrostatic repulsion is stored as potential energy.
• Potential energy difference: ΔU = U_P - U_R = W_RP
• Work done by electrostatic field depends only on initial and final points (path independent).
• Only potential energy difference is physically significant, not absolute values.

Solved Examples

• Ex 1: Calculate work done to move q₀ = 2 μC from R to P against repulsive force of q = 3 μC at origin, distance r = 0.5 m.
  • Use W_RP = ΔU = k qq₀/r
  • Substitute k = 9 × 10⁹, q = 3 × 10^-6, q₀ = 2 × 10^-6, r = 0.5
  • Calculate W_RP = 9 × 10⁹ × (3 × 10^-6 × 2 × 10^-6)/0.5 = 0.108 J

Practice Set

• Q1: Work done by external force moving positive charge against repulsive force is
  • A) Negative
  • B) Zero
  • C) Positive
  • D) Infinite
• Q2: Potential energy difference ΔU between two points depends on
  • A) Path taken
  • B) Initial and final points only
  • C) Speed of charge
  • D) Charge magnitude only
• Q3: Work done to move 1 μC charge from 1 m to 0.5 m near 2 μC charge is
  • A) 18 J
  • B) 0.018 J
  • C) 0.18 J
  • D) 1.8 J

Answer Key

• Q1: C - Work done by external force against repulsion is positive.
• Q2: B - Work depends only on initial and final points (path independent).
• Q3: B - Use W = k qq₀/r, calculate difference for given distances.

Memory Aids

• Work done = Potential energy difference.
• Electrostatic work is path independent.
• Only potential energy difference matters, not absolute values.

Concepts and Formulas

• Test charge ₀ > 0 moved slowly from point R to P against repulsive force by fixed charge > 0 at origin.
• Work done by external force: W_RP = ∫_R^P vector F_ext · dvector r = - ∫_R^P vector F_E · dvector r
• Work done against electrostatic repulsion is stored as potential energy.
• Potential energy difference: ΔU = U_P - U_R = W_RP
• Work done by electrostatic field depends only on initial and final points (path independent).
• Only potential energy difference is physically significant, not absolute values.

Solved Examples

• Ex 1: Calculate work done to move q₀ = 2 μC from R to P against repulsive force of q = 3 μC at origin, distance r = 0.5 m.
  • Use W_RP = ΔU = k qq₀/r
  • Substitute k = 9 × 10⁹, q = 3 × 10^-6, q₀ = 2 × 10^-6, r = 0.5
  • Calculate W_RP = 9 × 10⁹ × (3 × 10^-6 × 2 × 10^-6)/0.5 = 0.108 J

Practice Set

• Q1: Work done by external force moving positive charge against repulsive force is
  • A) Negative
  • B) Zero
  • C) Positive
  • D) Infinite
• Q2: Potential energy difference ΔU between two points depends on
  • A) Path taken
  • B) Initial and final points only
  • C) Speed of charge
  • D) Charge magnitude only
• Q3: Work done to move 1 μC charge from 1 m to 0.5 m near 2 μC charge is
  • A) 18 J
  • B) 0.018 J
  • C) 0.18 J
  • D) 1.8 J

Answer Key

• Q1: C - Work done by external force against repulsion is positive.
• Q2: B - Work depends only on initial and final points (path independent).
• Q3: B - Use W = k qq₀/r, calculate difference for given distances.

Memory Aids

• Work done = Potential energy difference.
• Electrostatic work is path independent.
• Only potential energy difference matters, not absolute values.

Concepts and Formulas

• Test charge \( q_0 > 0 \) moved slowly from point R to P against repulsive force by fixed charge \( q > 0 \) at origin.
• Work done by external force: \( W_{RP} = \int_R^P \vec{F}_{ext} \cdot d\vec{r} = - \int_R^P \vec{F}_E \cdot d\vec{r} \)
• Work done against electrostatic repulsion is stored as potential energy.
• Potential energy difference: \( \Delta U = U_P - U_R = W_{RP} \)
• Work done by electrostatic field depends only on initial and final points (path independent).
• Only potential energy difference is physically significant, not absolute values.

Solved Examples

• Ex 1: Calculate work done to move \( q_0 = 2 \mu C \) from \( R \) to \( P \) against repulsive force of \( q = 3 \mu C \) at origin, distance \( r = 0.5 \) m.
  • Use \( W_{RP} = \Delta U = k \frac{q q_0}{r} \)
  • Substitute \( k = 9 \times 10^9 \), \( q = 3 \times 10^{-6} \), \( q_0 = 2 \times 10^{-6} \), \( r = 0.5 \)
  • Calculate \( W_{RP} = 9 \times 10^9 \times \frac{3 \times 10^{-6} \times 2 \times 10^{-6}}{0.5} = 0.108 \text{ J} \)

Practice Set

• Q1: Work done by external force moving positive charge against repulsive force is
  • A) Negative
  • B) Zero
  • C) Positive
  • D) Infinite
• Q2: Potential energy difference \( \Delta U \) between two points depends on
  • A) Path taken
  • B) Initial and final points only
  • C) Speed of charge
  • D) Charge magnitude only
• Q3: Work done to move \( 1 \mu C \) charge from 1 m to 0.5 m near \( 2 \mu C \) charge is
  • A) 18 J
  • B) 0.018 J
  • C) 0.18 J
  • D) 1.8 J

Answer Key

• Q1: C - Work done by external force against repulsion is positive.
• Q2: B - Work depends only on initial and final points (path independent).
• Q3: B - Use \( W = k \frac{q q_0}{r} \), calculate difference for given distances.

Memory Aids

• Work done = Potential energy difference.
• Electrostatic work is path independent.
• Only potential energy difference matters, not absolute values.

Electrostatic Potential

Concepts and Formulas

• Electrostatic potential V at a point = work done by external force to bring unit positive charge from infinity to that point.
• Formula: V = V_P - V_∞ = W_P/q₀, with V_∞ = 0.
• Potential at a point is potential difference between that point and infinity (reference).
• Potential difference between points A and B:
  • V_B - V_A = W_AB/q₀ = - ∫_A^B vector E · dvector l

Solved Examples

• Ex 1: Calculate potential at 0.5 m from charge +q = 4 μC.
  • Use V = kq/r
  • Substitute k = 9 × 10⁹, q = 4 × 10^-6, r = 0.5
  • Calculate V = 9 × 10⁹ × (4 × 10^-6)/0.5 = 72,000 V

Practice Set

• Q1: Electrostatic potential at infinity is
  • A) Zero
  • B) Infinite
  • C) Negative
  • D) Undefined
• Q2: Potential difference between two points is work done per
  • A) Unit negative charge
  • B) Unit positive charge
  • C) Total charge
  • D) None
• Q3: Potential at 1 m from 3 μC charge is
  • A) 27,000 V
  • B) 3,000 V
  • C) 9,000 V
  • D) 90,000 V

Answer Key

• Q1: A - Potential at infinity is zero by definition.
• Q2: B - Potential difference is work done per unit positive charge.
• Q3: A - V = kq/r = 9 × 10⁹ × (3 × 10^-6/1) = 27,000 V.

Memory Aids

• Potential V = W/q₀.
• Potential at infinity = 0.
• Potential difference = work done per unit positive charge.

Concepts and Formulas

• Electrostatic potential \( V \) at a point = work done by external force to bring unit positive charge from infinity to that point.
• Formula: \( V = V_P - V_\infty = \frac{W_P}{q_0} \), with \( V_\infty = 0 \).
• Potential at a point is potential difference between that point and infinity (reference).
• Potential difference between points A and B:
  • \( V_B - V_A = \frac{W_{AB}}{q_0} = - \int_A^B \vec{E} \cdot d\vec{l} \)

Solved Examples

• Ex 1: Calculate potential at 0.5 m from charge \( +q = 4 \mu C \).
  • Use \( V = k \frac{q}{r} \)
  • Substitute \( k = 9 \times 10^9 \), \( q = 4 \times 10^{-6} \), \( r = 0.5 \)
  • Calculate \( V = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{0.5} = 72,000 \text{ V} \)

Practice Set

• Q1: Electrostatic potential at infinity is
  • A) Zero
  • B) Infinite
  • C) Negative
  • D) Undefined
• Q2: Potential difference between two points is work done per
  • A) Unit negative charge
  • B) Unit positive charge
  • C) Total charge
  • D) None
• Q3: Potential at 1 m from \( 3 \mu C \) charge is
  • A) 27,000 V
  • B) 3,000 V
  • C) 9,000 V
  • D) 90,000 V

Answer Key

• Q1: A - Potential at infinity is zero by definition.
• Q2: B - Potential difference is work done per unit positive charge.
• Q3: A - \( V = k \frac{q}{r} = 9 \times 10^9 \times \frac{3 \times 10^{-6}}{1} = 27,000 \text{ V} \).

Memory Aids

• Potential \( V = \frac{W}{q_0} \).
• Potential at infinity = 0.
• Potential difference = work done per unit positive charge.

Potential due to an Electric Dipole

Axial Position

• Dipole consists of charges -q at A and +q at B, separated by 2a.
• Point C lies on the axis beyond B at distance r from A.
• Potential at C, V_C = V_CA + V_CB
• Formula: V_C = 1/(4πϵ₀) [ -q/(r+a) + q/(r-a) ] = 1/(4πϵ₀) (q · 2a)/(r² - a²)
• Using dipole moment p = q · 2a, V_C = 1/(4πϵ₀) p/(r² - a²)

Equatorial Position

• Charges -q and +q separated by 2a on horizontal line.
• Point C is above midpoint between charges at distance r.
• Potential at C, V_C = V_CA + V_CB
• Formula: V_C = 1/(4πϵ₀) [ -q/√(r² + a²) + q/√(r² + a²) ] = 0

Any Point due to Short Electric Dipole

• Distances: OP = r, PB = r - a cosθ, PA = r + a cosθ
• Potential at P, V_P = V_PA + V_PB
• Formula: V_P = 1/(4πϵ₀) [ -q/(r + a cosθ) + q/(r - a cosθ) ]
• Simplified: V_P = 1/(4πϵ₀) (q 2a cosθ)/(r² - a² cos²θ) = 1/(4πϵ₀) (p cosθ)/(r² - a² cos²θ)
• For r ≫ a, approximation: V_P = 1/(4πϵ₀) (p cosθ)/r² = 1/(4πϵ₀) (vector p · îr)/r²

Solved Examples

• Ex 1: Calculate potential at axial point C, r = 10a, dipole moment p = 2 × 10^-29 C·m.
  • Use V_C = 1/(4πϵ₀) p/(r² - a²) ≈ 1/(4πϵ₀) p/r²
  • Substitute values and calculate.
• Ex 2: Find potential at equatorial point r = 5a.
  • Use V_C = 0 (due to symmetry).
• Ex 3: Potential at point P at angle θ = 60°, distance r = 20a, dipole moment p = 3 × 10^-29 C·m.
  • Use V_P = 1/(4πϵ₀) (p cosθ)/r²
  • Calculate cos 60° = 0.5 and substitute.

Practice Set

• Q1: Potential at axial point for dipole moment p and distance r is:
  • A) 1/(4πϵ₀) p/r
  • B) 1/(4πϵ₀) p/r²
  • C) 1/(4πϵ₀) p/r³
  • D) Zero
• Q2: Potential at equatorial point is:
  • A) 1/(4πϵ₀) p/r²
  • B) Zero
  • C) 1/(4πϵ₀) p/r³
  • D) 1/(4πϵ₀) p cosθ/r²
• Q3: Potential at point P at angle 45°, distance r, dipole moment p is:
  • A) 1/(4πϵ₀) p sinθ/r²
  • B) 1/(4πϵ₀) p cosθ/r²
  • C) 1/(4πϵ₀) p/r³
  • D) Zero

Answer Key

• Q1: B) 1/(4πϵ₀) p/r² - Potential at axial point varies as 1/r².
• Q2: B) Zero - Potential at equatorial point cancels out due to equal and opposite charges.
• Q3: B) 1/(4πϵ₀) p cosθ/r² - Potential depends on cosθ at any point.

Memory Aids

• Axial potential ∝ p/r²
• Equatorial potential = 0
• General potential V = 1/(4πϵ₀) (vector p · îr)/r²
• Dipole moment p = q × 2a

Axial Position

• Dipole consists of charges -q at A and +q at B, separated by 2a.
• Point C lies on the axis beyond B at distance r from A.
• Potential at C, \( V_C = V_{CA} + V_{CB} \)
• Formula: \( V_C = \frac{1}{4\pi\epsilon_0} \left[ \frac{-q}{r+a} + \frac{q}{r-a} \right] = \frac{1}{4\pi\epsilon_0} \frac{q \cdot 2a}{r^2 - a^2} \)
• Using dipole moment \( p = q \cdot 2a \), \( V_C = \frac{1}{4\pi\epsilon_0} \frac{p}{r^2 - a^2} \)

Equatorial Position

• Charges -q and +q separated by 2a on horizontal line.
• Point C is above midpoint between charges at distance r.
• Potential at C, \( V_C = V_{CA} + V_{CB} \)
• Formula: \( V_C = \frac{1}{4\pi\epsilon_0} \left[ \frac{-q}{\sqrt{r^2 + a^2}} + \frac{q}{\sqrt{r^2 + a^2}} \right] = 0 \)

Any Point due to Short Electric Dipole

• Distances: \( OP = r \), \( PB = r - a \cos\theta \), \( PA = r + a \cos\theta \)
• Potential at P, \( V_P = V_{PA} + V_{PB} \)
• Formula: \( V_P = \frac{1}{4\pi\epsilon_0} \left[ \frac{-q}{r + a \cos\theta} + \frac{q}{r - a \cos\theta} \right] \)
• Simplified: \( V_P = \frac{1}{4\pi\epsilon_0} \frac{q 2a \cos\theta}{r^2 - a^2 \cos^2\theta} = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2 - a^2 \cos^2\theta} \)
• For \( r \gg a \), approximation: \( V_P = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \hat{r}}{r^2} \)

Solved Examples

• Ex 1: Calculate potential at axial point C, \( r = 10a \), dipole moment \( p = 2 \times 10^{-29} \) C·m.
  • Use \( V_C = \frac{1}{4\pi\epsilon_0} \frac{p}{r^2 - a^2} \approx \frac{1}{4\pi\epsilon_0} \frac{p}{r^2} \)
  • Substitute values and calculate.
• Ex 2: Find potential at equatorial point \( r = 5a \).
  • Use \( V_C = 0 \) (due to symmetry).
• Ex 3: Potential at point P at angle \( \theta = 60^\circ \), distance \( r = 20a \), dipole moment \( p = 3 \times 10^{-29} \) C·m.
  • Use \( V_P = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2} \)
  • Calculate \( \cos 60^\circ = 0.5 \) and substitute.

Practice Set

• Q1: Potential at axial point for dipole moment \( p \) and distance \( r \) is:
  • A) \( \frac{1}{4\pi\epsilon_0} \frac{p}{r} \)
  • B) \( \frac{1}{4\pi\epsilon_0} \frac{p}{r^2} \)
  • C) \( \frac{1}{4\pi\epsilon_0} \frac{p}{r^3} \)
  • D) Zero
• Q2: Potential at equatorial point is:
  • A) \( \frac{1}{4\pi\epsilon_0} \frac{p}{r^2} \)
  • B) Zero
  • C) \( \frac{1}{4\pi\epsilon_0} \frac{p}{r^3} \)
  • D) \( \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2} \)
• Q3: Potential at point P at angle \( 45^\circ \), distance \( r \), dipole moment \( p \) is:
  • A) \( \frac{1}{4\pi\epsilon_0} \frac{p \sin\theta}{r^2} \)
  • B) \( \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2} \)
  • C) \( \frac{1}{4\pi\epsilon_0} \frac{p}{r^3} \)
  • D) Zero

Answer Key

• Q1: B) \( \frac{1}{4\pi\epsilon_0} \frac{p}{r^2} \) - Potential at axial point varies as \( 1/r^2 \).
• Q2: B) Zero - Potential at equatorial point cancels out due to equal and opposite charges.
• Q3: B) \( \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2} \) - Potential depends on \( \cos\theta \) at any point.

Memory Aids

• Axial potential \( \propto \frac{p}{r^2} \)
• Equatorial potential = 0
• General potential \( V = \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \hat{r}}{r^2} \)
• Dipole moment \( p = q \times 2a \)

Potential due to a system of charges

Concepts and Formulas

• Electric potential (V) is the potential energy per unit charge at a point (unit: Volt, V = J/C).
• Electric potential is a scalar quantity (no direction).
• Total potential at a point due to multiple charges is the algebraic sum of potentials from each charge.
• Formula: V = k (Q₁/r₁ + Q₂/r₂ + ... + Q_n/r_n), where k = 1/(4 π ε₀).
• Q_i = charge, r_i = distance from charge to point.

Solved Examples

Ex 1: Calculate total potential at point P due to charges +q₁ = 2 μC at r₁ = 0.5 m and -q₂ = 3 μC at r₂ = 0.3 m. Use k = 9 × 10⁹ Nm²/C².

• Calculate individual potentials: V₁ = kq₁/r₁ = 9 × 10⁹ × (2 × 10^-6)/0.5 = 36000 V.
• V₂ = k(-q₂)/r₂ = 9 × 10⁹ × (-3 × 10^-6)/0.3 = -90000 V.
• Total potential: V_p = V₁ + V₂ = 36000 - 90000 = -54000 V.

Practice Set

Q1: What is the unit of electric potential?

• A) Coulomb
• B) Volt
• C) Newton
• D) Joule

Q2: Two charges +4 μC and -2 μC are placed 0.2 m and 0.1 m from point P respectively. What is the total potential at P? (Use k = 9 × 10⁹)

• A) 0 V
• B) 1.8 × 10⁵ V
• C) 3.6 × 10⁵ V
• D) -1.8 × 10⁵ V

Q3: For three charges q₁ = 1 μC, q₂ = 2 μC, q₃ = -3 μC at distances r₁ = 0.1 m, r₂ = 0.2 m, r₃ = 0.3 m from point P, find the total potential.

• A) 0 V
• B) 9 × 10⁴ V
• C) -4.5 × 10⁴ V
• D) 1.35 × 10⁵ V

Answer Key

Q1: B) Volt — Electric potential is measured in volts (Joule per Coulomb).

Q2: B) 1.8 × 10⁵ V — Calculate potentials: V₁ = 9 × 10⁹ × (4 × 10^-6)/0.2 = 1.8 × 10⁵ V, V₂ = 9 × 10⁹ × (-2 × 10^-6)/0.1 = -1.8 × 10⁵ V, total = 0 V (Check carefully: sum is zero, so correct answer is 0 V, option A. Correction: The sum is zero, so correct answer is A).

Q3: C) -4.5 × 10⁴ V — Calculate potentials: V₁ = 9 × 10⁹ × (1 × 10^-6)/0.1 = 9 × 10⁴ V, V₂ = 9 × 10⁹ × (2 × 10^-6)/0.2 = 9 × 10⁴ V, V₃ = 9 × 10⁹ × (-3 × 10^-6)/0.3 = -9 × 10⁴ V, total = 9 × 10⁴ + 9 × 10⁴ - 9 × 10⁴ = 9 × 10⁴ V (Correction: sum is 9 × 10⁴ V, so correct answer is B).

Memory Aids

• Potential is scalar: sum potentials algebraically.
• Use V = k ∑ Q_i/r_i for multiple charges.
• Unit of potential: Volt (V) = Joule/Coulomb.

Concepts and Formulas

• Electric potential (V) is the potential energy per unit charge at a point (unit: Volt, V = J/C).
• Electric potential is a scalar quantity (no direction).
• Total potential at a point due to multiple charges is the algebraic sum of potentials from each charge.
• Formula: V = k (Q₁/r₁ + Q₂/r₂ + ... + Q_n/r_n), where k = 1/(4 π ε₀).
• Q_i = charge, r_i = distance from charge to point.

Solved Examples

Ex 1: Calculate total potential at point P due to charges +q₁ = 2 μC at r₁ = 0.5 m and -q₂ = 3 μC at r₂ = 0.3 m. Use k = 9 × 10⁹ Nm²/C².

• Calculate individual potentials: V₁ = kq₁/r₁ = 9 × 10⁹ × (2 × 10^-6)/0.5 = 36000 V.
• V₂ = k(-q₂)/r₂ = 9 × 10⁹ × (-3 × 10^-6)/0.3 = -90000 V.
• Total potential: V_p = V₁ + V₂ = 36000 - 90000 = -54000 V.

Practice Set

Q1: What is the unit of electric potential?

• A) Coulomb
• B) Volt
• C) Newton
• D) Joule

Q2: Two charges +4 μC and -2 μC are placed 0.2 m and 0.1 m from point P respectively. What is the total potential at P? (Use k = 9 × 10⁹)

• A) 0 V
• B) 1.8 × 10⁵ V
• C) 3.6 × 10⁵ V
• D) -1.8 × 10⁵ V

Q3: For three charges q₁ = 1 μC, q₂ = 2 μC, q₃ = -3 μC at distances r₁ = 0.1 m, r₂ = 0.2 m, r₃ = 0.3 m from point P, find the total potential.

• A) 0 V
• B) 9 × 10⁴ V
• C) -4.5 × 10⁴ V
• D) 1.35 × 10⁵ V

Answer Key

Q1: B) Volt — Electric potential is measured in volts (Joule per Coulomb).

Q2: B) 1.8 × 10⁵ V — Calculate potentials: V₁ = 9 × 10⁹ × (4 × 10^-6)/0.2 = 1.8 × 10⁵ V, V₂ = 9 × 10⁹ × (-2 × 10^-6)/0.1 = -1.8 × 10⁵ V, total = 0 V (Check carefully: sum is zero, so correct answer is 0 V, option A. Correction: The sum is zero, so correct answer is A).

Q3: C) -4.5 × 10⁴ V — Calculate potentials: V₁ = 9 × 10⁹ × (1 × 10^-6)/0.1 = 9 × 10⁴ V, V₂ = 9 × 10⁹ × (2 × 10^-6)/0.2 = 9 × 10⁴ V, V₃ = 9 × 10⁹ × (-3 × 10^-6)/0.3 = -9 × 10⁴ V, total = 9 × 10⁴ + 9 × 10⁴ - 9 × 10⁴ = 9 × 10⁴ V (Correction: sum is 9 × 10⁴ V, so correct answer is B).

Memory Aids

• Potential is scalar: sum potentials algebraically.
• Use V = k ∑ Q_i/r_i for multiple charges.
• Unit of potential: Volt (V) = Joule/Coulomb.

Concepts and Formulas

• Electric potential (V) is the potential energy per unit charge at a point (unit: Volt, V = J/C).
• Electric potential is a scalar quantity (no direction).
• Total potential at a point due to multiple charges is the algebraic sum of potentials from each charge.
• Formula: V = k (Q₁/r₁ + Q₂/r₂ + ... + Q_n/r_n), where k = 1/(4 π ε₀).
• Q_i = charge, r_i = distance from charge to point.

Solved Examples

Ex 1: Calculate total potential at point P due to charges +q₁ = 2 μC at r₁ = 0.5 m and -q₂ = 3 μC at r₂ = 0.3 m. Use k = 9 × 10⁹ Nm²/C².

• Calculate individual potentials: V₁ = kq₁/r₁ = 9 × 10⁹ × (2 × 10^-6)/0.5 = 36000 V.
• V₂ = k(-q₂)/r₂ = 9 × 10⁹ × (-3 × 10^-6)/0.3 = -90000 V.
• Total potential: V_p = V₁ + V₂ = 36000 - 90000 = -54000 V.

Practice Set

Q1: What is the unit of electric potential?

• A) Coulomb
• B) Volt
• C) Newton
• D) Joule

Q2: Two charges +4 μC and -2 μC are placed 0.2 m and 0.1 m from point P respectively. What is the total potential at P? (Use k = 9 × 10⁹)

• A) 0 V
• B) 1.8 × 10⁵ V
• C) 3.6 × 10⁵ V
• D) -1.8 × 10⁵ V

Q3: For three charges q₁ = 1 μC, q₂ = 2 μC, q₃ = -3 μC at distances r₁ = 0.1 m, r₂ = 0.2 m, r₃ = 0.3 m from point P, find the total potential.

• A) 0 V
• B) 9 × 10⁴ V
• C) -4.5 × 10⁴ V
• D) 1.35 × 10⁵ V

Answer Key

Q1: B) Volt — Electric potential is measured in volts (Joule per Coulomb).

Q2: B) 1.8 × 10⁵ V — Calculate potentials: V₁ = 9 × 10⁹ × (4 × 10^-6)/0.2 = 1.8 × 10⁵ V, V₂ = 9 × 10⁹ × (-2 × 10^-6)/0.1 = -1.8 × 10⁵ V, total = 0 V (Check carefully: sum is zero, so correct answer is 0 V, option A. Correction: The sum is zero, so correct answer is A).

Q3: C) -4.5 × 10⁴ V — Calculate potentials: V₁ = 9 × 10⁹ × (1 × 10^-6)/0.1 = 9 × 10⁴ V, V₂ = 9 × 10⁹ × (2 × 10^-6)/0.2 = 9 × 10⁴ V, V₃ = 9 × 10⁹ × (-3 × 10^-6)/0.3 = -9 × 10⁴ V, total = 9 × 10⁴ + 9 × 10⁴ - 9 × 10⁴ = 9 × 10⁴ V (Correction: sum is 9 × 10⁴ V, so correct answer is B).

Memory Aids

• Potential is scalar: sum potentials algebraically.
• Use V = k ∑ Q_i/r_i for multiple charges.
• Unit of potential: Volt (V) = Joule/Coulomb.

Concepts and Formulas

• Electric potential (V) is the potential energy per unit charge at a point (unit: Volt, V = J/C).
• Electric potential is a scalar quantity (no direction).
• Total potential at a point due to multiple charges is the algebraic sum of potentials from each charge.
• Formula: \( V = k \left( \frac{Q_1}{r_1} + \frac{Q_2}{r_2} + \ldots + \frac{Q_n}{r_n} \right) \), where \( k = \frac{1}{4 \pi \epsilon_0} \).
• \( Q_i \) = charge, \( r_i \) = distance from charge to point.

Solved Examples

Ex 1: Calculate total potential at point P due to charges \( +q_1 = 2 \mu C \) at \( r_1 = 0.5 m \) and \( -q_2 = 3 \mu C \) at \( r_2 = 0.3 m \). Use \( k = 9 \times 10^9 \ Nm^2/C^2 \).

• Calculate individual potentials: \( V_1 = k \frac{q_1}{r_1} = 9 \times 10^9 \times \frac{2 \times 10^{-6}}{0.5} = 36000 V \).
• \( V_2 = k \frac{-q_2}{r_2} = 9 \times 10^9 \times \frac{-3 \times 10^{-6}}{0.3} = -90000 V \).
• Total potential: \( V_p = V_1 + V_2 = 36000 - 90000 = -54000 V \).

Practice Set

Q1: What is the unit of electric potential?

• A) Coulomb
• B) Volt
• C) Newton
• D) Joule

Q2: Two charges \( +4 \mu C \) and \( -2 \mu C \) are placed 0.2 m and 0.1 m from point P respectively. What is the total potential at P? (Use \( k = 9 \times 10^9 \))

• A) 0 V
• B) 1.8 \times 10^5 V
• C) 3.6 \times 10^5 V
• D) -1.8 \times 10^5 V

Q3: For three charges \( q_1 = 1 \mu C, q_2 = 2 \mu C, q_3 = -3 \mu C \) at distances \( r_1 = 0.1 m, r_2 = 0.2 m, r_3 = 0.3 m \) from point P, find the total potential.

• A) 0 V
• B) 9 \times 10^4 V
• C) -4.5 \times 10^4 V
• D) 1.35 \times 10^5 V

Answer Key

Q1: B) Volt — Electric potential is measured in volts (Joule per Coulomb).

Q2: B) 1.8 \times 10^5 V — Calculate potentials: \( V_1 = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{0.2} = 1.8 \times 10^5 V \), \( V_2 = 9 \times 10^9 \times \frac{-2 \times 10^{-6}}{0.1} = -1.8 \times 10^5 V \), total \( = 0 V \) (Check carefully: sum is zero, so correct answer is 0 V, option A. Correction: The sum is zero, so correct answer is A).

Q3: C) -4.5 \times 10^4 V — Calculate potentials: \( V_1 = 9 \times 10^9 \times \frac{1 \times 10^{-6}}{0.1} = 9 \times 10^4 V \), \( V_2 = 9 \times 10^9 \times \frac{2 \times 10^{-6}}{0.2} = 9 \times 10^4 V \), \( V_3 = 9 \times 10^9 \times \frac{-3 \times 10^{-6}}{0.3} = -9 \times 10^4 V \), total \( = 9 \times 10^4 + 9 \times 10^4 - 9 \times 10^4 = 9 \times 10^4 V \) (Correction: sum is \( 9 \times 10^4 V \), so correct answer is B).

Memory Aids

• Potential is scalar: sum potentials algebraically.
• Use \( V = k \sum \frac{Q_i}{r_i} \) for multiple charges.
• Unit of potential: Volt (V) = Joule/Coulomb.

Potential Energy of a System of Charges

Two Point Charges

• Work done to bring charge q₁ from infinity to point A: W₁ = q₁ V(A) = 0 (since V(A) = 0)
• Potential at point B due to q₁: V(B) = 1/(4 π ε₀) q₁/r₁₂
• Work done to bring charge q₂ from infinity to point B: W₂ = q₂ V(B) = 1/(4 π ε₀) q₁q₂/r₁₂
• Total potential energy of two charges: U = W₁ + W₂ = 1/(4 π ε₀) q₁q₂/r₁₂

Three Point Charges

• Work done to bring q₃ from infinity to point C: W₃ = q₃ [V(C) - V(∞)] = q₃ (1/(4 π ε₀) (q₁/r₁₃ + q₂/r₂₃)) = 1/(4 π ε₀) (q₁q₃/r₁₃ + q₂q₃/r₂₃)
• Total potential energy of three charges: U = 1/(4 π ε₀) (q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₁q₃/r₁₃)
• Potential energy is distributed over the entire field, not localized.

System of Multiple Charges

• Potential energy of a system of N point charges: U = 1/(4 π ε₀) ∑_i=1^N ∑_j>i^N q_iq_j/r_ij
• Summation is over all unique pairs to avoid double counting.

Solved Examples

• Ex 1: Calculate potential energy of two charges q₁ = 2 μC, q₂ = 3 μC separated by 0.5 m.
  • Use U = 1/(4 π ε₀) q₁q₂/r
  • U = 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/0.5 = 0.108 J
• Ex 2: Find potential energy of three charges q₁ = q₂ = q₃ = 1 μC at vertices of equilateral triangle of side 1 m.
  • U = 1/(4 π ε₀) ((q₁q₂/r) + (q₂q₃/r) + (q₁q₃/r)) = 3 × (9 × 10⁹ × (1 × 10^-6)²/1) = 0.027 J

Practice Set

• Q1: Potential energy of two charges 1 μC and 2 μC separated by 1 m is?
  • A) 0.018 J
  • B) 0.009 J
  • C) 0.036 J
  • D) 0.027 J
• Q2: For three charges each 1 μC at vertices of equilateral triangle side 2 m, potential energy is?
  • A) 0.0135 J
  • B) 0.027 J
  • C) 0.054 J
  • D) 0.009 J
• Q3: Total potential energy of 4 charges each 1 μC placed at corners of square side 1 m is?
  • A) 0.027 J
  • B) 0.036 J
  • C) 0.045 J
  • D) 0.054 J

Answer Key

• Q1: A) 0.018 J - Use U = 1/(4 π ε₀) q₁q₂/r = 9 × 10⁹ × (1 × 10^-6 × 2 × 10^-6)/1 = 0.018 J
• Q2: A) 0.0135 J - U = 3 × (9 × 10⁹ × (1 × 10^-6)²/2) = 0.0135 J
• Q3: B) 0.036 J - Four charges at square corners have 6 pairs, U = 6 × (9 × 10⁹ × (1 × 10^-6)²/1) = 0.036 J

Memory Aids

• Potential energy of two charges: U = k q₁q₂/r
• For multiple charges, sum over all unique pairs.
• Potential energy is stored in the electric field around charges.

Two Point Charges

• Work done to bring charge \( q_1 \) from infinity to point A: \( W_1 = q_1 V(A) = 0 \) (since \( V(A) = 0 \))
• Potential at point B due to \( q_1 \): \( V(B) = \frac{1}{4 \pi \epsilon_0} \frac{q_1}{r_{12}} \)
• Work done to bring charge \( q_2 \) from infinity to point B: \( W_2 = q_2 V(B) = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r_{12}} \)
• Total potential energy of two charges: \( U = W_1 + W_2 = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r_{12}} \)

Three Point Charges

• Work done to bring \( q_3 \) from infinity to point C: \( W_3 = q_3 [V(C) - V(\infty)] = q_3 \left( \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{r_{13}} + \frac{q_2}{r_{23}} \right) \right) = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}} \right) \)
• Total potential energy of three charges: \( U = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_1 q_3}{r_{13}} \right) \)
• Potential energy is distributed over the entire field, not localized.

System of Multiple Charges

• Potential energy of a system of \( N \) point charges: \( U = \frac{1}{4 \pi \epsilon_0} \sum_{i=1}^N \sum_{j>i}^N \frac{q_i q_j}{r_{ij}} \)
• Summation is over all unique pairs to avoid double counting.

Solved Examples

• Ex 1: Calculate potential energy of two charges \( q_1 = 2 \mu C \), \( q_2 = 3 \mu C \) separated by \( 0.5 \ m \).
  • Use \( U = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r} \)
  • \( U = 9 \times 10^9 \times \frac{2 \times 10^{-6} \times 3 \times 10^{-6}}{0.5} = 0.108 \ J \)
• Ex 2: Find potential energy of three charges \( q_1 = q_2 = q_3 = 1 \mu C \) at vertices of equilateral triangle of side \( 1 \ m \).
  • \( U = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1 q_2}{r} + \frac{q_2 q_3}{r} + \frac{q_1 q_3}{r} \right) = 3 \times \frac{9 \times 10^9 \times (1 \times 10^{-6})^2}{1} = 0.027 \ J \)

Practice Set

• Q1: Potential energy of two charges \( 1 \mu C \) and \( 2 \mu C \) separated by \( 1 \ m \) is?
  • A) 0.018 J
  • B) 0.009 J
  • C) 0.036 J
  • D) 0.027 J
• Q2: For three charges each \( 1 \mu C \) at vertices of equilateral triangle side \( 2 \ m \), potential energy is?
  • A) 0.0135 J
  • B) 0.027 J
  • C) 0.054 J
  • D) 0.009 J
• Q3: Total potential energy of 4 charges each \( 1 \mu C \) placed at corners of square side \( 1 \ m \) is?
  • A) 0.027 J
  • B) 0.036 J
  • C) 0.045 J
  • D) 0.054 J

Answer Key

• Q1: A) 0.018 J - Use \( U = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r} = 9 \times 10^9 \times \frac{1 \times 10^{-6} \times 2 \times 10^{-6}}{1} = 0.018 J \)
• Q2: A) 0.0135 J - \( U = 3 \times \frac{9 \times 10^9 \times (1 \times 10^{-6})^2}{2} = 0.0135 J \)
• Q3: B) 0.036 J - Four charges at square corners have 6 pairs, \( U = 6 \times \frac{9 \times 10^9 \times (1 \times 10^{-6})^2}{1} = 0.036 J \)

Memory Aids

• Potential energy of two charges: \( U = k \frac{q_1 q_2}{r} \)
• For multiple charges, sum over all unique pairs.
• Potential energy is stored in the electric field around charges.

Electric Potential due to a Continuous Charge Distribution

General Concept

• Electric potential due to small charge element dq at point P: dV = kdq/r, where k = 1/(4 π ε₀)
• Total potential: V = ∫dV = 1/(4 π ε₀) ∫dq/r
• Used for irregular or continuous charge distributions by integration.

Uniformly Charged Ring

• Potential at point P on axis at distance x: V = kq/√(R² + x²)
• At center (x=0): V₀ = kq/R

Uniformly Charged Disc

• Disc divided into thin rings of radius r and thickness dr
• Charge on ring: dq = σ(2πrdr), σ = surface charge density
• Potential at point on axis: V = ∫₀^R kdq/√(r² + x²)
• After integration, V = 2πkσ(√(R² + x²) - x)
• At center (x=0): V = 2πkσR

Electric Potential Difference due to Uniformly Charged Wire

• Electric field at distance x from wire: E = λ/(2 π ε₀x)
• Relation: E = - dV/dx
• Potential difference between points x₁ and x₂: [ V₂ - V₁ = - ∫_x1^x₂ E dx = λ/(2 π ε₀) ln x₁/x₂ ]

Solved Examples

• Ex 1: Potential at point on axis of ring with q = 4 μC, R = 0.5 m, x = 0.5 m
  • V = kq/√(R² + x²) = (9 × 10⁹ × 4 × 10^-6)/√(0.5² + 0.5²) = 5.09 × 10⁴ V
• Ex 2: Potential difference between x₁ = 0.1 m and x₂ = 0.5 m from charged wire with λ = 2 × 10^-6C/m
  • V₂ - V₁ = λ/(2 π ε₀) ln x₁/x₂ = (2 × 10^-6)/(2 π × 8.85 × 10^-12) ln 0.1/0.5 = -1.02 × 10⁴ V

Practice Set

• Q1: Potential at center of ring with charge 3 μC and radius 0.4 m is?
  • A) 6.75 × 10⁴ V
  • B) 1.35 × 10⁵ V
  • C) 9.0 × 10⁴ V
  • D) 4.5 × 10⁴ V
• Q2: Potential difference between 0.2 m and 0.5 m from charged wire with λ = 1 × 10^-6C/m is?
  • A) 4.56 × 10³ V
  • B) 3.22 × 10³ V
  • C) 2.78 × 10³ V
  • D) 1.89 × 10³ V
• Q3: Potential at point x = 0.3 m on axis of ring with q = 5 μC, R = 0.4 m is?
  • A) 7.5 × 10⁴ V
  • B) 8.5 × 10⁴ V
  • C) 9.0 × 10⁴ V
  • D) 6.0 × 10⁴ V

Answer Key

• Q1: A) 6.75 × 10⁴ V - V₀ = kq/R = (9 × 10⁹ × 3 × 10^-6)/0.4 = 6.75 × 10⁴ V
• Q2: B) 3.22 × 10³ V - V₂ - V₁ = λ/(2 π ε₀) ln 0.2/0.5 = 3.22 × 10³ V
• Q3: C) 9.0 × 10⁴ V - V = kq/√(R² + x²) = (9 × 10⁹ × 5 × 10^-6)/√(0.4² + 0.3²) = 9.0 × 10⁴ V

Memory Aids

• Potential due to charge element: dV = kdq/r
• Ring potential: V = kq/√(R² + x²)
• Wire potential difference: logarithmic function of distances.

General Concept

• Electric potential due to small charge element dq at point P: dV = kdq/r, where k = 1/(4 π ε₀)
• Total potential: V = ∫dV = 1/(4 π ε₀) ∫dq/r
• Used for irregular or continuous charge distributions by integration.

Uniformly Charged Ring

• Potential at point P on axis at distance x: V = kq/√(R² + x²)
• At center (x=0): V₀ = kq/R

Uniformly Charged Disc

• Disc divided into thin rings of radius r and thickness dr
• Charge on ring: dq = σ(2πrdr), σ = surface charge density
• Potential at point on axis: V = ∫₀^R kdq/√(r² + x²)
• After integration, V = 2πkσ(√(R² + x²) - x)
• At center (x=0): V = 2πkσR

Electric Potential Difference due to Uniformly Charged Wire

• Electric field at distance x from wire: E = λ/(2 π ε₀x)
• Relation: E = - dV/dx
• Potential difference between points x₁ and x₂: [ V₂ - V₁ = - ∫_x1^x₂ E dx = λ/(2 π ε₀) ln x₁/x₂ ]

Solved Examples

• Ex 1: Potential at point on axis of ring with q = 4 μC, R = 0.5 m, x = 0.5 m
  • V = kq/√(R² + x²) = (9 × 10⁹ × 4 × 10^-6)/√(0.5² + 0.5²) = 5.09 × 10⁴ V
• Ex 2: Potential difference between x₁ = 0.1 m and x₂ = 0.5 m from charged wire with λ = 2 × 10^-6C/m
  • V₂ - V₁ = λ/(2 π ε₀) ln x₁/x₂ = (2 × 10^-6)/(2 π × 8.85 × 10^-12) ln 0.1/0.5 = -1.02 × 10⁴ V

Practice Set

• Q1: Potential at center of ring with charge 3 μC and radius 0.4 m is?
  • A) 6.75 × 10⁴ V
  • B) 1.35 × 10⁵ V
  • C) 9.0 × 10⁴ V
  • D) 4.5 × 10⁴ V
• Q2: Potential difference between 0.2 m and 0.5 m from charged wire with λ = 1 × 10^-6C/m is?
  • A) 4.56 × 10³ V
  • B) 3.22 × 10³ V
  • C) 2.78 × 10³ V
  • D) 1.89 × 10³ V
• Q3: Potential at point x = 0.3 m on axis of ring with q = 5 μC, R = 0.4 m is?
  • A) 7.5 × 10⁴ V
  • B) 8.5 × 10⁴ V
  • C) 9.0 × 10⁴ V
  • D) 6.0 × 10⁴ V

Answer Key

• Q1: A) 6.75 × 10⁴ V - V₀ = kq/R = (9 × 10⁹ × 3 × 10^-6)/0.4 = 6.75 × 10⁴ V
• Q2: B) 3.22 × 10³ V - V₂ - V₁ = λ/(2 π ε₀) ln 0.2/0.5 = 3.22 × 10³ V
• Q3: C) 9.0 × 10⁴ V - V = kq/√(R² + x²) = (9 × 10⁹ × 5 × 10^-6)/√(0.4² + 0.3²) = 9.0 × 10⁴ V

Memory Aids

• Potential due to charge element: dV = kdq/r
• Ring potential: V = kq/√(R² + x²)
• Wire potential difference: logarithmic function of distances.

General Concept

• Electric potential due to small charge element dq at point P: dV = kdq/r, where k = 1/(4 π ε₀)
• Total potential: V = ∫dV = 1/(4 π ε₀) ∫dq/r
• Used for irregular or continuous charge distributions by integration.

Uniformly Charged Ring

• Potential at point P on axis at distance x: V = kq/√(R² + x²)
• At center (x=0): V₀ = kq/R

Uniformly Charged Disc

• Disc divided into thin rings of radius r and thickness dr
• Charge on ring: dq = σ(2πrdr), σ = surface charge density
• Potential at point on axis: V = ∫₀^R kdq/√(r² + x²)
• After integration, V = 2πkσ(√(R² + x²) - x)
• At center (x=0): V = 2πkσR

Electric Potential Difference due to Uniformly Charged Wire

• Electric field at distance x from wire: E = λ/(2 π ε₀x)
• Relation: E = - dV/dx
• Potential difference between points x₁ and x₂: [ V₂ - V₁ = - ∫_x1^x₂ E dx = λ/(2 π ε₀) ln x₁/x₂ ]

Solved Examples

• Ex 1: Potential at point on axis of ring with q = 4 μC, R = 0.5 m, x = 0.5 m
  • V = kq/√(R² + x²) = (9 × 10⁹ × 4 × 10^-6)/√(0.5² + 0.5²) = 5.09 × 10⁴ V
• Ex 2: Potential difference between x₁ = 0.1 m and x₂ = 0.5 m from charged wire with λ = 2 × 10^-6C/m
  • V₂ - V₁ = λ/(2 π ε₀) ln x₁/x₂ = (2 × 10^-6)/(2 π × 8.85 × 10^-12) ln 0.1/0.5 = -1.02 × 10⁴ V

Practice Set

• Q1: Potential at center of ring with charge 3 μC and radius 0.4 m is?
  • A) 6.75 × 10⁴ V
  • B) 1.35 × 10⁵ V
  • C) 9.0 × 10⁴ V
  • D) 4.5 × 10⁴ V
• Q2: Potential difference between 0.2 m and 0.5 m from charged wire with λ = 1 × 10^-6C/m is?
  • A) 4.56 × 10³ V
  • B) 3.22 × 10³ V
  • C) 2.78 × 10³ V
  • D) 1.89 × 10³ V
• Q3: Potential at point x = 0.3 m on axis of ring with q = 5 μC, R = 0.4 m is?
  • A) 7.5 × 10⁴ V
  • B) 8.5 × 10⁴ V
  • C) 9.0 × 10⁴ V
  • D) 6.0 × 10⁴ V

Answer Key

• Q1: A) 6.75 × 10⁴ V - V₀ = kq/R = (9 × 10⁹ × 3 × 10^-6)/0.4 = 6.75 × 10⁴ V
• Q2: B) 3.22 × 10³ V - V₂ - V₁ = λ/(2 π ε₀) ln 0.2/0.5 = 3.22 × 10³ V
• Q3: C) 9.0 × 10⁴ V - V = kq/√(R² + x²) = (9 × 10⁹ × 5 × 10^-6)/√(0.4² + 0.3²) = 9.0 × 10⁴ V

Memory Aids

• Potential due to charge element: dV = kdq/r
• Ring potential: V = kq/√(R² + x²)
• Wire potential difference: logarithmic function of distances.

General Concept

• Electric potential due to small charge element \( dq \) at point P: \( dV = \frac{k dq}{r} \), where \( k = \frac{1}{4 \pi \epsilon_0} \)
• Total potential: \( V = \int dV = \frac{1}{4 \pi \epsilon_0} \int \frac{dq}{r} \)
• Used for irregular or continuous charge distributions by integration.

Uniformly Charged Ring

• Potential at point P on axis at distance \( x \): \( V = \frac{k q}{\sqrt{R^2 + x^2}} \)
• At center (\( x=0 \)): \( V_0 = \frac{k q}{R} \)

Uniformly Charged Disc

• Disc divided into thin rings of radius \( r \) and thickness \( dr \)
• Charge on ring: \( dq = \sigma (2 \pi r dr) \), \( \sigma \) = surface charge density
• Potential at point on axis: \( V = \int_0^R \frac{k dq}{\sqrt{r^2 + x^2}} \)
• After integration, \( V = 2 \pi k \sigma \left( \sqrt{R^2 + x^2} - x \right) \)
• At center (\( x=0 \)): \( V = 2 \pi k \sigma R \)

Electric Potential Difference due to Uniformly Charged Wire

• Electric field at distance \( x \) from wire: \( E = \frac{\lambda}{2 \pi \epsilon_0 x} \)
• Relation: \( E = - \frac{dV}{dx} \)
• Potential difference between points \( x_1 \) and \( x_2 \): \[ V_2 - V_1 = - \int_{x_1}^{x_2} E dx = \frac{\lambda}{2 \pi \epsilon_0} \ln \frac{x_1}{x_2} \]

Solved Examples

• Ex 1: Potential at point on axis of ring with \( q = 4 \mu C \), \( R = 0.5 \ m \), \( x = 0.5 \ m \)
  • \( V = \frac{k q}{\sqrt{R^2 + x^2}} = \frac{9 \times 10^9 \times 4 \times 10^{-6}}{\sqrt{0.5^2 + 0.5^2}} = 5.09 \times 10^4 \ V \)
• Ex 2: Potential difference between \( x_1 = 0.1 \ m \) and \( x_2 = 0.5 \ m \) from charged wire with \( \lambda = 2 \times 10^{-6} C/m \)
  • \( V_2 - V_1 = \frac{\lambda}{2 \pi \epsilon_0} \ln \frac{x_1}{x_2} = \frac{2 \times 10^{-6}}{2 \pi \times 8.85 \times 10^{-12}} \ln \frac{0.1}{0.5} = -1.02 \times 10^4 \ V \)

Practice Set

• Q1: Potential at center of ring with charge \( 3 \mu C \) and radius \( 0.4 \ m \) is?
  • A) 6.75 × 10^4 V
  • B) 1.35 × 10^5 V
  • C) 9.0 × 10^4 V
  • D) 4.5 × 10^4 V
• Q2: Potential difference between \( 0.2 \ m \) and \( 0.5 \ m \) from charged wire with \( \lambda = 1 \times 10^{-6} C/m \) is?
  • A) 4.56 × 10^3 V
  • B) 3.22 × 10^3 V
  • C) 2.78 × 10^3 V
  • D) 1.89 × 10^3 V
• Q3: Potential at point \( x = 0.3 \ m \) on axis of ring with \( q = 5 \mu C \), \( R = 0.4 \ m \) is?
  • A) 7.5 × 10^4 V
  • B) 8.5 × 10^4 V
  • C) 9.0 × 10^4 V
  • D) 6.0 × 10^4 V

Answer Key

• Q1: A) 6.75 × 10^4 V - \( V_0 = \frac{k q}{R} = \frac{9 \times 10^9 \times 3 \times 10^{-6}}{0.4} = 6.75 \times 10^4 V \)
• Q2: B) 3.22 × 10^3 V - \( V_2 - V_1 = \frac{\lambda}{2 \pi \epsilon_0} \ln \frac{0.2}{0.5} = 3.22 \times 10^3 V \)
• Q3: C) 9.0 × 10^4 V - \( V = \frac{k q}{\sqrt{R^2 + x^2}} = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{\sqrt{0.4^2 + 0.3^2}} = 9.0 \times 10^4 V \)

Memory Aids

• Potential due to charge element: \( dV = \frac{k dq}{r} \)
• Ring potential: \( V = \frac{k q}{\sqrt{R^2 + x^2}} \)
• Wire potential difference: logarithmic function of distances.

Electric Potential due to a Uniformly Charged Spherical Shell

Outside the Shell (r > R)

• Potential at distance r: V = kq/r = 1/(4 π ε₀) q/r
• Electric field: E = - dV/dr = kq/r²

On the Surface (r = R)

• Potential: V = kq/R = 1/(4 π ε₀) q/R
• Potential is constant inside the shell.

Inside the Shell (r < R)

• Electric field E = 0 inside shell
• Potential V = constant = kq/R

Solved Examples

• Ex 1: Potential at r = 2R outside shell with charge q = 5 μC, radius R = 0.1 m
  • V = kq/r = (9 × 10⁹ × 5 × 10^-6)/0.2 = 2.25 × 10⁵ V
• Ex 2: Potential inside shell (r < R) is constant: V = kq/R

Practice Set

• Q1: Potential at surface of shell with q = 3 μC, R = 0.2 m is?
  • A) 1.35 × 10⁵ V
  • B) 6.75 × 10⁴ V
  • C) 9.0 × 10⁴ V
  • D) 4.5 × 10⁴ V
• Q2: Electric field inside shell is?
  • A) Zero
  • B) Constant non-zero
  • C) Varies with r
  • D) Infinite
• Q3: Potential at r = 3R outside shell with charge q = 2 μC, R = 0.1 m is?
  • A) 6 × 10⁴ V
  • B) 3 × 10⁴ V
  • C) 9 × 10⁴ V
  • D) 1.5 × 10⁵ V

Answer Key

• Q1: A) 1.35 × 10⁵ V - V = kq/R = (9 × 10⁹ × 3 × 10^-6)/0.2 = 1.35 × 10⁵ V
• Q2: A) Zero - Electric field inside a charged shell is zero.
• Q3: B) 3 × 10⁴ V - V = kq/r = (9 × 10⁹ × 2 × 10^-6)/0.3 = 3 × 10⁴ V

Memory Aids

• Outside shell: V ∝ 1/r
• Inside shell: V = constant, E = 0
• Potential continuous at surface r = R

Outside the Shell (r > R)

• Potential at distance r: V = kq/r = 1/(4 π ε₀) q/r
• Electric field: E = - dV/dr = kq/r²

On the Surface (r = R)

• Potential: V = kq/R = 1/(4 π ε₀) q/R
• Potential is constant inside the shell.

Inside the Shell (r < R)

• Electric field E = 0 inside shell
• Potential V = constant = kq/R

Solved Examples

• Ex 1: Potential at r = 2R outside shell with charge q = 5 μC, radius R = 0.1 m
  • V = kq/r = (9 × 10⁹ × 5 × 10^-6)/0.2 = 2.25 × 10⁵ V
• Ex 2: Potential inside shell (r < R) is constant: V = kq/R

Practice Set

• Q1: Potential at surface of shell with q = 3 μC, R = 0.2 m is?
  • A) 1.35 × 10⁵ V
  • B) 6.75 × 10⁴ V
  • C) 9.0 × 10⁴ V
  • D) 4.5 × 10⁴ V
• Q2: Electric field inside shell is?
  • A) Zero
  • B) Constant non-zero
  • C) Varies with r
  • D) Infinite
• Q3: Potential at r = 3R outside shell with charge q = 2 μC, R = 0.1 m is?
  • A) 6 × 10⁴ V
  • B) 3 × 10⁴ V
  • C) 9 × 10⁴ V
  • D) 1.5 × 10⁵ V

Answer Key

• Q1: A) 1.35 × 10⁵ V - V = kq/R = (9 × 10⁹ × 3 × 10^-6)/0.2 = 1.35 × 10⁵ V
• Q2: A) Zero - Electric field inside a charged shell is zero.
• Q3: B) 3 × 10⁴ V - V = kq/r = (9 × 10⁹ × 2 × 10^-6)/0.3 = 3 × 10⁴ V

Memory Aids

• Outside shell: V ∝ 1/r
• Inside shell: V = constant, E = 0
• Potential continuous at surface r = R

Outside the Shell (r > R)

• Potential at distance r: V = kq/r = 1/(4 π ε₀) q/r
• Electric field: E = - dV/dr = kq/r²

On the Surface (r = R)

• Potential: V = kq/R = 1/(4 π ε₀) q/R
• Potential is constant inside the shell.

Inside the Shell (r < R)

• Electric field E = 0 inside shell
• Potential V = constant = kq/R

Solved Examples

• Ex 1: Potential at r = 2R outside shell with charge q = 5 μC, radius R = 0.1 m
  • V = kq/r = (9 × 10⁹ × 5 × 10^-6)/0.2 = 2.25 × 10⁵ V
• Ex 2: Potential inside shell (r < R) is constant: V = kq/R

Practice Set

• Q1: Potential at surface of shell with q = 3 μC, R = 0.2 m is?
  • A) 1.35 × 10⁵ V
  • B) 6.75 × 10⁴ V
  • C) 9.0 × 10⁴ V
  • D) 4.5 × 10⁴ V
• Q2: Electric field inside shell is?
  • A) Zero
  • B) Constant non-zero
  • C) Varies with r
  • D) Infinite
• Q3: Potential at r = 3R outside shell with charge q = 2 μC, R = 0.1 m is?
  • A) 6 × 10⁴ V
  • B) 3 × 10⁴ V
  • C) 9 × 10⁴ V
  • D) 1.5 × 10⁵ V

Answer Key

• Q1: A) 1.35 × 10⁵ V - V = kq/R = (9 × 10⁹ × 3 × 10^-6)/0.2 = 1.35 × 10⁵ V
• Q2: A) Zero - Electric field inside a charged shell is zero.
• Q3: B) 3 × 10⁴ V - V = kq/r = (9 × 10⁹ × 2 × 10^-6)/0.3 = 3 × 10⁴ V

Memory Aids

• Outside shell: V ∝ 1/r
• Inside shell: V = constant, E = 0
• Potential continuous at surface r = R

Outside the Shell (\( r > R \))

• Potential at distance \( r \): \( V = \frac{k q}{r} = \frac{1}{4 \pi \epsilon_0} \frac{q}{r} \)
• Electric field: \( E = - \frac{dV}{dr} = \frac{k q}{r^2} \)

On the Surface (\( r = R \))

• Potential: \( V = \frac{k q}{R} = \frac{1}{4 \pi \epsilon_0} \frac{q}{R} \)
• Potential is constant inside the shell.

Inside the Shell (\( r < R \))

• Electric field \( E = 0 \) inside shell
• Potential \( V = \) constant = \( \frac{k q}{R} \)

Solved Examples

• Ex 1: Potential at \( r = 2R \) outside shell with charge \( q = 5 \mu C \), radius \( R = 0.1 \ m \)
  • \( V = \frac{k q}{r} = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{0.2} = 2.25 \times 10^5 \ V \)
• Ex 2: Potential inside shell (\( r < R \)) is constant: \( V = \frac{k q}{R} \)

Practice Set

• Q1: Potential at surface of shell with \( q = 3 \mu C \), \( R = 0.2 \ m \) is?
  • A) 1.35 × 10^5 V
  • B) 6.75 × 10^4 V
  • C) 9.0 × 10^4 V
  • D) 4.5 × 10^4 V
• Q2: Electric field inside shell is?
  • A) Zero
  • B) Constant non-zero
  • C) Varies with r
  • D) Infinite
• Q3: Potential at \( r = 3R \) outside shell with charge \( q = 2 \mu C \), \( R = 0.1 \ m \) is?
  • A) 6 × 10^4 V
  • B) 3 × 10^4 V
  • C) 9 × 10^4 V
  • D) 1.5 × 10^5 V

Answer Key

• Q1: A) 1.35 × 10^5 V - \( V = \frac{k q}{R} = \frac{9 \times 10^9 \times 3 \times 10^{-6}}{0.2} = 1.35 \times 10^5 V \)
• Q2: A) Zero - Electric field inside a charged shell is zero.
• Q3: B) 3 × 10^4 V - \( V = \frac{k q}{r} = \frac{9 \times 10^9 \times 2 \times 10^{-6}}{0.3} = 3 \times 10^4 V \)

Memory Aids

• Outside shell: \( V \propto \frac{1}{r} \)
• Inside shell: \( V = \) constant, \( E = 0 \)
• Potential continuous at surface \( r = R \)

Electric Potential due to a Uniformly Charged Dielectric Solid Sphere

Charge Distribution and Density

• Total charge: q, radius: R
• Charge density: ρ = q/(4/3 πR³)

Potential Outside Sphere (r > R)

• V_B = kq/r

Potential On Surface (r = R)

• V_C = kq/R

Potential Inside Sphere (r < R)

• V_D = kq/(2R) (3 - r²/R²)
• Potential at center V₀ = 3kq/(2R)

Solved Examples

• Ex 1: Calculate potential at center of sphere with q = 6 μC, R = 0.1 m
  • V₀ = 3kq/(2R) = (3 × 9 × 10⁹ × 6 × 10^-6)/(2 × 0.1) = 8.1 × 10⁵ V
• Ex 2: Potential at r = 0.05 m inside sphere
  • V_D = kq/(2R) (3 - r²/R²) = (9 × 10⁹ × 6 × 10^-6)/(2 × 0.1) (3 - 0.05²/0.1²) = 7.29 × 10⁵ V

Practice Set

• Q1: Charge density of sphere with q = 4 μC, R = 0.2 m is?
  • A) 1.43 × 10^-4 C/m³
  • B) 2.39 × 10^-4 C/m³
  • C) 3.82 × 10^-4 C/m³
  • D) 4.77 × 10^-4 C/m³
• Q2: Potential at center of sphere with q = 3 μC, R = 0.1 m is?
  • A) 4.05 × 10⁵ V
  • B) 6.75 × 10⁵ V
  • C) 9.0 × 10⁵ V
  • D) 1.35 × 10⁵ V
• Q3: Potential at r = 0.05 m inside sphere with q = 5 μC, R = 0.1 m is?
  • A) 3.75 × 10⁵ V
  • B) 4.5 × 10⁵ V
  • C) 5.06 × 10⁵ V
  • D) 6.0 × 10⁵ V

Answer Key

• Q1: A) 1.43 × 10^-4 C/m³ - ρ = q/(4/3 πR³) = (4 × 10^-6)/(4/3 π(0.2)³) = 1.43 × 10^-4 C/m³
• Q2: B) 6.75 × 10⁵ V - V₀ = 3kq/(2R) = (3 × 9 × 10⁹ × 3 × 10^-6)/(2 × 0.1) = 6.75 × 10⁵ V
• Q3: C) 5.06 × 10⁵ V - V_D = kq/(2R) (3 - r²/R²) = (9 × 10⁹ × 5 × 10^-6)/(2 × 0.1) (3 - 0.05²/0.1²) = 5.06 × 10⁵ V

Memory Aids

• Charge density: ρ = q/(4/3 πR³)
• Potential inside sphere varies as V_D = kq/(2R) (3 - r²/R²)
• Potential at center is maximum.

Charge Distribution and Density

• Total charge: q, radius: R
• Charge density: ρ = q/(4/3 πR³)

Potential Outside Sphere (r > R)

• V_B = kq/r

Potential On Surface (r = R)

• V_C = kq/R

Potential Inside Sphere (r < R)

• V_D = kq/(2R) (3 - r²/R²)
• Potential at center V₀ = 3kq/(2R)

Solved Examples

• Ex 1: Calculate potential at center of sphere with q = 6 μC, R = 0.1 m
  • V₀ = 3kq/(2R) = (3 × 9 × 10⁹ × 6 × 10^-6)/(2 × 0.1) = 8.1 × 10⁵ V
• Ex 2: Potential at r = 0.05 m inside sphere
  • V_D = kq/(2R) (3 - r²/R²) = (9 × 10⁹ × 6 × 10^-6)/(2 × 0.1) (3 - 0.05²/0.1²) = 7.29 × 10⁵ V

Practice Set

• Q1: Charge density of sphere with q = 4 μC, R = 0.2 m is?
  • A) 1.43 × 10^-4 C/m³
  • B) 2.39 × 10^-4 C/m³
  • C) 3.82 × 10^-4 C/m³
  • D) 4.77 × 10^-4 C/m³
• Q2: Potential at center of sphere with q = 3 μC, R = 0.1 m is?
  • A) 4.05 × 10⁵ V
  • B) 6.75 × 10⁵ V
  • C) 9.0 × 10⁵ V
  • D) 1.35 × 10⁵ V
• Q3: Potential at r = 0.05 m inside sphere with q = 5 μC, R = 0.1 m is?
  • A) 3.75 × 10⁵ V
  • B) 4.5 × 10⁵ V
  • C) 5.06 × 10⁵ V
  • D) 6.0 × 10⁵ V

Answer Key

• Q1: A) 1.43 × 10^-4 C/m³ - ρ = q/(4/3 πR³) = (4 × 10^-6)/(4/3 π(0.2)³) = 1.43 × 10^-4 C/m³
• Q2: B) 6.75 × 10⁵ V - V₀ = 3kq/(2R) = (3 × 9 × 10⁹ × 3 × 10^-6)/(2 × 0.1) = 6.75 × 10⁵ V
• Q3: C) 5.06 × 10⁵ V - V_D = kq/(2R) (3 - r²/R²) = (9 × 10⁹ × 5 × 10^-6)/(2 × 0.1) (3 - 0.05²/0.1²) = 5.06 × 10⁵ V

Memory Aids

• Charge density: ρ = q/(4/3 πR³)
• Potential inside sphere varies as V_D = kq/(2R) (3 - r²/R²)
• Potential at center is maximum.

Charge Distribution and Density

• Total charge: q, radius: R
• Charge density: ρ = q/(4/3 πR³)

Potential Outside Sphere (r > R)

• V_B = kq/r

Potential On Surface (r = R)

• V_C = kq/R

Potential Inside Sphere (r < R)

• V_D = kq/(2R) (3 - r²/R²)
• Potential at center V₀ = 3kq/(2R)

Solved Examples

• Ex 1: Calculate potential at center of sphere with q = 6 μC, R = 0.1 m
  • V₀ = 3kq/(2R) = (3 × 9 × 10⁹ × 6 × 10^-6)/(2 × 0.1) = 8.1 × 10⁵ V
• Ex 2: Potential at r = 0.05 m inside sphere
  • V_D = kq/(2R) (3 - r²/R²) = (9 × 10⁹ × 6 × 10^-6)/(2 × 0.1) (3 - 0.05²/0.1²) = 7.29 × 10⁵ V

Practice Set

• Q1: Charge density of sphere with q = 4 μC, R = 0.2 m is?
  • A) 1.43 × 10^-4 C/m³
  • B) 2.39 × 10^-4 C/m³
  • C) 3.82 × 10^-4 C/m³
  • D) 4.77 × 10^-4 C/m³
• Q2: Potential at center of sphere with q = 3 μC, R = 0.1 m is?
  • A) 4.05 × 10⁵ V
  • B) 6.75 × 10⁵ V
  • C) 9.0 × 10⁵ V
  • D) 1.35 × 10⁵ V
• Q3: Potential at r = 0.05 m inside sphere with q = 5 μC, R = 0.1 m is?
  • A) 3.75 × 10⁵ V
  • B) 4.5 × 10⁵ V
  • C) 5.06 × 10⁵ V
  • D) 6.0 × 10⁵ V

Answer Key

• Q1: A) 1.43 × 10^-4 C/m³ - ρ = q/(4/3 πR³) = (4 × 10^-6)/(4/3 π(0.2)³) = 1.43 × 10^-4 C/m³
• Q2: B) 6.75 × 10⁵ V - V₀ = 3kq/(2R) = (3 × 9 × 10⁹ × 3 × 10^-6)/(2 × 0.1) = 6.75 × 10⁵ V
• Q3: C) 5.06 × 10⁵ V - V_D = kq/(2R) (3 - r²/R²) = (9 × 10⁹ × 5 × 10^-6)/(2 × 0.1) (3 - 0.05²/0.1²) = 5.06 × 10⁵ V

Memory Aids

• Charge density: ρ = q/(4/3 πR³)
• Potential inside sphere varies as V_D = kq/(2R) (3 - r²/R²)
• Potential at center is maximum.

Charge Distribution and Density

• Total charge: \( q \), radius: \( R \)
• Charge density: \( \rho = \frac{q}{\frac{4}{3} \pi R^3} \)

Potential Outside Sphere (\( r > R \))

• \( V_B = \frac{k q}{r} \)

Potential On Surface (\( r = R \))

• \( V_C = \frac{k q}{R} \)

Potential Inside Sphere (\( r < R \))

• \( V_D = \frac{k q}{2 R} \left( 3 - \frac{r^2}{R^2} \right) \)
• Potential at center \( V_0 = \frac{3 k q}{2 R} \)

Solved Examples

• Ex 1: Calculate potential at center of sphere with \( q = 6 \mu C \), \( R = 0.1 \ m \)
  • \( V_0 = \frac{3 k q}{2 R} = \frac{3 \times 9 \times 10^9 \times 6 \times 10^{-6}}{2 \times 0.1} = 8.1 \times 10^5 \ V \)
• Ex 2: Potential at \( r = 0.05 \ m \) inside sphere
  • \( V_D = \frac{k q}{2 R} \left( 3 - \frac{r^2}{R^2} \right) = \frac{9 \times 10^9 \times 6 \times 10^{-6}}{2 \times 0.1} \left( 3 - \frac{0.05^2}{0.1^2} \right) = 7.29 \times 10^5 \ V \)

Practice Set

• Q1: Charge density of sphere with \( q = 4 \mu C \), \( R = 0.2 \ m \) is?
  • A) 1.43 × 10^{-4} C/m³
  • B) 2.39 × 10^{-4} C/m³
  • C) 3.82 × 10^{-4} C/m³
  • D) 4.77 × 10^{-4} C/m³
• Q2: Potential at center of sphere with \( q = 3 \mu C \), \( R = 0.1 \ m \) is?
  • A) 4.05 × 10^5 V
  • B) 6.75 × 10^5 V
  • C) 9.0 × 10^5 V
  • D) 1.35 × 10^5 V
• Q3: Potential at \( r = 0.05 \ m \) inside sphere with \( q = 5 \mu C \), \( R = 0.1 \ m \) is?
  • A) 3.75 × 10^5 V
  • B) 4.5 × 10^5 V
  • C) 5.06 × 10^5 V
  • D) 6.0 × 10^5 V

Answer Key

• Q1: A) 1.43 × 10^{-4} C/m³ - \( \rho = \frac{q}{\frac{4}{3} \pi R^3} = \frac{4 \times 10^{-6}}{\frac{4}{3} \pi (0.2)^3} = 1.43 \times 10^{-4} C/m^3 \)
• Q2: B) 6.75 × 10^5 V - \( V_0 = \frac{3 k q}{2 R} = \frac{3 \times 9 \times 10^9 \times 3 \times 10^{-6}}{2 \times 0.1} = 6.75 \times 10^5 V \)
• Q3: C) 5.06 × 10^5 V - \( V_D = \frac{k q}{2 R} \left( 3 - \frac{r^2}{R^2} \right) = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{2 \times 0.1} \left( 3 - \frac{0.05^2}{0.1^2} \right) = 5.06 \times 10^5 V \)

Memory Aids

• Charge density: \( \rho = \frac{q}{\frac{4}{3} \pi R^3} \)
• Potential inside sphere varies as \( V_D = \frac{k q}{2 R} \left( 3 - \frac{r^2}{R^2} \right) \)
• Potential at center is maximum.

Potential Energy of a Single Charge

Concepts

• Potential energy (U) of charge q at point P in external field with potential V: U = qV
• Work done by external agent to bring charge from infinity (where potential = 0) to P: W = q(V - 0) = qV
• Potential at infinity is zero by definition

Solved Examples

• Ex 1: Calculate potential energy of charge q = 2 μC at point with potential V = 5 V.
  • Use U = qV = 2 × 10^-6 × 5 = 10 × 10^-6 J = 10 μJ

Practice Set

• Q1: Potential energy of a charge 3 μC at a point with potential 10 V is?
  • A) 30 μJ
  • B) 3 J
  • C) 0
  • D) 300 μJ
• Q2: Work done to bring 1 C charge from infinity to a point at 12 V is?
  • A) 12 J
  • B) 1 J
  • C) 0
  • D) 24 J
• Q3: If potential at point P is zero, potential energy of any charge at P is?
  • A) Zero
  • B) Infinite
  • C) Depends on charge
  • D) Negative

Answer Key

• Q1: A) 30 μJ - U = qV = 3 × 10^-6 × 10 = 30 × 10^-6 J
• Q2: A) 12 J - Work done W = qV = 1 × 12 = 12 J
• Q3: A) Zero - Potential energy U = qV = q × 0 = 0

Memory Aids

• Potential energy U = qV
• Potential at infinity = 0
• Work done = potential energy

Concepts

• Potential energy (U) of charge q at point P in external field with potential V: U = qV
• Work done by external agent to bring charge from infinity (where potential = 0) to P: W = q(V - 0) = qV
• Potential at infinity is zero by definition

Solved Examples

• Ex 1: Calculate potential energy of charge q = 2 μC at point with potential V = 5 V.
  • Use U = qV = 2 × 10^-6 × 5 = 10 × 10^-6 J = 10 μJ

Practice Set

• Q1: Potential energy of a charge 3 μC at a point with potential 10 V is?
  • A) 30 μJ
  • B) 3 J
  • C) 0
  • D) 300 μJ
• Q2: Work done to bring 1 C charge from infinity to a point at 12 V is?
  • A) 12 J
  • B) 1 J
  • C) 0
  • D) 24 J
• Q3: If potential at point P is zero, potential energy of any charge at P is?
  • A) Zero
  • B) Infinite
  • C) Depends on charge
  • D) Negative

Answer Key

• Q1: A) 30 μJ - U = qV = 3 × 10^-6 × 10 = 30 × 10^-6 J
• Q2: A) 12 J - Work done W = qV = 1 × 12 = 12 J
• Q3: A) Zero - Potential energy U = qV = q × 0 = 0

Memory Aids

• Potential energy U = qV
• Potential at infinity = 0
• Work done = potential energy

Concepts

• Potential energy (U) of charge q at point P in external field with potential V: U = qV
• Work done by external agent to bring charge from infinity (where potential = 0) to P: W = q(V - 0) = qV
• Potential at infinity is zero by definition

Solved Examples

• Ex 1: Calculate potential energy of charge q = 2 μC at point with potential V = 5 V.
  • Use U = qV = 2 × 10^-6 × 5 = 10 × 10^-6 J = 10 μJ

Practice Set

• Q1: Potential energy of a charge 3 μC at a point with potential 10 V is?
  • A) 30 μJ
  • B) 3 J
  • C) 0
  • D) 300 μJ
• Q2: Work done to bring 1 C charge from infinity to a point at 12 V is?
  • A) 12 J
  • B) 1 J
  • C) 0
  • D) 24 J
• Q3: If potential at point P is zero, potential energy of any charge at P is?
  • A) Zero
  • B) Infinite
  • C) Depends on charge
  • D) Negative

Answer Key

• Q1: A) 30 μJ - U = qV = 3 × 10^-6 × 10 = 30 × 10^-6 J
• Q2: A) 12 J - Work done W = qV = 1 × 12 = 12 J
• Q3: A) Zero - Potential energy U = qV = q × 0 = 0

Memory Aids

• Potential energy U = qV
• Potential at infinity = 0
• Work done = potential energy

Concepts

• Potential energy (U) of charge \( q \) at point \( P \) in external field with potential \( V \): \( U = qV \)
• Work done by external agent to bring charge from infinity (where potential = 0) to \( P \): \( W = q(V - 0) = qV \)
• Potential at infinity is zero by definition

Solved Examples

• Ex 1: Calculate potential energy of charge \( q = 2 \mu C \) at point with potential \( V = 5 \text{ V} \).
  • Use \( U = qV = 2 \times 10^{-6} \times 5 = 10 \times 10^{-6} \text{ J} = 10 \mu \text{J} \)

Practice Set

• Q1: Potential energy of a charge \( 3 \mu C \) at a point with potential \( 10 \text{ V} \) is?
  • A) \( 30 \mu \text{J} \)
  • B) \( 3 \text{ J} \)
  • C) \( 0 \)
  • D) \( 300 \mu \text{J} \)
• Q2: Work done to bring \( 1 \text{ C} \) charge from infinity to a point at \( 12 \text{ V} \) is?
  • A) \( 12 \text{ J} \)
  • B) \( 1 \text{ J} \)
  • C) \( 0 \)
  • D) \( 24 \text{ J} \)
• Q3: If potential at point \( P \) is zero, potential energy of any charge at \( P \) is?
  • A) Zero
  • B) Infinite
  • C) Depends on charge
  • D) Negative

Answer Key

• Q1: A) \( 30 \mu \text{J} \) - \( U = qV = 3 \times 10^{-6} \times 10 = 30 \times 10^{-6} \text{ J} \)
• Q2: A) \( 12 \text{ J} \) - Work done \( W = qV = 1 \times 12 = 12 \text{ J} \)
• Q3: A) Zero - Potential energy \( U = qV = q \times 0 = 0 \)

Memory Aids

• Potential energy \( U = qV \)
• Potential at infinity = 0
• Work done = potential energy

Potential Energy of Two Charges

Concepts

• Work done to bring q₁ from infinity to point A with potential V_A: W₁ = q₁V_A
• Potential at B due to q₁ at A: V_B = 1/(4 π ε₀) q₁/x, where x = AB
• Work done to bring q₂ from infinity to B: W₂ = q₂(V_B + 1/(4 π ε₀) q₁/x) = q₂V_B + 1/(4 π ε₀) q₁q₂/x
• Total work done (potential energy) in assembling two charges in external field: [ W = W₁ + W₂ = q₁V_A + q₂V_B + 1/(4 π ε₀) q₁q₂/x ]

Solved Examples

• Ex 1: Calculate total potential energy for q₁ = 2 μC, q₂ = 3 μC, V_A = 5 V, V_B = 4 V, and x = 0.1 m.
  • Use W = q₁V_A + q₂V_B + 1/(4 π ε₀) q₁q₂/x
  • Calculate 1/(4 π ε₀) = 9 × 10⁹ Nm²/C²
  • Calculate interaction term: 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/0.1 = 0.54 J
  • Total W = 2 × 10^-6 × 5 + 3 × 10^-6 × 4 + 0.54 = 0.00001 + 0.000012 + 0.54 ≈ 0.54 J

Practice Set

• Q1: Work done to bring q₁ = 1 μC to point A with V_A = 3 V is?
  • A) 3 μJ
  • B) 3 J
  • C) 0
  • D) 30 μJ
• Q2: Potential at B due to q₁ = 2 μC at distance x = 0.2 m is?
  • A) 4.5 × 10⁴ V
  • B) 9 × 10⁴ V
  • C) 1.8 × 10⁵ V
  • D) 9 × 10³ V
• Q3: Total work done to assemble q₁ = 1 μC, q₂ = 1 μC, V_A = 2 V, V_B = 3 V, x = 0.1 m is?
  • A) 0.09 J
  • B) 0.1 J
  • C) 0.12 J
  • D) 0.15 J

Answer Key

• Q1: A) 3 μJ - W₁ = q₁V_A = 1 × 10^-6 × 3 = 3 × 10^-6 J
• Q2: D) 9 × 10³ V - V_B = 1/(4 π ε₀) q₁/x = 9 × 10⁹ × (2 × 10^-6/0.2) = 9 × 10³ V
• Q3: C) 0.12 J - Calculate total work using formula, interaction term dominates

Memory Aids

• Total work = sum of individual works + interaction energy
• Potential due to point charge: V = 1/(4 π ε₀) q/r
• Use 9 × 10⁹ Nm²/C² for 1/(4 π ε₀)

Concepts

• Work done to bring q₁ from infinity to point A with potential V_A: W₁ = q₁V_A
• Potential at B due to q₁ at A: V_B = 1/(4 π ε₀) q₁/x, where x = AB
• Work done to bring q₂ from infinity to B: W₂ = q₂(V_B + 1/(4 π ε₀) q₁/x) = q₂V_B + 1/(4 π ε₀) q₁q₂/x
• Total work done (potential energy) in assembling two charges in external field: [ W = W₁ + W₂ = q₁V_A + q₂V_B + 1/(4 π ε₀) q₁q₂/x ]

Solved Examples

• Ex 1: Calculate total potential energy for q₁ = 2 μC, q₂ = 3 μC, V_A = 5 V, V_B = 4 V, and x = 0.1 m.
  • Use W = q₁V_A + q₂V_B + 1/(4 π ε₀) q₁q₂/x
  • Calculate 1/(4 π ε₀) = 9 × 10⁹ Nm²/C²
  • Calculate interaction term: 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/0.1 = 0.54 J
  • Total W = 2 × 10^-6 × 5 + 3 × 10^-6 × 4 + 0.54 = 0.00001 + 0.000012 + 0.54 ≈ 0.54 J

Practice Set

• Q1: Work done to bring q₁ = 1 μC to point A with V_A = 3 V is?
  • A) 3 μJ
  • B) 3 J
  • C) 0
  • D) 30 μJ
• Q2: Potential at B due to q₁ = 2 μC at distance x = 0.2 m is?
  • A) 4.5 × 10⁴ V
  • B) 9 × 10⁴ V
  • C) 1.8 × 10⁵ V
  • D) 9 × 10³ V
• Q3: Total work done to assemble q₁ = 1 μC, q₂ = 1 μC, V_A = 2 V, V_B = 3 V, x = 0.1 m is?
  • A) 0.09 J
  • B) 0.1 J
  • C) 0.12 J
  • D) 0.15 J

Answer Key

• Q1: A) 3 μJ - W₁ = q₁V_A = 1 × 10^-6 × 3 = 3 × 10^-6 J
• Q2: D) 9 × 10³ V - V_B = 1/(4 π ε₀) q₁/x = 9 × 10⁹ × (2 × 10^-6/0.2) = 9 × 10³ V
• Q3: C) 0.12 J - Calculate total work using formula, interaction term dominates

Memory Aids

• Total work = sum of individual works + interaction energy
• Potential due to point charge: V = 1/(4 π ε₀) q/r
• Use 9 × 10⁹ Nm²/C² for 1/(4 π ε₀)

Concepts

• Work done to bring q₁ from infinity to point A with potential V_A: W₁ = q₁V_A
• Potential at B due to q₁ at A: V_B = 1/(4 π ε₀) q₁/x, where x = AB
• Work done to bring q₂ from infinity to B: W₂ = q₂(V_B + 1/(4 π ε₀) q₁/x) = q₂V_B + 1/(4 π ε₀) q₁q₂/x
• Total work done (potential energy) in assembling two charges in external field: [ W = W₁ + W₂ = q₁V_A + q₂V_B + 1/(4 π ε₀) q₁q₂/x ]

Solved Examples

• Ex 1: Calculate total potential energy for q₁ = 2 μC, q₂ = 3 μC, V_A = 5 V, V_B = 4 V, and x = 0.1 m.
  • Use W = q₁V_A + q₂V_B + 1/(4 π ε₀) q₁q₂/x
  • Calculate 1/(4 π ε₀) = 9 × 10⁹ Nm²/C²
  • Calculate interaction term: 9 × 10⁹ × (2 × 10^-6 × 3 × 10^-6)/0.1 = 0.54 J
  • Total W = 2 × 10^-6 × 5 + 3 × 10^-6 × 4 + 0.54 = 0.00001 + 0.000012 + 0.54 ≈ 0.54 J

Practice Set

• Q1: Work done to bring q₁ = 1 μC to point A with V_A = 3 V is?
  • A) 3 μJ
  • B) 3 J
  • C) 0
  • D) 30 μJ
• Q2: Potential at B due to q₁ = 2 μC at distance x = 0.2 m is?
  • A) 4.5 × 10⁴ V
  • B) 9 × 10⁴ V
  • C) 1.8 × 10⁵ V
  • D) 9 × 10³ V
• Q3: Total work done to assemble q₁ = 1 μC, q₂ = 1 μC, V_A = 2 V, V_B = 3 V, x = 0.1 m is?
  • A) 0.09 J
  • B) 0.1 J
  • C) 0.12 J
  • D) 0.15 J

Answer Key

• Q1: A) 3 μJ - W₁ = q₁V_A = 1 × 10^-6 × 3 = 3 × 10^-6 J
• Q2: D) 9 × 10³ V - V_B = 1/(4 π ε₀) q₁/x = 9 × 10⁹ × (2 × 10^-6/0.2) = 9 × 10³ V
• Q3: C) 0.12 J - Calculate total work using formula, interaction term dominates

Memory Aids

• Total work = sum of individual works + interaction energy
• Potential due to point charge: V = 1/(4 π ε₀) q/r
• Use 9 × 10⁹ Nm²/C² for 1/(4 π ε₀)

Concepts

• Work done to bring \( q_1 \) from infinity to point A with potential \( V_A \): \( W_1 = q_1 V_A \)
• Potential at B due to \( q_1 \) at A: \( V_B = \frac{1}{4 \pi \epsilon_0} \frac{q_1}{x} \), where \( x = AB \)
• Work done to bring \( q_2 \) from infinity to B: \( W_2 = q_2 \left(V_B + \frac{1}{4 \pi \epsilon_0} \frac{q_1}{x}\right) = q_2 V_B + \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{x} \)
• Total work done (potential energy) in assembling two charges in external field: \[ W = W_1 + W_2 = q_1 V_A + q_2 V_B + \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{x} \]

Solved Examples

• Ex 1: Calculate total potential energy for \( q_1 = 2 \mu C \), \( q_2 = 3 \mu C \), \( V_A = 5 \text{ V} \), \( V_B = 4 \text{ V} \), and \( x = 0.1 \text{ m} \).
  • Use \( W = q_1 V_A + q_2 V_B + \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{x} \)
  • Calculate \( \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2 \)
  • Calculate interaction term: \( 9 \times 10^9 \times \frac{2 \times 10^{-6} \times 3 \times 10^{-6}}{0.1} = 0.54 \text{ J} \)
  • Total \( W = 2 \times 10^{-6} \times 5 + 3 \times 10^{-6} \times 4 + 0.54 = 0.00001 + 0.000012 + 0.54 \approx 0.54 \text{ J} \)

Practice Set

• Q1: Work done to bring \( q_1 = 1 \mu C \) to point A with \( V_A = 3 \text{ V} \) is?
  • A) \( 3 \mu \text{J} \)
  • B) \( 3 \text{ J} \)
  • C) \( 0 \)
  • D) \( 30 \mu \text{J} \)
• Q2: Potential at B due to \( q_1 = 2 \mu C \) at distance \( x = 0.2 \text{ m} \) is?
  • A) \( 4.5 \times 10^4 \text{ V} \)
  • B) \( 9 \times 10^4 \text{ V} \)
  • C) \( 1.8 \times 10^5 \text{ V} \)
  • D) \( 9 \times 10^3 \text{ V} \)
• Q3: Total work done to assemble \( q_1 = 1 \mu C \), \( q_2 = 1 \mu C \), \( V_A = 2 \text{ V} \), \( V_B = 3 \text{ V} \), \( x = 0.1 \text{ m} \) is?
  • A) \( 0.09 \text{ J} \)
  • B) \( 0.1 \text{ J} \)
  • C) \( 0.12 \text{ J} \)
  • D) \( 0.15 \text{ J} \)

Answer Key

• Q1: A) \( 3 \mu \text{J} \) - \( W_1 = q_1 V_A = 1 \times 10^{-6} \times 3 = 3 \times 10^{-6} \text{ J} \)
• Q2: D) \( 9 \times 10^3 \text{ V} \) - \( V_B = \frac{1}{4 \pi \epsilon_0} \frac{q_1}{x} = 9 \times 10^9 \times \frac{2 \times 10^{-6}}{0.2} = 9 \times 10^3 \text{ V} \)
• Q3: C) \( 0.12 \text{ J} \) - Calculate total work using formula, interaction term dominates

Memory Aids

• Total work = sum of individual works + interaction energy
• Potential due to point charge: \( V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r} \)
• Use \( 9 \times 10^9 \text{ Nm}^2/\text{C}^2 \) for \( \frac{1}{4 \pi \epsilon_0} \)

Potential Energy of a Dipole

Concepts

• Potential energy of dipole in external field: U = - vector p · vector E = -pE cos θ
• θ is angle between dipole moment vector p and external electric field vector E
• Work done to rotate dipole from θ₁ to θ₂: W_{θ1 → θ2} = U_θ2 - U_θ1 = pE(cos θ₁ - cos θ₂)
• Stable equilibrium at θ = 0° with minimum potential energy U = -pE
• Unstable equilibrium at θ = 180° with maximum potential energy U = pE

Solved Examples

• Ex 1: Calculate potential energy of dipole with p = 2 × 10^-29 C m, E = 10⁵ V/m, at θ = 60°.
  • Use U = -pE cos θ = -2 × 10^-29 × 10⁵ × 0.5 = -1 × 10^-24 J

Practice Set

• Q1: Potential energy of dipole with p = 1 × 10^-29 C m, E = 2 × 10⁵ V/m, at θ = 0° is?
  • A) -2 × 10^-24 J
  • B) 2 × 10^-24 J
  • C) 0
  • D) -1 × 10^-24 J
• Q2: Work done to rotate dipole from 0° to 180° with p = 3 × 10^-29 C m, E = 10⁵ V/m is?
  • A) 6 × 10^-24 J
  • B) 0
  • C) -6 × 10^-24 J
  • D) 3 × 10^-24 J
• Q3: At which angle is dipole in stable equilibrium?
  • A) 0°
  • B) 90°
  • C) 180°
  • D) 45°

Answer Key

• Q1: A) -2 × 10^-24 J - U = -pE cos 0° = -pE
• Q2: A) 6 × 10^-24 J - Work done = pE(cos 0° - cos 180°) = pE(1 - (-1)) = 2pE
• Q3: A) 0° - Minimum potential energy, stable equilibrium

Memory Aids

• Dipole energy: U = -pE cos θ
• Stable at 0°, unstable at 180°
• Work done = change in potential energy

Concepts

• Potential energy of dipole in external field: U = - vector p · vector E = -pE cos θ
• θ is angle between dipole moment vector p and external electric field vector E
• Work done to rotate dipole from θ₁ to θ₂: W_{θ1 → θ2} = U_θ2 - U_θ1 = pE(cos θ₁ - cos θ₂)
• Stable equilibrium at θ = 0° with minimum potential energy U = -pE
• Unstable equilibrium at θ = 180° with maximum potential energy U = pE

Solved Examples

• Ex 1: Calculate potential energy of dipole with p = 2 × 10^-29 C m, E = 10⁵ V/m, at θ = 60°.
  • Use U = -pE cos θ = -2 × 10^-29 × 10⁵ × 0.5 = -1 × 10^-24 J

Practice Set

• Q1: Potential energy of dipole with p = 1 × 10^-29 C m, E = 2 × 10⁵ V/m, at θ = 0° is?
  • A) -2 × 10^-24 J
  • B) 2 × 10^-24 J
  • C) 0
  • D) -1 × 10^-24 J
• Q2: Work done to rotate dipole from 0° to 180° with p = 3 × 10^-29 C m, E = 10⁵ V/m is?
  • A) 6 × 10^-24 J
  • B) 0
  • C) -6 × 10^-24 J
  • D) 3 × 10^-24 J
• Q3: At which angle is dipole in stable equilibrium?
  • A) 0°
  • B) 90°
  • C) 180°
  • D) 45°

Answer Key

• Q1: A) -2 × 10^-24 J - U = -pE cos 0° = -pE
• Q2: A) 6 × 10^-24 J - Work done = pE(cos 0° - cos 180°) = pE(1 - (-1)) = 2pE
• Q3: A) 0° - Minimum potential energy, stable equilibrium

Memory Aids

• Dipole energy: U = -pE cos θ
• Stable at 0°, unstable at 180°
• Work done = change in potential energy

Concepts

• Potential energy of dipole in external field: U = - vector p · vector E = -pE cos θ
• θ is angle between dipole moment vector p and external electric field vector E
• Work done to rotate dipole from θ₁ to θ₂: W_{θ1 → θ2} = U_θ2 - U_θ1 = pE(cos θ₁ - cos θ₂)
• Stable equilibrium at θ = 0° with minimum potential energy U = -pE
• Unstable equilibrium at θ = 180° with maximum potential energy U = pE

Solved Examples

• Ex 1: Calculate potential energy of dipole with p = 2 × 10^-29 C m, E = 10⁵ V/m, at θ = 60°.
  • Use U = -pE cos θ = -2 × 10^-29 × 10⁵ × 0.5 = -1 × 10^-24 J

Practice Set

• Q1: Potential energy of dipole with p = 1 × 10^-29 C m, E = 2 × 10⁵ V/m, at θ = 0° is?
  • A) -2 × 10^-24 J
  • B) 2 × 10^-24 J
  • C) 0
  • D) -1 × 10^-24 J
• Q2: Work done to rotate dipole from 0° to 180° with p = 3 × 10^-29 C m, E = 10⁵ V/m is?
  • A) 6 × 10^-24 J
  • B) 0
  • C) -6 × 10^-24 J
  • D) 3 × 10^-24 J
• Q3: At which angle is dipole in stable equilibrium?
  • A) 0°
  • B) 90°
  • C) 180°
  • D) 45°

Answer Key

• Q1: A) -2 × 10^-24 J - U = -pE cos 0° = -pE
• Q2: A) 6 × 10^-24 J - Work done = pE(cos 0° - cos 180°) = pE(1 - (-1)) = 2pE
• Q3: A) 0° - Minimum potential energy, stable equilibrium

Memory Aids

• Dipole energy: U = -pE cos θ
• Stable at 0°, unstable at 180°
• Work done = change in potential energy

Concepts

• Potential energy of dipole in external field: U = - vector p · vector E = -pE cos θ
• θ is angle between dipole moment vector p and external electric field vector E
• Work done to rotate dipole from θ₁ to θ₂: W_{θ1 → θ2} = U_θ2 - U_θ1 = pE(cos θ₁ - cos θ₂)
• Stable equilibrium at θ = 0° with minimum potential energy U = -pE
• Unstable equilibrium at θ = 180° with maximum potential energy U = pE

Solved Examples

• Ex 1: Calculate potential energy of dipole with p = 2 × 10^-29 C m, E = 10⁵ V/m, at θ = 60°.
  • Use U = -pE cos θ = -2 × 10^-29 × 10⁵ × 0.5 = -1 × 10^-24 J

Practice Set

• Q1: Potential energy of dipole with p = 1 × 10^-29 C m, E = 2 × 10⁵ V/m, at θ = 0° is?
  • A) -2 × 10^-24 J
  • B) 2 × 10^-24 J
  • C) 0
  • D) -1 × 10^-24 J
• Q2: Work done to rotate dipole from 0° to 180° with p = 3 × 10^-29 C m, E = 10⁵ V/m is?
  • A) 6 × 10^-24 J
  • B) 0
  • C) -6 × 10^-24 J
  • D) 3 × 10^-24 J
• Q3: At which angle is dipole in stable equilibrium?
  • A) 0°
  • B) 90°
  • C) 180°
  • D) 45°

Answer Key

• Q1: A) -2 × 10^-24 J - U = -pE cos 0° = -pE
• Q2: A) 6 × 10^-24 J - Work done = pE(cos 0° - cos 180°) = pE(1 - (-1)) = 2pE
• Q3: A) 0° - Minimum potential energy, stable equilibrium

Memory Aids

• Dipole energy: U = -pE cos θ
• Stable at 0°, unstable at 180°
• Work done = change in potential energy

Concepts

• Potential energy of dipole in external field: U = - vector p · vector E = -pE cos θ
• θ is angle between dipole moment vector p and external electric field vector E
• Work done to rotate dipole from θ₁ to θ₂: W_{θ1 → θ2} = U_θ2 - U_θ1 = pE(cos θ₁ - cos θ₂)
• Stable equilibrium at θ = 0° with minimum potential energy U = -pE
• Unstable equilibrium at θ = 180° with maximum potential energy U = pE

Solved Examples

• Ex 1: Calculate potential energy of dipole with p = 2 × 10^-29 C m, E = 10⁵ V/m, at θ = 60°.
  • Use U = -pE cos θ = -2 × 10^-29 × 10⁵ × 0.5 = -1 × 10^-24 J

Practice Set

• Q1: Potential energy of dipole with p = 1 × 10^-29 C m, E = 2 × 10⁵ V/m, at θ = 0° is?
  • A) -2 × 10^-24 J
  • B) 2 × 10^-24 J
  • C) 0
  • D) -1 × 10^-24 J
• Q2: Work done to rotate dipole from 0° to 180° with p = 3 × 10^-29 C m, E = 10⁵ V/m is?
  • A) 6 × 10^-24 J
  • B) 0
  • C) -6 × 10^-24 J
  • D) 3 × 10^-24 J
• Q3: At which angle is dipole in stable equilibrium?
  • A) 0°
  • B) 90°
  • C) 180°
  • D) 45°

Answer Key

• Q1: A) -2 × 10^-24 J - U = -pE cos 0° = -pE
• Q2: A) 6 × 10^-24 J - Work done = pE(cos 0° - cos 180°) = pE(1 - (-1)) = 2pE
• Q3: A) 0° - Minimum potential energy, stable equilibrium

Memory Aids

• Dipole energy: U = -pE cos θ
• Stable at 0°, unstable at 180°
• Work done = change in potential energy

Concepts

• Potential energy of dipole in external field: U = - p · E = -pE cos θ
• θ is angle between dipole moment p and external electric field E
• Work done to rotate dipole from θ₁ to θ₂: W_{θ1 → θ2} = U_θ2 - U_θ1 = pE(cos θ₁ - cos θ₂)
• Stable equilibrium at θ = 0° with minimum potential energy U = -pE
• Unstable equilibrium at θ = 180° with maximum potential energy U = pE

Solved Examples

• Ex 1: Calculate potential energy of dipole with p = 2 × 10^-29 C m, E = 10⁵ V/m, at θ = 60°.
  • Use U = -pE cos θ = -2 × 10^-29 × 10⁵ × 0.5 = -1 × 10^-24 J

Practice Set

• Q1: Potential energy of dipole with p = 1 × 10^-29 C m, E = 2 × 10⁵ V/m, at θ = 0° is?
  • A) -2 × 10^-24 J
  • B) 2 × 10^-24 J
  • C) 0
  • D) -1 × 10^-24 J
• Q2: Work done to rotate dipole from 0° to 180° with p = 3 × 10^-29 C m, E = 10⁵ V/m is?
  • A) 6 × 10^-24 J
  • B) 0
  • C) -6 × 10^-24 J
  • D) 3 × 10^-24 J
• Q3: At which angle is dipole in stable equilibrium?
  • A) 0°
  • B) 90°
  • C) 180°
  • D) 45°

Answer Key

• Q1: A) -2 × 10^-24 J - U = -pE cos 0° = -pE
• Q2: A) 6 × 10^-24 J - Work done = pE(cos 0° - cos 180°) = pE(1 - (-1)) = 2pE
• Q3: A) 0° - Minimum potential energy, stable equilibrium

Memory Aids

• Dipole energy: U = -pE cos θ
• Stable at 0°, unstable at 180°
• Work done = change in potential energy

Concepts

• Potential energy of dipole in external field: U = - vector p · vector E = -pE cos θ
• θ is angle between dipole moment vector p and external electric field vector E
• Work done to rotate dipole from θ₁ to θ₂: W_{θ1 → θ2} = U_θ2 - U_θ1 = pE(cos θ₁ - cos θ₂)
• Stable equilibrium at θ = 0° with minimum potential energy U = -pE
• Unstable equilibrium at θ = 180° with maximum potential energy U = pE

Solved Examples

• Ex 1: Calculate potential energy of dipole with p = 2 × 10^-29 C m, E = 10⁵ V/m, at θ = 60°.
  • Use U = -pE cos θ = -2 × 10^-29 × 10⁵ × 0.5 = -1 × 10^-24 J

Practice Set

• Q1: Potential energy of dipole with p = 1 × 10^-29 C m, E = 2 × 10⁵ V/m, at θ = 0° is?
  • A) -2 × 10^-24 J
  • B) 2 × 10^-24 J
  • C) 0
  • D) -1 × 10^-24 J
• Q2: Work done to rotate dipole from 0° to 180° with p = 3 × 10^-29 C m, E = 10⁵ V/m is?
  • A) 6 × 10^-24 J
  • B) 0
  • C) -6 × 10^-24 J
  • D) 3 × 10^-24 J
• Q3: At which angle is dipole in stable equilibrium?
  • A) 0°
  • B) 90°
  • C) 180°
  • D) 45°

Answer Key

• Q1: A) -2 × 10^-24 J - U = -pE cos 0° = -pE
• Q2: A) 6 × 10^-24 J - Work done = pE(cos 0° - cos 180°) = pE(1 - (-1)) = 2pE
• Q3: A) 0° - Minimum potential energy, stable equilibrium

Memory Aids

• Dipole energy: U = -pE cos θ
• Stable at 0°, unstable at 180°
• Work done = change in potential energy

Concepts

• Potential energy of dipole in external field: \( U = - \mathbf{p} \cdot \mathbf{E} = -pE \cos \theta \)
• \( \theta \) is angle between dipole moment \( \mathbf{p} \) and external electric field \( \mathbf{E} \)
• Work done to rotate dipole from \( \theta_1 \) to \( \theta_2 \): \( W_{\theta_1 \to \theta_2} = U_{\theta_2} - U_{\theta_1} = pE (\cos \theta_1 - \cos \theta_2) \)
• Stable equilibrium at \( \theta = 0^\circ \) with minimum potential energy \( U = -pE \)
• Unstable equilibrium at \( \theta = 180^\circ \) with maximum potential energy \( U = pE \)

Solved Examples

• Ex 1: Calculate potential energy of dipole with \( p = 2 \times 10^{-29} \text{ C m} \), \( E = 10^5 \text{ V/m} \), at \( \theta = 60^\circ \).
  • Use \( U = -pE \cos \theta = -2 \times 10^{-29} \times 10^5 \times 0.5 = -1 \times 10^{-24} \text{ J} \)

Practice Set

• Q1: Potential energy of dipole with \( p = 1 \times 10^{-29} \text{ C m} \), \( E = 2 \times 10^5 \text{ V/m} \), at \( \theta = 0^\circ \) is?
  • A) \( -2 \times 10^{-24} \text{ J} \)
  • B) \( 2 \times 10^{-24} \text{ J} \)
  • C) 0
  • D) \( -1 \times 10^{-24} \text{ J} \)
• Q2: Work done to rotate dipole from \( 0^\circ \) to \( 180^\circ \) with \( p = 3 \times 10^{-29} \text{ C m} \), \( E = 10^5 \text{ V/m} \) is?
  • A) \( 6 \times 10^{-24} \text{ J} \)
  • B) 0
  • C) \( -6 \times 10^{-24} \text{ J} \)
  • D) \( 3 \times 10^{-24} \text{ J} \)
• Q3: At which angle is dipole in stable equilibrium?
  • A) \( 0^\circ \)
  • B) \( 90^\circ \)
  • C) \( 180^\circ \)
  • D) \( 45^\circ \)

Answer Key

• Q1: A) \( -2 \times 10^{-24} \text{ J} \) - \( U = -pE \cos 0^\circ = -pE \)
• Q2: A) \( 6 \times 10^{-24} \text{ J} \) - Work done \( = pE (\cos 0^\circ - \cos 180^\circ) = pE (1 - (-1)) = 2pE \)
• Q3: A) \( 0^\circ \) - Minimum potential energy, stable equilibrium

Memory Aids

• Dipole energy: \( U = -pE \cos \theta \)
• Stable at \( 0^\circ \), unstable at \( 180^\circ \)
• Work done = change in potential energy

Electrostatics of Conductors

Properties of Conductors in Electrostatics

• Conductors have many free electrons that move freely inside.
• In electrostatics, conductors are equipotential surfaces (constant potential everywhere on the surface).
• Charge resides only on the outer surface of a conductor.
• Potential inside a charged conducting sphere is constant.
• If a cavity inside a charged conductor has no charge, all charge stays on the outer surface.
• Electric field is always perpendicular to the conductor's surface.
• Electric field near the surface: vector E = σ/ε₀ în, where σ = surface charge density, în = outward normal unit vector.

Solved Examples

Ex 1: Calculate electric field intensity near a conductor surface with surface charge density σ = 5 × 10^-6 \ mathrm{C/m^2}. Given ε₀ = 8.85 × 10^-12 \ mathrm{F/m}.

• Use formula: E = σ/ε₀
• E = (5 × 10^-6)/(8.85 × 10^-12) = 5.65 × 10⁵ \ mathrm{N/C}

Practice Set

• Q1: What is true about charge distribution on a charged conductor?
  • A. Charge is inside the conductor
  • B. Charge is uniformly distributed inside
  • C. Charge resides only on the outer surface
  • D. Charge is zero on the surface
• Q2: Electric field near a charged conductor surface depends on?
  • A. Volume charge density
  • B. Surface charge density
  • C. Potential inside conductor
  • D. Electric potential energy
• Q3: If surface charge density σ = 1 × 10^-6 \ mathrm{C/m^2}, what is electric field near surface? (ε₀ = 8.85 × 10^-12 \ mathrm{F/m})
  • A. 1.13 × 10⁵ \ mathrm{N/C}
  • B. 8.85 × 10⁵ \ mathrm{N/C}
  • C. 1.0 × 10⁶ \ mathrm{N/C}
  • D. 1.13 × 10⁶ \ mathrm{N/C}

Answer Key

• Q1: C - Charge resides only on the outer surface of a conductor in electrostatics.
• Q2: B - Electric field near surface depends on surface charge density σ.
• Q3: A - E = σ/ε₀ = (1 × 10^-6)/(8.85 × 10^-12) = 1.13 × 10⁵ \ mathrm{N/C}.

Memory Aids

• Conductors = Equipotential surfaces.
• Charge only on outer surface.
• Electric field E = σ/ε₀ perpendicular outward.
• Potential inside conductor = constant.

Properties of Conductors in Electrostatics

• Conductors have many free electrons that move freely inside.
• In electrostatics, conductors are equipotential surfaces (constant potential everywhere on the surface).
• Charge resides only on the outer surface of a conductor.
• Potential inside a charged conducting sphere is constant.
• If a cavity inside a charged conductor has no charge, all charge stays on the outer surface.
• Electric field is always perpendicular to the conductor's surface.
• Electric field near the surface: vector E = σ/ε₀ î, where σ = surface charge density, î = outward normal unit vector.

Solved Examples

Ex 1: Calculate electric field intensity near a conductor surface with surface charge density σ = 5 × 10^-6 mathrm{C/m^2}. Given ε₀ = 8.85 × 10^-12 mathrm{F/m}.

• Use formula: E = σ/ε₀
• E = (5 × 10^-6)/(8.85 × 10^-12) = 5.65 × 10⁵ mathrm{N/C}

Practice Set

• Q1: What is true about charge distribution on a charged conductor?
  • A. Charge is inside the conductor
  • B. Charge is uniformly distributed inside
  • C. Charge resides only on the outer surface
  • D. Charge is zero on the surface
• Q2: Electric field near a charged conductor surface depends on?
  • A. Volume charge density
  • B. Surface charge density
  • C. Potential inside conductor
  • D. Electric potential energy
• Q3: If surface charge density σ = 1 × 10^-6 mathrm{C/m^2}, what is electric field near surface? (ε₀ = 8.85 × 10^-12 mathrm{F/m})
  • A. 1.13 × 10⁵ mathrm{N/C}
  • B. 8.85 × 10⁵ mathrm{N/C}
  • C. 1.0 × 10⁶ mathrm{N/C}
  • D. 1.13 × 10⁶ mathrm{N/C}

Answer Key

• Q1: C - Charge resides only on the outer surface of a conductor in electrostatics.
• Q2: B - Electric field near surface depends on surface charge density σ.
• Q3: A - E = σ/ε₀ = (1 × 10^-6)/(8.85 × 10^-12) = 1.13 × 10⁵ mathrm{N/C}.

Memory Aids

• Conductors = Equipotential surfaces.
• Charge only on outer surface.
• Electric field E = σ/ε₀ perpendicular outward.
• Potential inside conductor = constant.

Properties of Conductors in Electrostatics

• Conductors have many free electrons that move freely inside.
• In electrostatics, conductors are equipotential surfaces (constant potential everywhere on the surface).
• Charge resides only on the outer surface of a conductor.
• Potential inside a charged conducting sphere is constant.
• If a cavity inside a charged conductor has no charge, all charge stays on the outer surface.
• Electric field is always perpendicular to the conductor's surface.
• Electric field near the surface: vector E = σ/ε₀ în, where σ = surface charge density, în = outward normal unit vector.

Solved Examples

Ex 1: Calculate electric field intensity near a conductor surface with surface charge density σ = 5 × 10^-6 \ mathrm{C/m^2}. Given ε₀ = 8.85 × 10^-12 \ mathrm{F/m}.

• Use formula: E = σ/ε₀
• E = (5 × 10^-6)/(8.85 × 10^-12) = 5.65 × 10⁵ \ mathrm{N/C}

Practice Set

• Q1: What is true about charge distribution on a charged conductor?
  • A. Charge is inside the conductor
  • B. Charge is uniformly distributed inside
  • C. Charge resides only on the outer surface
  • D. Charge is zero on the surface
• Q2: Electric field near a charged conductor surface depends on?
  • A. Volume charge density
  • B. Surface charge density
  • C. Potential inside conductor
  • D. Electric potential energy
• Q3: If surface charge density σ = 1 × 10^-6 \ mathrm{C/m^2}, what is electric field near surface? (ε₀ = 8.85 × 10^-12 \ mathrm{F/m})
  • A. 1.13 × 10⁵ \ mathrm{N/C}
  • B. 8.85 × 10⁵ \ mathrm{N/C}
  • C. 1.0 × 10⁶ \ mathrm{N/C}
  • D. 1.13 × 10⁶ \ mathrm{N/C}

Answer Key

• Q1: C - Charge resides only on the outer surface of a conductor in electrostatics.
• Q2: B - Electric field near surface depends on surface charge density σ.
• Q3: A - E = σ/ε₀ = (1 × 10^-6)/(8.85 × 10^-12) = 1.13 × 10⁵ \ mathrm{N/C}.

Memory Aids

• Conductors = Equipotential surfaces.
• Charge only on outer surface.
• Electric field E = σ/ε₀ perpendicular outward.
• Potential inside conductor = constant.

Properties of Conductors in Electrostatics

• Conductors have many free electrons that move freely inside.
• In electrostatics, conductors are equipotential surfaces (constant potential everywhere on the surface).
• Charge resides only on the outer surface of a conductor.
• Potential inside a charged conducting sphere is constant.
• If a cavity inside a charged conductor has no charge, all charge stays on the outer surface.
• Electric field is always perpendicular to the conductor's surface.
• Electric field near the surface: \( \vec{E} = \frac{\sigma}{\epsilon_0} \hat{n} \), where \( \sigma \) = surface charge density, \( \hat{n} \) = outward normal unit vector.

Solved Examples

Ex 1: Calculate electric field intensity near a conductor surface with surface charge density \( \sigma = 5 \times 10^{-6} \ \mathrm{C/m^2} \). Given \( \epsilon_0 = 8.85 \times 10^{-12} \ \mathrm{F/m} \).

• Use formula: \( E = \frac{\sigma}{\epsilon_0} \)
• \( E = \frac{5 \times 10^{-6}}{8.85 \times 10^{-12}} = 5.65 \times 10^{5} \ \mathrm{N/C} \)

Practice Set

• Q1: What is true about charge distribution on a charged conductor?
  • A. Charge is inside the conductor
  • B. Charge is uniformly distributed inside
  • C. Charge resides only on the outer surface
  • D. Charge is zero on the surface
• Q2: Electric field near a charged conductor surface depends on?
  • A. Volume charge density
  • B. Surface charge density
  • C. Potential inside conductor
  • D. Electric potential energy
• Q3: If surface charge density \( \sigma = 1 \times 10^{-6} \ \mathrm{C/m^2} \), what is electric field near surface? (\( \epsilon_0 = 8.85 \times 10^{-12} \ \mathrm{F/m} \))
  • A. \( 1.13 \times 10^{5} \ \mathrm{N/C} \)
  • B. \( 8.85 \times 10^{5} \ \mathrm{N/C} \)
  • C. \( 1.0 \times 10^{6} \ \mathrm{N/C} \)
  • D. \( 1.13 \times 10^{6} \ \mathrm{N/C} \)

Answer Key

• Q1: C - Charge resides only on the outer surface of a conductor in electrostatics.
• Q2: B - Electric field near surface depends on surface charge density \( \sigma \).
• Q3: A - \( E = \frac{\sigma}{\epsilon_0} = \frac{1 \times 10^{-6}}{8.85 \times 10^{-12}} = 1.13 \times 10^{5} \ \mathrm{N/C} \).

Memory Aids

• Conductors = Equipotential surfaces.
• Charge only on outer surface.
• Electric field \( E = \sigma/\epsilon_0 \) perpendicular outward.
• Potential inside conductor = constant.

Electrostatic Shielding

Concept and Application

• Electrostatic shielding protects a region from external electric fields.
• A cavity inside a conductor with no internal charge is field-free (electric field = 0).
• Field inside cavity is always zero regardless of external charges.
• Used to protect sensitive instruments by enclosing them in hollow conductors.

Solved Examples

Ex 1: Why is the electric field zero inside a hollow conductor cavity?

• Because charges reside on outer surface, no field penetrates inside.
• Electric field inside cavity = 0 due to electrostatic shielding.

Practice Set

• Q1: What happens to electric field inside a hollow conductor cavity?
  • A. It is maximum
  • B. It is zero
  • C. It depends on external charges
  • D. It varies with time
• Q2: Electrostatic shielding is used to protect?
  • A. Magnetic fields
  • B. Sensitive electronic instruments
  • C. Mechanical parts
  • D. Heat
• Q3: If a cavity inside a conductor has no charge, where does the charge reside?
  • A. Inside cavity
  • B. On inner surface of cavity
  • C. On outer surface of conductor
  • D. Uniformly inside conductor

Answer Key

• Q1: B - Electric field inside hollow cavity is zero due to shielding.
• Q2: B - Shielding protects sensitive electronic instruments from electric fields.
• Q3: C - Charge resides only on outer surface if cavity has no charge.

Memory Aids

• Shielding = zero field inside cavity.
• Charge stays outside cavity.
• Protect instruments by hollow conductor enclosure.

Concept and Application

• Electrostatic shielding protects a region from external electric fields.
• A cavity inside a conductor with no internal charge is field-free (electric field = 0).
• Field inside cavity is always zero regardless of external charges.
• Used to protect sensitive instruments by enclosing them in hollow conductors.

Solved Examples

Ex 1: Why is the electric field zero inside a hollow conductor cavity?

• Because charges reside on outer surface, no field penetrates inside.
• Electric field inside cavity = 0 due to electrostatic shielding.

Practice Set

• Q1: What happens to electric field inside a hollow conductor cavity?
  • A. It is maximum
  • B. It is zero
  • C. It depends on external charges
  • D. It varies with time
• Q2: Electrostatic shielding is used to protect?
  • A. Magnetic fields
  • B. Sensitive electronic instruments
  • C. Mechanical parts
  • D. Heat
• Q3: If a cavity inside a conductor has no charge, where does the charge reside?
  • A. Inside cavity
  • B. On inner surface of cavity
  • C. On outer surface of conductor
  • D. Uniformly inside conductor

Answer Key

• Q1: B - Electric field inside hollow cavity is zero due to shielding.
• Q2: B - Shielding protects sensitive electronic instruments from electric fields.
• Q3: C - Charge resides only on outer surface if cavity has no charge.

Memory Aids

• Shielding = zero field inside cavity.
• Charge stays outside cavity.
• Protect instruments by hollow conductor enclosure.

Summary of Electrostatic Properties

Key Points

• Inside neutral conductor: Electric field E = 0, charge density ρ = 0.
• On charged surface: Electric field E = σ/ε₀ în, points outward for positive charge.
• Inside charged conductor: Potential V = V₀ constant.
• Inside hollow conductor: Electric field and surface charge density are zero; charges on outer surface.

Solved Examples

Ex 1: State electric field inside a neutral conductor.

• Electric field inside neutral conductor is zero.

Practice Set

• Q1: Electric field inside a neutral conductor is?
  • A. Zero
  • B. Maximum
  • C. Depends on charge
  • D. Infinite
• Q2: Potential inside a charged conductor is?
  • A. Zero
  • B. Constant
  • C. Varies with position
  • D. Negative
• Q3: Where do charges reside in a hollow conductor?
  • A. Inside cavity
  • B. On inner surface
  • C. On outer surface
  • D. Uniformly inside

Answer Key

• Q1: A - Electric field inside neutral conductor is zero.
• Q2: B - Potential inside charged conductor is constant.
• Q3: C - Charges reside on outer surface of hollow conductor.

Memory Aids

• Neutral conductor: E=0, ρ=0.
• Charged surface: E=σ/ε₀ în.
• Inside conductor: Potential constant.
• Hollow conductor: Charges outside, field zero inside.

Key Points

• Inside neutral conductor: Electric field E = 0, charge density ρ = 0.
• On charged surface: Electric field E = σ/ε₀ î, points outward for positive charge.
• Inside charged conductor: Potential V = V₀ constant.
• Inside hollow conductor: Electric field and surface charge density are zero; charges on outer surface.

Solved Examples

Ex 1: State electric field inside a neutral conductor.

• Electric field inside neutral conductor is zero.

Practice Set

• Q1: Electric field inside a neutral conductor is?
  • A. Zero
  • B. Maximum
  • C. Depends on charge
  • D. Infinite
• Q2: Potential inside a charged conductor is?
  • A. Zero
  • B. Constant
  • C. Varies with position
  • D. Negative
• Q3: Where do charges reside in a hollow conductor?
  • A. Inside cavity
  • B. On inner surface
  • C. On outer surface
  • D. Uniformly inside

Answer Key

• Q1: A - Electric field inside neutral conductor is zero.
• Q2: B - Potential inside charged conductor is constant.
• Q3: C - Charges reside on outer surface of hollow conductor.

Memory Aids

• Neutral conductor: E=0, ρ=0.
• Charged surface: E=σ/ε₀ î.
• Inside conductor: Potential constant.
• Hollow conductor: Charges outside, field zero inside.

Key Points

• Inside neutral conductor: Electric field E = 0, charge density ρ = 0.
• On charged surface: Electric field E = σ/ε₀ în, points outward for positive charge.
• Inside charged conductor: Potential V = V₀ constant.
• Inside hollow conductor: Electric field and surface charge density are zero; charges on outer surface.

Solved Examples

Ex 1: State electric field inside a neutral conductor.

• Electric field inside neutral conductor is zero.

Practice Set

• Q1: Electric field inside a neutral conductor is?
  • A. Zero
  • B. Maximum
  • C. Depends on charge
  • D. Infinite
• Q2: Potential inside a charged conductor is?
  • A. Zero
  • B. Constant
  • C. Varies with position
  • D. Negative
• Q3: Where do charges reside in a hollow conductor?
  • A. Inside cavity
  • B. On inner surface
  • C. On outer surface
  • D. Uniformly inside

Answer Key

• Q1: A - Electric field inside neutral conductor is zero.
• Q2: B - Potential inside charged conductor is constant.
• Q3: C - Charges reside on outer surface of hollow conductor.

Memory Aids

• Neutral conductor: E=0, ρ=0.
• Charged surface: E=σ/ε₀ în.
• Inside conductor: Potential constant.
• Hollow conductor: Charges outside, field zero inside.

Key Points

• Inside neutral conductor: Electric field \( E = 0 \), charge density \( \rho = 0 \).
• On charged surface: Electric field \( E = \frac{\sigma}{\epsilon_0} \hat{n} \), points outward for positive charge.
• Inside charged conductor: Potential \( V = V_0 \) constant.
• Inside hollow conductor: Electric field and surface charge density are zero; charges on outer surface.

Solved Examples

Ex 1: State electric field inside a neutral conductor.

• Electric field inside neutral conductor is zero.

Practice Set

• Q1: Electric field inside a neutral conductor is?
  • A. Zero
  • B. Maximum
  • C. Depends on charge
  • D. Infinite
• Q2: Potential inside a charged conductor is?
  • A. Zero
  • B. Constant
  • C. Varies with position
  • D. Negative
• Q3: Where do charges reside in a hollow conductor?
  • A. Inside cavity
  • B. On inner surface
  • C. On outer surface
  • D. Uniformly inside

Answer Key

• Q1: A - Electric field inside neutral conductor is zero.
• Q2: B - Potential inside charged conductor is constant.
• Q3: C - Charges reside on outer surface of hollow conductor.

Memory Aids

• Neutral conductor: \( E=0, \rho=0 \).
• Charged surface: \( E=\frac{\sigma}{\epsilon_0} \hat{n} \).
• Inside conductor: Potential constant.
• Hollow conductor: Charges outside, field zero inside.

Dielectrics and Polarisation

Types of Dielectrics

• Dielectrics are insulating materials that support electric fields but do not conduct electricity.
• Polar dielectrics: positive and negative charges are separated; centers do not coincide.
• Non-polar dielectrics: positive and negative charges overlap; centers coincide.
• Polarisation occurs when dipoles align under an external electric field.
• Polarisation field vector E_i is induced inside the dielectric opposing the external field vector E₀.
• Net electric field inside dielectric: vector E = vector E₀ + vector E_i.
• Polarisation vector vector P = dipole moment per unit volume.
• For linear isotropic dielectrics: vector P = χ_e vector E_i, where χ_e is electrical susceptibility.

Solved Examples

• Ex 1: Calculate induced electric field E_i if E₀ = 1000 mathrm{V/m} and relative permittivity ε_r = 5.
  • Use E_i = (1 - 1/ε_r)E₀ = (1 - 1/5) × 1000 = 800 mathrm{V/m}.
• Ex 2: Find induced charge q_i if free charge q = 10 mu C and ε_r = 4.
  • Use q_i = q(1 - 1/ε_r) = 10 × (1 - 1/4) = 7.5 mu C.

Practice Set

• Q1: What is the polarisation vector vector P in a dielectric?
  • A. Electric field inside dielectric
  • B. Dipole moment per unit volume
  • C. Induced charge density
  • D. External electric field
• Q2: If ε_r = 3 and E₀ = 600 mathrm{V/m}, what is E_i?
  • A. 200 V/m
  • B. 400 V/m
  • C. 600 V/m
  • D. 800 V/m
• Q3: For a dielectric with χ_e = 2 and E_i = 500 mathrm{V/m}, find P.
  • A. 1000 C/m³
  • B. 250 C/m³
  • C. 500 C/m³
  • D. 2000 C/m³

Answer Key

• Q1: B - Polarisation vector vector P is the dipole moment per unit volume.
• Q2: B - E_i = (1 - 1/3) × 600 = 400 mathrm{V/m}.
• Q3: A - P = χ_eE_i = 2 × 500 = 1000 mathrm{C/m^3}.

Memory Aids

• Polarisation vector vector P = dipole moment density.
• Induced field E_i = (1 - 1/ε_r)E₀.
• Net field inside dielectric is reduced by E_i.

Types of Dielectrics

• Dielectrics are insulating materials that support electric fields but do not conduct electricity.
• Polar dielectrics: positive and negative charges are separated; centers do not coincide.
• Non-polar dielectrics: positive and negative charges overlap; centers coincide.
• Polarisation occurs when dipoles align under an external electric field.
• Polarisation field E_i is induced inside the dielectric opposing the external field E₀.
• Net electric field inside dielectric: E = E₀ + E_i.
• Polarisation vector P = dipole moment per unit volume.
• For linear isotropic dielectrics: P = χ_e E_i, where χ_e is electrical susceptibility.

Solved Examples

• Ex 1: Calculate induced electric field E_i if E₀ = 1000 mathrm{V/m} and relative permittivity ε_r = 5.
  • Use E_i = (1 - 1/ε_r)E₀ = (1 - 1/5) × 1000 = 800 mathrm{V/m}.
• Ex 2: Find induced charge q_i if free charge q = 10 mu C and ε_r = 4.
  • Use q_i = q(1 - 1/ε_r) = 10 × (1 - 1/4) = 7.5 mu C.

Practice Set

• Q1: What is the polarisation vector P in a dielectric?
  • A. Electric field inside dielectric
  • B. Dipole moment per unit volume
  • C. Induced charge density
  • D. External electric field
• Q2: If ε_r = 3 and E₀ = 600 mathrm{V/m}, what is E_i?
  • A. 200 V/m
  • B. 400 V/m
  • C. 600 V/m
  • D. 800 V/m
• Q3: For a dielectric with χ_e = 2 and E_i = 500 mathrm{V/m}, find P.
  • A. 1000 C/m³
  • B. 250 C/m³
  • C. 500 C/m³
  • D. 2000 C/m³

Answer Key

• Q1: B - Polarisation P is the dipole moment per unit volume.
• Q2: B - E_i = (1 - 1/3) × 600 = 400 mathrm{V/m}.
• Q3: A - P = χ_eE_i = 2 × 500 = 1000 mathrm{C/m^3}.

Memory Aids

• Polarisation P = dipole moment density.
• Induced field E_i = (1 - 1/ε_r)E₀.
• Net field inside dielectric is reduced by E_i.

Types of Dielectrics

• Dielectrics are insulating materials that support electric fields but do not conduct electricity.
• Polar dielectrics: positive and negative charges are separated; centers do not coincide.
• Non-polar dielectrics: positive and negative charges overlap; centers coincide.
• Polarisation occurs when dipoles align under an external electric field.
• Polarisation field vector E_i is induced inside the dielectric opposing the external field vector E₀.
• Net electric field inside dielectric: vector E = vector E₀ + vector E_i.
• Polarisation vector vector P = dipole moment per unit volume.
• For linear isotropic dielectrics: vector P = χ_e vector E_i, where χ_e is electrical susceptibility.

Solved Examples

• Ex 1: Calculate induced electric field E_i if E₀ = 1000 \ mathrm{V/m} and relative permittivity ε_r = 5.
  • Use E_i = (1 - 1/ε_r)E₀ = (1 - 1/5) × 1000 = 800 \ mathrm{V/m}.
• Ex 2: Find induced charge q_i if free charge q = 10 \ mu C and ε_r = 4.
  • Use q_i = q(1 - 1/ε_r) = 10 × (1 - 1/4) = 7.5 \ mu C.

Practice Set

• Q1: What is the polarisation vector vector P in a dielectric?
  • A. Electric field inside dielectric
  • B. Dipole moment per unit volume
  • C. Induced charge density
  • D. External electric field
• Q2: If ε_r = 3 and E₀ = 600 \ mathrm{V/m}, what is E_i?
  • A. 200 V/m
  • B. 400 V/m
  • C. 600 V/m
  • D. 800 V/m
• Q3: For a dielectric with χ_e = 2 and E_i = 500 \ mathrm{V/m}, find P.
  • A. 1000 C/m³
  • B. 250 C/m³
  • C. 500 C/m³
  • D. 2000 C/m³

Answer Key

• Q1: B - Polarisation vector vector P is the dipole moment per unit volume.
• Q2: B - E_i = (1 - 1/3) × 600 = 400 \ mathrm{V/m}.
• Q3: A - P = χ_eE_i = 2 × 500 = 1000 \ mathrm{C/m^3}.

Memory Aids

• Polarisation vector vector P = dipole moment density.
• Induced field E_i = (1 - 1/ε_r)E₀.
• Net field inside dielectric is reduced by E_i.

Types of Dielectrics

• Dielectrics are insulating materials that support electric fields but do not conduct electricity.
• Polar dielectrics: positive and negative charges are separated; centers do not coincide.
• Non-polar dielectrics: positive and negative charges overlap; centers coincide.
• Polarisation occurs when dipoles align under an external electric field.
• Polarisation field \( \vec{E_i} \) is induced inside the dielectric opposing the external field \( \vec{E_0} \).
• Net electric field inside dielectric: \( \vec{E} = \vec{E_0} + \vec{E_i} \).
• Polarisation vector \( \vec{P} \) = dipole moment per unit volume.
• For linear isotropic dielectrics: \( \vec{P} = \chi_e \vec{E_i} \), where \( \chi_e \) is electrical susceptibility.

Solved Examples

• Ex 1: Calculate induced electric field \( E_i \) if \( E_0 = 1000 \ \mathrm{V/m} \) and relative permittivity \( \varepsilon_r = 5 \).
  • Use \( E_i = \left(1 - \frac{1}{\varepsilon_r}\right) E_0 = \left(1 - \frac{1}{5}\right) \times 1000 = 800 \ \mathrm{V/m} \).
• Ex 2: Find induced charge \( q_i \) if free charge \( q = 10 \ \mu C \) and \( \varepsilon_r = 4 \).
  • Use \( q_i = q \left(1 - \frac{1}{\varepsilon_r}\right) = 10 \times \left(1 - \frac{1}{4}\right) = 7.5 \ \mu C \).

Practice Set

• Q1: What is the polarisation vector \( \vec{P} \) in a dielectric?
  • A. Electric field inside dielectric
  • B. Dipole moment per unit volume
  • C. Induced charge density
  • D. External electric field
• Q2: If \( \varepsilon_r = 3 \) and \( E_0 = 600 \ \mathrm{V/m} \), what is \( E_i \)?
  • A. 200 V/m
  • B. 400 V/m
  • C. 600 V/m
  • D. 800 V/m
• Q3: For a dielectric with \( \chi_e = 2 \) and \( E_i = 500 \ \mathrm{V/m} \), find \( P \).
  • A. 1000 C/m³
  • B. 250 C/m³
  • C. 500 C/m³
  • D. 2000 C/m³

Answer Key

• Q1: B - Polarisation \( \vec{P} \) is the dipole moment per unit volume.
• Q2: B - \( E_i = (1 - 1/3) \times 600 = 400 \ \mathrm{V/m} \).
• Q3: A - \( P = \chi_e E_i = 2 \times 500 = 1000 \ \mathrm{C/m^3} \).

Memory Aids

• Polarisation \( \vec{P} \) = dipole moment density.
• Induced field \( E_i = (1 - 1/\varepsilon_r) E_0 \).
• Net field inside dielectric is reduced by \( E_i \).

Induced Electric Charge and Net Field

Key Concepts

• Net electric field inside dielectric: E = E₀ - E_i.
• Relation: E₀ / ε_r = E₀ - E_i.
• Induced field: E_i = (1 - 1/ε_r) E₀.
• Induced surface charge density: σ_i = (1 - 1/ε_r) σ.
• Induced charge: q_i = q (1 - 1/ε_r).

Solved Examples

• Ex 1: Calculate induced charge q_i if q = 20 μC and ε_r = 5.
  • q_i = 20 × (1 - 1/5) = 16 μC.
• Ex 2: Find net field E if E₀ = 1000 V/m and ε_r = 4.
  • E = E₀ - E_i = E₀ - (1 - 1/4) E₀ = E₀/4 = 250 V/m.

Practice Set

• Q1: If ε_r = 2 and q = 10 μC, find induced charge q_i.
  • A. 5 μC
  • B. 10 μC
  • C. 15 μC
  • D. 20 μC
• Q2: Net electric field E inside dielectric is related to E₀ by?
  • A. E = E₀ + E_i
  • B. E = E₀ - E_i
  • C. E = E_i - E₀
  • D. E = E₀ / ε_r
• Q3: For ε_r = 3 and E₀ = 900 V/m, find E_i.
  • A. 300 V/m
  • B. 600 V/m
  • C. 900 V/m
  • D. 1200 V/m

Answer Key

• Q1: A - q_i = q (1 - 1/2) = 5 μC.
• Q2: B - Net field E = E₀ - E_i.
• Q3: B - E_i = (1 - 1/3) × 900 = 600 V/m.

Memory Aids

• Induced charge q_i = q (1 - 1/ε_r).
• Net field reduced by induced field inside dielectric.
• Use E = E₀ - E_i for net field.

Key Concepts

• Net electric field inside dielectric: E = E₀ - E_i.
• Relation: E₀ / ε_r = E₀ - E_i.
• Induced field: E_i = (1 - 1/ε_r) E₀.
• Induced surface charge density: σ_i = (1 - 1/ε_r) σ.
• Induced charge: q_i = q (1 - 1/ε_r).

Solved Examples

• Ex 1: Calculate induced charge q_i if q = 20 μC and ε_r = 5.
  • q_i = 20 × (1 - 1/5) = 16 μC.
• Ex 2: Find net field E if E₀ = 1000 V/m and ε_r = 4.
  • E = E₀ - E_i = E₀ - (1 - 1/4) E₀ = E₀/4 = 250 V/m.

Practice Set

• Q1: If ε_r = 2 and q = 10 μC, find induced charge q_i.
  • A. 5 μC
  • B. 10 μC
  • C. 15 μC
  • D. 20 μC
• Q2: Net electric field E inside dielectric is related to E₀ by?
  • A. E = E₀ + E_i
  • B. E = E₀ - E_i
  • C. E = E_i - E₀
  • D. E = E₀ / ε_r
• Q3: For ε_r = 3 and E₀ = 900 V/m, find E_i.
  • A. 300 V/m
  • B. 600 V/m
  • C. 900 V/m
  • D. 1200 V/m

Answer Key

• Q1: A - q_i = q (1 - 1/2) = 5 μC.
• Q2: B - Net field E = E₀ - E_i.
• Q3: B - E_i = (1 - 1/3) × 900 = 600 V/m.

Memory Aids

• Induced charge q_i = q (1 - 1/ε_r).
• Net field reduced by induced field inside dielectric.
• Use E = E₀ - E_i for net field.

Key Concepts

• Net electric field inside dielectric: E = E₀ - E_i.
• Relation: E₀ / ε_r = E₀ - E_i.
• Induced field: E_i = (1 - 1/ε_r) E₀.
• Induced surface charge density: σ_i = (1 - 1/ε_r) σ.
• Induced charge: q_i = q (1 - 1/ε_r).

Solved Examples

• Ex 1: Calculate induced charge q_i if q = 20 μC and ε_r = 5.
  • q_i = 20 × (1 - 1/5) = 16 μC.
• Ex 2: Find net field E if E₀ = 1000 V/m and ε_r = 4.
  • E = E₀ - E_i = E₀ - (1 - 1/4) E₀ = E₀/4 = 250 V/m.

Practice Set

• Q1: If ε_r = 2 and q = 10 μC, find induced charge q_i.
  • A. 5 μC
  • B. 10 μC
  • C. 15 μC
  • D. 20 μC
• Q2: Net electric field E inside dielectric is related to E₀ by?
  • A. E = E₀ + E_i
  • B. E = E₀ - E_i
  • C. E = E_i - E₀
  • D. E = E₀ / ε_r
• Q3: For ε_r = 3 and E₀ = 900 V/m, find E_i.
  • A. 300 V/m
  • B. 600 V/m
  • C. 900 V/m
  • D. 1200 V/m

Answer Key

• Q1: A - q_i = q (1 - 1/2) = 5 μC.
• Q2: B - Net field E = E₀ - E_i.
• Q3: B - E_i = (1 - 1/3) × 900 = 600 V/m.

Memory Aids

• Induced charge q_i = q (1 - 1/ε_r).
• Net field reduced by induced field inside dielectric.
• Use E = E₀ - E_i for net field.

Key Concepts

• Net electric field inside dielectric: E = E₀ - E_i.
• Relation: E₀ / ε_r = E₀ - E_i.
• Induced field: E_i = (1 - 1/ε_r) E₀.
• Induced surface charge density: σ_i = (1 - 1/ε_r) σ.
• Induced charge: q_i = q (1 - 1/ε_r).

Solved Examples

• Ex 1: Calculate induced charge q_i if q = 20 μC and ε_r = 5.
  • q_i = 20 × (1 - 1/5) = 16 μC.
• Ex 2: Find net field E if E₀ = 1000 V/m and ε_r = 4.
  • E = E₀ - E_i = E₀ - (1 - 1/4) E₀ = E₀/4 = 250 V/m.

Practice Set

• Q1: If ε_r = 2 and q = 10 μC, find induced charge q_i.
  • A. 5 μC
  • B. 10 μC
  • C. 15 μC
  • D. 20 μC
• Q2: Net electric field E inside dielectric is related to E₀ by?
  • A. E = E₀ + E_i
  • B. E = E₀ - E_i
  • C. E = E_i - E₀
  • D. E = E₀ / ε_r
• Q3: For ε_r = 3 and E₀ = 900 V/m, find E_i.
  • A. 300 V/m
  • B. 600 V/m
  • C. 900 V/m
  • D. 1200 V/m

Answer Key

• Q1: A - q_i = q (1 - 1/2) = 5 μC.
• Q2: B - Net field E = E₀ - E_i.
• Q3: B - E_i = (1 - 1/3) × 900 = 600 V/m.

Memory Aids

• Induced charge q_i = q (1 - 1/ε_r).
• Net field reduced by induced field inside dielectric.
• Use E = E₀ - E_i for net field.

Key Concepts

• Net electric field inside dielectric: E = E₀ - E_i.
• Relation: E₀ / ε_r = E₀ - E_i.
• Induced field: E_i = (1 - 1/ε_r) E₀.
• Induced surface charge density: σ_i = (1 - 1/ε_r) σ.
• Induced charge: q_i = q (1 - 1/ε_r).

Solved Examples

• Ex 1: Calculate induced charge q_i if q = 20 μC and ε_r = 5.
  • q_i = 20 × (1 - 1/5) = 16 μC.
• Ex 2: Find net field E if E₀ = 1000 V/m and ε_r = 4.
  • E = E₀ - E_i = E₀ - (1 - 1/4) E₀ = E₀/4 = 250 V/m.

Practice Set

• Q1: If ε_r = 2 and q = 10 μC, find induced charge q_i.
  • A. 5 μC
  • B. 10 μC
  • C. 15 μC
  • D. 20 μC
• Q2: Net electric field E inside dielectric is related to E₀ by?
  • A. E = E₀ + E_i
  • B. E = E₀ - E_i
  • C. E = E_i - E₀
  • D. E = E₀ / ε_r
• Q3: For ε_r = 3 and E₀ = 900 V/m, find E_i.
  • A. 300 V/m
  • B. 600 V/m
  • C. 900 V/m
  • D. 1200 V/m

Answer Key

• Q1: A - q_i = q (1 - 1/2) = 5 μC.
• Q2: B - Net field E = E₀ - E_i.
• Q3: B - E_i = (1 - 1/3) × 900 = 600 V/m.

Memory Aids

• Induced charge q_i = q (1 - 1/ε_r).
• Net field reduced by induced field inside dielectric.
• Use E = E₀ - E_i for net field.

Key Concepts

• Net electric field inside dielectric: \( E = E_0 - E_i \).
• Relation: \( E_0 / \varepsilon_r = E_0 - E_i \).
• Induced field: \( E_i = \left(1 - \frac{1}{\varepsilon_r}\right) E_0 \).
• Induced surface charge density: \( \sigma_i = \left(1 - \frac{1}{\varepsilon_r}\right) \sigma \).
• Induced charge: \( q_i = q \left(1 - \frac{1}{\varepsilon_r}\right) \).

Solved Examples

• Ex 1: Calculate induced charge \( q_i \) if \( q = 20 \ \mu C \) and \( \varepsilon_r = 5 \).
  • \( q_i = 20 \times (1 - 1/5) = 16 \ \mu C \).
• Ex 2: Find net field \( E \) if \( E_0 = 1000 \ \mathrm{V/m} \) and \( \varepsilon_r = 4 \).
  • \( E = E_0 - E_i = E_0 - (1 - 1/4) E_0 = E_0/4 = 250 \ \mathrm{V/m} \).

Practice Set

• Q1: If \( \varepsilon_r = 2 \) and \( q = 10 \ \mu C \), find induced charge \( q_i \).
  • A. 5 \mu C
  • B. 10 \mu C
  • C. 15 \mu C
  • D. 20 \mu C
• Q2: Net electric field \( E \) inside dielectric is related to \( E_0 \) by?
  • A. \( E = E_0 + E_i \)
  • B. \( E = E_0 - E_i \)
  • C. \( E = E_i - E_0 \)
  • D. \( E = E_0 / \varepsilon_r \)
• Q3: For \( \varepsilon_r = 3 \) and \( E_0 = 900 \ \mathrm{V/m} \), find \( E_i \).
  • A. 300 V/m
  • B. 600 V/m
  • C. 900 V/m
  • D. 1200 V/m

Answer Key

• Q1: A - \( q_i = q (1 - 1/2) = 5 \ \mu C \).
• Q2: B - Net field \( E = E_0 - E_i \).
• Q3: B - \( E_i = (1 - 1/3) \times 900 = 600 \ \mathrm{V/m} \).

Memory Aids

• Induced charge \( q_i = q (1 - 1/\varepsilon_r) \).
• Net field reduced by induced field inside dielectric.
• Use \( E = E_0 - E_i \) for net field.

Electrostatic Pressure on Charged Conductor

Key Points

• Force on surface element dA: dF = (σ dA) E₂ = σ² dA/2 ε₀.
• Electrostatic pressure P = dF/dA = σ²/2 ε₀ = 1/2 ε₀ E² using E = σ/ε₀.
• In presence of dielectric, pressure P = 1/2 ε₀ K E², where K is dielectric constant.

Solved Examples

• Ex 1: Calculate electrostatic pressure if σ = 2 × 10^-6 C/m² and ε₀ = 8.85 × 10^-12 F/m.
  • P = σ²/2 ε₀ = (2 × 10^-6)²/2 × 8.85 × 10^-12 = 226 N/m².
• Ex 2: Find pressure in dielectric with K = 4, E = 10⁴ V/m, and ε₀ = 8.85 × 10^-12 F/m.
  • P = 1/2 ε₀ K E² = 0.5 × 8.85 × 10^-12 × 4 × (10⁴)² = 1.77 N/m².

Practice Set

• Q1: Electrostatic pressure on charged conductor surface is given by?
  • A. σ/ε₀
  • B. σ²/2 ε₀
  • C. σ²/ε₀
  • D. σ/2 ε₀
• Q2: If E = 5 × 10³ V/m, ε₀ = 8.85 × 10^-12 F/m, find pressure.
  • A. 1.1 N/m²
  • B. 0.5 N/m²
  • C. 0.25 N/m²
  • D. 2.2 N/m²
• Q3: In dielectric with dielectric constant K = 3, pressure is?
  • A. 1/2 ε₀ E²
  • B. ε₀ E²
  • C. 3/2 ε₀ E²
  • D. 1/6 ε₀ E²

Answer Key

• Q1: B - Pressure P = σ²/2 ε₀ from surface charge and field.
• Q2: A - P = 1/2 ε₀ E² = 0.5 × 8.85 × 10^-12 × (5 × 10³)² = 1.1 N/m².
• Q3: C - Pressure in dielectric P = 1/2 ε₀ K E² = 3/2 ε₀ E².

Memory Aids

• Pressure P = σ²/2 ε₀ = 1/2 ε₀ E².
• Dielectric increases pressure by factor K.
• Use P = σ/ε₀ to relate field and charge density.

Key Points

• Force on surface element dA: dF = (σ dA) E₂ = σ² dA/2 ε₀.
• Electrostatic pressure P = dF/dA = σ²/2 ε₀ = 1/2 ε₀ E² using E = σ/ε₀.
• In presence of dielectric, pressure P = 1/2 ε₀ K E², where K is dielectric constant.

Solved Examples

• Ex 1: Calculate electrostatic pressure if σ = 2 × 10^-6 C/m² and ε₀ = 8.85 × 10^-12 F/m.
  • P = σ²/2 ε₀ = (2 × 10^-6)²/2 × 8.85 × 10^-12 = 226 N/m².
• Ex 2: Find pressure in dielectric with K = 4, E = 10⁴ V/m, and ε₀ = 8.85 × 10^-12 F/m.
  • P = 1/2 ε₀ K E² = 0.5 × 8.85 × 10^-12 × 4 × (10⁴)² = 1.77 N/m².

Practice Set

• Q1: Electrostatic pressure on charged conductor surface is given by?
  • A. σ/ε₀
  • B. σ²/2 ε₀
  • C. σ²/ε₀
  • D. σ/2 ε₀
• Q2: If E = 5 × 10³ V/m, ε₀ = 8.85 × 10^-12 F/m, find pressure.
  • A. 1.1 N/m²
  • B. 0.5 N/m²
  • C. 0.25 N/m²
  • D. 2.2 N/m²
• Q3: In dielectric with dielectric constant K = 3, pressure is?
  • A. 1/2 ε₀ E²
  • B. ε₀ E²
  • C. 3/2 ε₀ E²
  • D. 1/6 ε₀ E²

Answer Key

• Q1: B - Pressure P = σ²/2 ε₀ from surface charge and field.
• Q2: A - P = 1/2 ε₀ E² = 0.5 × 8.85 × 10^-12 × (5 × 10³)² = 1.1 N/m².
• Q3: C - Pressure in dielectric P = 1/2 ε₀ K E² = 3/2 ε₀ E².

Memory Aids

• Pressure P = σ²/2 ε₀ = 1/2 ε₀ E².
• Dielectric increases pressure by factor K.
• Use P = σ/ε₀ to relate field and charge density.

Key Points

• Force on surface element dA: dF = (σ dA) E₂ = σ² dA/2 ε₀.
• Electrostatic pressure P = dF/dA = σ²/2 ε₀ = 1/2 ε₀ E² using E = σ/ε₀.
• In presence of dielectric, pressure P = 1/2 ε₀ K E², where K is dielectric constant.

Solved Examples

• Ex 1: Calculate electrostatic pressure if σ = 2 × 10^-6 C/m² and ε₀ = 8.85 × 10^-12 F/m.
  • P = σ²/2 ε₀ = (2 × 10^-6)²/2 × 8.85 × 10^-12 = 226 N/m².
• Ex 2: Find pressure in dielectric with K = 4, E = 10⁴ V/m, and ε₀ = 8.85 × 10^-12 F/m.
  • P = 1/2 ε₀ K E² = 0.5 × 8.85 × 10^-12 × 4 × (10⁴)² = 1.77 N/m².

Practice Set

• Q1: Electrostatic pressure on charged conductor surface is given by?
  • A. σ/ε₀
  • B. σ²/2 ε₀
  • C. σ²/ε₀
  • D. σ/2 ε₀
• Q2: If E = 5 × 10³ V/m, ε₀ = 8.85 × 10^-12 F/m, find pressure.
  • A. 1.1 N/m²
  • B. 0.5 N/m²
  • C. 0.25 N/m²
  • D. 2.2 N/m²
• Q3: In dielectric with dielectric constant K = 3, pressure is?
  • A. 1/2 ε₀ E²
  • B. ε₀ E²
  • C. 3/2 ε₀ E²
  • D. 1/6 ε₀ E²

Answer Key

• Q1: B - Pressure P = σ²/2 ε₀ from surface charge and field.
• Q2: A - P = 1/2 ε₀ E² = 0.5 × 8.85 × 10^-12 × (5 × 10³)² = 1.1 N/m².
• Q3: C - Pressure in dielectric P = 1/2 ε₀ K E² = 3/2 ε₀ E².

Memory Aids

• Pressure P = σ²/2 ε₀ = 1/2 ε₀ E².
• Dielectric increases pressure by factor K.
• Use P = σ/ε₀ to relate field and charge density.

Key Points

• Force on surface element : = (σ dA) E₂ = σ² dA/2 ε₀.
• Electrostatic pressure ₌ = σ²/2 ε₀ = 1/2 ε₀ E² using ₌ = σ/ε₀.
• In presence of dielectric, pressure ₌ = 1/2 ε₀ K E², where is dielectric constant.

Solved Examples

• Ex 1: Calculate electrostatic pressure if = 2 × 10^-6 C/m² and ε₀ = 8.85 × 10^-12 F/m.
  • ₌ = σ²/2 ε₀ = (2 × 10^-6)²/2 × 8.85 × 10^-12 = 226 N/m².
• Ex 2: Find pressure in dielectric with = 4, = 10⁴ V/m, and ε₀ = 8.85 × 10^-12 F/m.
  • ₌ = 1/2 ε₀ K E² = 0.5 × 8.85 × 10^-12 × 4 × (10⁴)² = 1.77 N/m².

Practice Set

• Q1: Electrostatic pressure on charged conductor surface is given by?
  • A. = σ/ε₀
  • B. = σ²/2 ε₀
  • C. = σ²/ε₀
  • D. = σ/2 ε₀
• Q2: If = 5 × 10³ V/m, ε₀ = 8.85 × 10^-12 F/m, find pressure.
  • A. 1.1 N/m²
  • B. 0.5 N/m²
  • C. 0.25 N/m²
  • D. 2.2 N/m²
• Q3: In dielectric with dielectric constant = 3, pressure is?
  • A. 1/2 ε₀ E²
  • B. ε₀ E²
  • C. 3/2 ε₀ E²
  • D. 1/6 ε₀ E²

Answer Key

• Q1: B - Pressure ₌ = σ²/2 ε₀ from surface charge and field.
• Q2: A - ₌ = 1/2 ε₀ E² = 0.5 × 8.85 × 10^-12 × (5 × 10³)² = 1.1 N/m².
• Q3: C - Pressure in dielectric ₌ = 1/2 ε₀ K E² = 3/2 ε₀ E².

Memory Aids

• Pressure ₌ = σ²/2 ε₀ = 1/2 ε₀ E².
• Dielectric increases pressure by factor .
• Use ₌ = σ/ε₀ to relate field and charge density.

Key Points

• Force on surface element \( dA \): \( dF = (\sigma dA) E_2 = \frac{\sigma^2 dA}{2 \varepsilon_0} \).
• Electrostatic pressure \( P = \frac{dF}{dA} = \frac{\sigma^2}{2 \varepsilon_0} = \frac{1}{2} \varepsilon_0 E^2 \) using \( E = \frac{\sigma}{\varepsilon_0} \).
• In presence of dielectric, pressure \( P = \frac{1}{2} \varepsilon_0 K E^2 \), where \( K \) is dielectric constant.

Solved Examples

• Ex 1: Calculate electrostatic pressure if \( \sigma = 2 \times 10^{-6} \ \mathrm{C/m^2} \) and \( \varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{F/m} \).
  • \( P = \frac{\sigma^2}{2 \varepsilon_0} = \frac{(2 \times 10^{-6})^2}{2 \times 8.85 \times 10^{-12}} = 226 \ \mathrm{N/m^2} \).
• Ex 2: Find pressure in dielectric with \( K = 4 \), \( E = 10^4 \ \mathrm{V/m} \), and \( \varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{F/m} \).
  • \( P = \frac{1}{2} \varepsilon_0 K E^2 = 0.5 \times 8.85 \times 10^{-12} \times 4 \times (10^4)^2 = 1.77 \ \mathrm{N/m^2} \).

Practice Set

• Q1: Electrostatic pressure on charged conductor surface is given by?
  • A. \( \frac{\sigma}{\varepsilon_0} \)
  • B. \( \frac{\sigma^2}{2 \varepsilon_0} \)
  • C. \( \frac{\sigma^2}{\varepsilon_0} \)
  • D. \( \frac{\sigma}{2 \varepsilon_0} \)
• Q2: If \( E = 5 \times 10^3 \ \mathrm{V/m} \), \( \varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{F/m} \), find pressure.
  • A. 1.1 N/m²
  • B. 0.5 N/m²
  • C. 0.25 N/m²
  • D. 2.2 N/m²
• Q3: In dielectric with dielectric constant \( K = 3 \), pressure is?
  • A. \( \frac{1}{2} \varepsilon_0 E^2 \)
  • B. \( \varepsilon_0 E^2 \)
  • C. \( \frac{3}{2} \varepsilon_0 E^2 \)
  • D. \( \frac{1}{6} \varepsilon_0 E^2 \)

Answer Key

• Q1: B - Pressure \( = \frac{\sigma^2}{2 \varepsilon_0} \) from surface charge and field.
• Q2: A - \( P = \frac{1}{2} \varepsilon_0 E^2 = 0.5 \times 8.85 \times 10^{-12} \times (5 \times 10^3)^2 = 1.1 \ \mathrm{N/m^2} \).
• Q3: C - Pressure in dielectric \( = \frac{1}{2} \varepsilon_0 K E^2 = \frac{3}{2} \varepsilon_0 E^2 \).

Memory Aids

• Pressure \( = \frac{\sigma^2}{2 \varepsilon_0} = \frac{1}{2} \varepsilon_0 E^2 \).
• Dielectric increases pressure by factor \( K \).
• Use \( E = \frac{\sigma}{\varepsilon_0} \) to relate field and charge density.

The Parallel Plate Capacitor

Basic Concept

• A capacitor consists of two conductors (plates) placed close but not touching.
• Plates are parallel with area (A) and separated by distance (d).
• One plate carries positive charge (+Q), the other negative charge (-Q).
• Voltage difference between plates is V = V₊ - V_-.
• Capacitance C relates charge and voltage: Q = CV.
• Electric field E is uniform between plates if plate size >> d, ignoring fringing.

Electric Field and Potential Difference

• Electric field between plates: E = Q/A ε₀, where ε₀ is permittivity of free space.
• Potential difference: V = Ed = Qd/A ε₀.
• Capacitance formula: C = Q/V = ε₀ A/d.
• Fringing effect causes field lines to spread near edges but is usually neglected.

Force Between Plates

• Plates exert attractive force due to opposite charges.
• Electric field from one plate acts on charge of the other.
• Electric field due to one plate: E = σ/2 ε₀ = Q/2 A ε₀, where σ is charge density.
• Force on a plate: F = QE = Q²/2 A ε₀.

Solved Examples

• Ex 1: Calculate capacitance of plates with area 0.01 m², separation 0.001 m, ε₀ = 8.85 × 10^-12 F/m.
  • Use C = ε₀ A/d = 8.85 × 10^-12 × 0.01/0.001 = 8.85 × 10^-11 F.
• Ex 2: Find force between plates if Q = 1 μC, A = 0.01 m², ε₀ = 8.85 × 10^-12 F/m.
  • Calculate F = Q²/2 A ε₀ = (1 × 10^-6)²/2 × 0.01 × 8.85 × 10^-12 = 5.65 N.

Practice Set

• Q1: What is the capacitance of a parallel plate capacitor with plate area 2 × 10^-4 m² and plate separation 0.5 mm?
  • A) 3.54 × 10^-12 F
  • B) 7.08 × 10^-12 F
  • C) 1.77 × 10^-11 F
  • D) 8.85 × 10^-12 F
• Q2: If charge on the positive plate is 2 μC and plate area is 0.02 m², what is the electric field between plates?
  • A) 1.13 × 10⁵ V/m
  • B) 5.65 × 10⁴ V/m
  • C) 2.26 × 10⁵ V/m
  • D) 4.52 × 10⁴ V/m
• Q3: Force between plates with charge 3 μC, area 0.01 m², and ε₀ = 8.85 × 10^-12 F/m is?
  • A) 5.1 N
  • B) 15.3 N
  • C) 25.5 N
  • D) 45.9 N

Answer Key

• Q1: B. C = ε₀ A/d = 8.85 × 10^-12 × 2 × 10^-4/0.5 × 10^-3 = 7.08 × 10^-12 F.
• Q2: A. E = Q/A ε₀ = 2 × 10^-6/0.02 × 8.85 × 10^-12 = 1.13 × 10⁵ V/m.
• Q3: B. F = Q²/2 A ε₀ = (3 × 10^-6)²/2 × 0.01 × 8.85 × 10^-12 = 15.3 N.

Memory Aids

• Capacitance C increases with plate area A and decreases with distance d.
• Electric field E = Q/A ε₀ between plates.
• Force between plates F = Q²/2 A ε₀ is attractive.
• Ignore fringing effect for quick calculations.

Basic Concept

• A capacitor consists of two conductors (plates) placed close but not touching.
• Plates are parallel with area (A) and separated by distance (d).
• One plate carries positive charge (+Q), the other negative charge (-Q).
• Voltage difference between plates is V = V₊ - V_-.
• Capacitance C relates charge and voltage: Q = CV.
• Electric field E is uniform between plates if plate size >> d, ignoring fringing.

Electric Field and Potential Difference

• Electric field between plates: E = Q/A ε₀, where ε₀ is permittivity of free space.
• Potential difference: V = Ed = Qd/A ε₀.
• Capacitance formula: C = Q/V = ε₀ A/d.
• Fringing effect causes field lines to spread near edges but is usually neglected.

Force Between Plates

• Plates exert attractive force due to opposite charges.
• Electric field from one plate acts on charge of the other.
• Electric field due to one plate: E = σ/2 ε₀ = Q/2 A ε₀, where σ is charge density.
• Force on a plate: F = QE = Q²/2 A ε₀.

Solved Examples

• Ex 1: Calculate capacitance of plates with area 0.01 m², separation 0.001 m, ε₀ = 8.85 × 10^-12 F/m.
  • Use C = ε₀ A/d = 8.85 × 10^-12 × 0.01/0.001 = 8.85 × 10^-11 F.
• Ex 2: Find force between plates if Q = 1 μC, A = 0.01 m², ε₀ = 8.85 × 10^-12 F/m.
  • Calculate F = Q²/2 A ε₀ = (1 × 10^-6)²/2 × 0.01 × 8.85 × 10^-12 = 5.65 N.

Practice Set

• Q1: What is the capacitance of a parallel plate capacitor with plate area 2 × 10^-4 m² and plate separation 0.5 mm?
  • A) 3.54 × 10^-12 F
  • B) 7.08 × 10^-12 F
  • C) 1.77 × 10^-11 F
  • D) 8.85 × 10^-12 F
• Q2: If charge on the positive plate is 2 μC and plate area is 0.02 m², what is the electric field between plates?
  • A) 1.13 × 10⁵ V/m
  • B) 5.65 × 10⁴ V/m
  • C) 2.26 × 10⁵ V/m
  • D) 4.52 × 10⁴ V/m
• Q3: Force between plates with charge 3 μC, area 0.01 m², and ε₀ = 8.85 × 10^-12 F/m is?
  • A) 5.1 N
  • B) 15.3 N
  • C) 25.5 N
  • D) 45.9 N

Answer Key

• Q1: B. C = ε₀ A/d = 8.85 × 10^-12 × 2 × 10^-4/0.5 × 10^-3 = 7.08 × 10^-12 F.
• Q2: A. E = Q/A ε₀ = 2 × 10^-6/0.02 × 8.85 × 10^-12 = 1.13 × 10⁵ V/m.
• Q3: B. F = Q²/2 A ε₀ = (3 × 10^-6)²/2 × 0.01 × 8.85 × 10^-12 = 15.3 N.

Memory Aids

• Capacitance C increases with plate area A and decreases with distance d.
• Electric field E = Q/A ε₀ between plates.
• Force between plates F = Q²/2 A ε₀ is attractive.
• Ignore fringing effect for quick calculations.

Basic Concept

• A capacitor consists of two conductors (plates) placed close but not touching.
• Plates are parallel with area (A) and separated by distance (d).
• One plate carries positive charge (+Q), the other negative charge (-Q).
• Voltage difference between plates is V = V₊ - V_-.
• Capacitance C relates charge and voltage: Q = CV.
• Electric field E is uniform between plates if plate size >> d, ignoring fringing.

Electric Field and Potential Difference

• Electric field between plates: E = Q/A ε₀, where ε₀ is permittivity of free space.
• Potential difference: V = Ed = Qd/A ε₀.
• Capacitance formula: C = Q/V = ε₀ A/d.
• Fringing effect causes field lines to spread near edges but is usually neglected.

Force Between Plates

• Plates exert attractive force due to opposite charges.
• Electric field from one plate acts on charge of the other.
• Electric field due to one plate: E = σ/2 ε₀ = Q/2 A ε₀, where σ is charge density.
• Force on a plate: F = QE = Q²/2 A ε₀.

Solved Examples

• Ex 1: Calculate capacitance of plates with area 0.01 m², separation 0.001 m, ε₀ = 8.85 × 10^-12 F/m.
  • Use C = ε₀ A/d = 8.85 × 10^-12 × 0.01/0.001 = 8.85 × 10^-11 F.
• Ex 2: Find force between plates if Q = 1 μC, A = 0.01 m², ε₀ = 8.85 × 10^-12 F/m.
  • Calculate F = Q²/2 A ε₀ = (1 × 10^-6)²/2 × 0.01 × 8.85 × 10^-12 = 5.65 N.

Practice Set

• Q1: What is the capacitance of a parallel plate capacitor with plate area 2 × 10^-4 m² and plate separation 0.5 mm?
  • A) 3.54 × 10^-12 F
  • B) 7.08 × 10^-12 F
  • C) 1.77 × 10^-11 F
  • D) 8.85 × 10^-12 F
• Q2: If charge on the positive plate is 2 μC and plate area is 0.02 m², what is the electric field between plates?
  • A) 1.13 × 10⁵ V/m
  • B) 5.65 × 10⁴ V/m
  • C) 2.26 × 10⁵ V/m
  • D) 4.52 × 10⁴ V/m
• Q3: Force between plates with charge 3 μC, area 0.01 m², and ε₀ = 8.85 × 10^-12 F/m is?
  • A) 5.1 N
  • B) 15.3 N
  • C) 25.5 N
  • D) 45.9 N

Answer Key

• Q1: B. C = ε₀ A/d = 8.85 × 10^-12 × 2 × 10^-4/0.5 × 10^-3 = 7.08 × 10^-12 F.
• Q2: A. E = Q/A ε₀ = 2 × 10^-6/0.02 × 8.85 × 10^-12 = 1.13 × 10⁵ V/m.
• Q3: B. F = Q²/2 A ε₀ = (3 × 10^-6)²/2 × 0.01 × 8.85 × 10^-12 = 15.3 N.

Memory Aids

• Capacitance C increases with plate area A and decreases with distance d.
• Electric field E = Q/A ε₀ between plates.
• Force between plates F = Q²/2 A ε₀ is attractive.
• Ignore fringing effect for quick calculations.

Basic Concept

• A capacitor consists of two conductors (plates) placed close but not touching.
• Plates are parallel with area (A) and separated by distance (d).
• One plate carries positive charge (+Q), the other negative charge (-Q).
• Voltage difference between plates is V = V₊ - V_-.
• Capacitance C relates charge and voltage: Q = CV.
• Electric field E is uniform between plates if plate size >> d, ignoring fringing.

Electric Field and Potential Difference

• Electric field between plates: E = Q/A ε₀, where ε₀ is permittivity of free space.
• Potential difference: V = Ed = Qd/A ε₀.
• Capacitance formula: C = Q/V = ε₀ A/d.
• Fringing effect causes field lines to spread near edges but is usually neglected.

Force Between Plates

• Plates exert attractive force due to opposite charges.
• Electric field from one plate acts on charge of the other.
• Electric field due to one plate: E = σ/2 ε₀ = Q/2 A ε₀, where σ is charge density.
• Force on a plate: F = QE = Q²/2 A ε₀.

Solved Examples

• Ex 1: Calculate capacitance of plates with area 0.01 m², separation 0.001 m, ε₀ = 8.85 × 10^-12 F/m.
  • Use C = ε₀ A/d = 8.85 × 10^-12 × 0.01/0.001 = 8.85 × 10^-11 F.
• Ex 2: Find force between plates if Q = 1 μC, A = 0.01 m², ε₀ = 8.85 × 10^-12 F/m.
  • Calculate F = Q²/2 A ε₀ = (1 × 10^-6)²/2 × 0.01 × 8.85 × 10^-12 = 5.65 N.

Practice Set

• Q1: What is the capacitance of a parallel plate capacitor with plate area 2 × 10^-4 m² and plate separation 0.5 mm?
  • A) 3.54 × 10^-12 F
  • B) 7.08 × 10^-12 F
  • C) 1.77 × 10^-11 F
  • D) 8.85 × 10^-12 F
• Q2: If charge on the positive plate is 2 μC and plate area is 0.02 m², what is the electric field between plates?
  • A) 1.13 × 10⁵ V/m
  • B) 5.65 × 10⁴ V/m
  • C) 2.26 × 10⁵ V/m
  • D) 4.52 × 10⁴ V/m
• Q3: Force between plates with charge 3 μC, area 0.01 m², and ε₀ = 8.85 × 10^-12 F/m is?
  • A) 5.1 N
  • B) 15.3 N
  • C) 25.5 N
  • D) 45.9 N

Answer Key

• Q1: B. C = ε₀ A/d = 8.85 × 10^-12 × 2 × 10^-4/0.5 × 10^-3 = 7.08 × 10^-12 F.
• Q2: A. E = Q/A ε₀ = 2 × 10^-6/0.02 × 8.85 × 10^-12 = 1.13 × 10⁵ V/m.
• Q3: B. F = Q²/2 A ε₀ = (3 × 10^-6)²/2 × 0.01 × 8.85 × 10^-12 = 15.3 N.

Memory Aids

• Capacitance C increases with plate area A and decreases with distance d.
• Electric field E = Q/A ε₀ between plates.
• Force between plates F = Q²/2 A ε₀ is attractive.
• Ignore fringing effect for quick calculations.

Basic Concept

• A capacitor consists of two conductors (plates) placed close but not touching.
• Plates are parallel with area \(A\) and separated by distance \(d\).
• One plate carries positive charge \(+Q\), the other negative charge \(-Q\).
• Voltage difference between plates is \(V = V_+ - V_-\).
• Capacitance \(C\) relates charge and voltage: \(Q = CV\).
• Electric field \(E\) is uniform between plates if plate size \(>> d\), ignoring fringing.

Electric Field and Potential Difference

• Electric field between plates: \(E = \frac{Q}{A \epsilon_0}\), where \(\epsilon_0\) is permittivity of free space.
• Potential difference: \(V = Ed = \frac{Qd}{A \epsilon_0}\).
• Capacitance formula: \(C = \frac{Q}{V} = \frac{\epsilon_0 A}{d}\).
• Fringing effect causes field lines to spread near edges but is usually neglected.

Force Between Plates

• Plates exert attractive force due to opposite charges.
• Electric field from one plate acts on charge of the other.
• Electric field due to one plate: \(E = \frac{\sigma}{2 \epsilon_0} = \frac{Q}{2 A \epsilon_0}\), where \(\sigma\) is charge density.
• Force on a plate: \(F = QE = \frac{Q^2}{2 A \epsilon_0}\).

Solved Examples

• Ex 1: Calculate capacitance of plates with area \(0.01\,m^2\), separation \(0.001\,m\), \(\epsilon_0 = 8.85 \times 10^{-12} F/m\).
  • Use \(C = \frac{\epsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 0.01}{0.001} = 8.85 \times 10^{-11} F\).
• Ex 2: Find force between plates if \(Q = 1 \mu C\), \(A = 0.01 m^2\), \(\epsilon_0 = 8.85 \times 10^{-12} F/m\).
  • Calculate \(F = \frac{Q^2}{2 A \epsilon_0} = \frac{(1 \times 10^{-6})^2}{2 \times 0.01 \times 8.85 \times 10^{-12}} = 5.65 N\).

Practice Set

• Q1: What is the capacitance of a parallel plate capacitor with plate area \(2 \times 10^{-4} m^2\) and plate separation \(0.5 mm\)?
  • A) \(3.54 \times 10^{-12} F\)
  • B) \(7.08 \times 10^{-12} F\)
  • C) \(1.77 \times 10^{-11} F\)
  • D) \(8.85 \times 10^{-12} F\)
• Q2: If charge on the positive plate is \(2 \mu C\) and plate area is \(0.02 m^2\), what is the electric field between plates?
  • A) \(1.13 \times 10^{5} V/m\)
  • B) \(5.65 \times 10^{4} V/m\)
  • C) \(2.26 \times 10^{5} V/m\)
  • D) \(4.52 \times 10^{4} V/m\)
• Q3: Force between plates with charge \(3 \mu C\), area \(0.01 m^2\), and \(\epsilon_0 = 8.85 \times 10^{-12} F/m\) is?
  • A) 5.1 N
  • B) 15.3 N
  • C) 25.5 N
  • D) 45.9 N

Answer Key

• Q1: B. \(C = \frac{\epsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 2 \times 10^{-4}}{0.5 \times 10^{-3}} = 7.08 \times 10^{-12} F\).
• Q2: A. \(E = \frac{Q}{A \epsilon_0} = \frac{2 \times 10^{-6}}{0.02 \times 8.85 \times 10^{-12}} = 1.13 \times 10^{5} V/m\).
• Q3: B. \(F = \frac{Q^2}{2 A \epsilon_0} = \frac{(3 \times 10^{-6})^2}{2 \times 0.01 \times 8.85 \times 10^{-12}} = 15.3 N\).

Memory Aids

• Capacitance \(C\) increases with plate area \(A\) and decreases with distance \(d\).
• Electric field \(E = \frac{Q}{A \epsilon_0}\) between plates.
• Force between plates \(F = \frac{Q^2}{2 A \epsilon_0}\) is attractive.
• Ignore fringing effect for quick calculations.

Effect of Dielectric on Capacitance

Basic Concepts

• Vacuum between plates: Electric field E₀ = σ/ε₀, potential difference V₀ = E₀ d
• Capacitance without dielectric: C₀ = Q/V₀ = ε₀ A/d
• Dielectric inserted fully between plates polarizes, creating surface charges σ_P and -σ_P
• Electric field inside dielectric: E = (σ - σ_P)/ε₀
• Potential difference with dielectric: V = E d = (σ - σ_P)/ε₀ d
• For linear dielectrics, σ_P proportional to σ, so σ - σ_P = σ/K, where K > 1
• Potential difference: V = σ d/ε₀ K = Q d/A ε₀ K
• Capacitance with dielectric: C = Q/V = ε₀ K A/d
• Permittivity of medium: ε = ε₀ K
• Dielectric constant: K = ε/ε₀ = C/C₀

Effect of Conducting Slab

• Conducting slab thickness t < d inserted between plates
• Electric field inside conductor is zero (E = 0)
• Potential difference reduces: V = E₀ (d - t) = Q/A ε₀ (d - t)
• Capacitance: C = ε₀ A/(d - t) = C₀/1 - t/d

Capacitance of Partially Filled Capacitor

• Dielectric thickness f partially fills the gap
• Capacitance: C = K ε₀ A/K(d - t) + t

Solved Examples

• Ex 1: Calculate capacitance with dielectric K=5, plate area A=0.01 m², plate separation d=1 mm, and ε₀ = 8.85 × 10^-12 F/m.
  • Use C = ε₀ K A/d
  • C = 8.85 × 10^-12 × 5 × 0.01/1 × 10^-3 = 4.425 × 10^-10 F
• Ex 2: Conducting slab thickness t=0.2 mm inserted in capacitor with d=1 mm, A=0.01 m², find new capacitance.
  • C = ε₀ A/d - t = 8.85 × 10^-12 × 0.01/1 × 10^-3 - 0.2 × 10^-3 = 1.106 × 10^-10 F

Practice Set

• Q1: What is the dielectric constant K of vacuum?
  • A. 0
  • B. 1
  • C. ε₀
  • D. Infinity
• Q2: Capacitance of a capacitor with dielectric K=4, plate area 2 m², plate separation 0.5 mm, and ε₀ = 8.85 × 10^-12 F/m is?
  • A. 1.42 × 10^-7 F
  • B. 7.08 × 10^-9 F
  • C. 7.08 × 10^-7 F
  • D. 1.42 × 10^-8 F
• Q3: A conducting slab of thickness 0.1 mm is inserted in a capacitor with plate separation 1 mm and capacitance without slab 10 pF. New capacitance is?
  • A. 11.1 pF
  • B. 9.0 pF
  • C. 10.1 pF
  • D. 8.9 pF

Answer Key

• Q1: B. 1 — Vacuum has dielectric constant 1 by definition.
• Q2: C. 7.08 × 10^-7 F — Use C = ε₀ K A/d.
• Q3: A. 11.1 pF — New capacitance C = C₀/1 - t/d = 10/1 - 0.1 = 11.1 pF.

Memory Aids

• Dielectric constant K = C/C₀
• Capacitance with dielectric C = ε₀ K A/d
• Conducting slab reduces effective distance: d - t

Basic Concepts

• Vacuum between plates: Electric field E₀ = σ/ε₀, potential difference V₀ = E₀ d
• Capacitance without dielectric: C₀ = Q/V₀ = ε₀ A/d
• Dielectric inserted fully between plates polarizes, creating surface charges σ_P and -σ_P
• Electric field inside dielectric: E = (σ - σ_P)/ε₀
• Potential difference with dielectric: V = E d = (σ - σ_P)/ε₀ d
• For linear dielectrics, σ_P proportional to σ, so σ - σ_P = σ/K, where K > 1
• Potential difference: V = σ d/ε₀ K = Q d/A ε₀ K
• Capacitance with dielectric: C = Q/V = ε₀ K A/d
• Permittivity of medium: ε = ε₀ K
• Dielectric constant: K = ε/ε₀ = C/C₀

Effect of Conducting Slab

• Conducting slab thickness t < d inserted between plates
• Electric field inside conductor is zero (E = 0)
• Potential difference reduces: V = E₀ (d - t) = Q/A ε₀ (d - t)
• Capacitance: C = ε₀ A/(d - t) = C₀/1 - t/d

Capacitance of Partially Filled Capacitor

• Dielectric thickness f partially fills the gap
• Capacitance: C = K ε₀ A/K(d - t) + t

Solved Examples

• Ex 1: Calculate capacitance with dielectric K=5, plate area A=0.01 m², plate separation d=1 mm, and ε₀ = 8.85 × 10^-12 F/m.
  • Use C = ε₀ K A/d
  • C = 8.85 × 10^-12 × 5 × 0.01/1 × 10^-3 = 4.425 × 10^-10 F
• Ex 2: Conducting slab thickness t=0.2 mm inserted in capacitor with d=1 mm, A=0.01 m², find new capacitance.
  • C = ε₀ A/d - t = 8.85 × 10^-12 × 0.01/1 × 10^-3 - 0.2 × 10^-3 = 1.106 × 10^-10 F

Practice Set

• Q1: What is the dielectric constant K of vacuum?
  • A. 0
  • B. 1
  • C. ε₀
  • D. Infinity
• Q2: Capacitance of a capacitor with dielectric K=4, plate area 2 m², plate separation 0.5 mm, and ε₀ = 8.85 × 10^-12 F/m is?
  • A. 1.42 × 10^-7 F
  • B. 7.08 × 10^-9 F
  • C. 7.08 × 10^-7 F
  • D. 1.42 × 10^-8 F
• Q3: A conducting slab of thickness 0.1 mm is inserted in a capacitor with plate separation 1 mm and capacitance without slab 10 pF. New capacitance is?
  • A. 11.1 pF
  • B. 9.0 pF
  • C. 10.1 pF
  • D. 8.9 pF

Answer Key

• Q1: B. 1 — Vacuum has dielectric constant 1 by definition.
• Q2: C. 7.08 × 10^-7 F — Use C = ε₀ K A/d.
• Q3: A. 11.1 pF — New capacitance C = C₀/1 - t/d = 10/1 - 0.1 = 11.1 pF.

Memory Aids

• Dielectric constant K = C/C₀
• Capacitance with dielectric C = ε₀ K A/d
• Conducting slab reduces effective distance: d - t

Basic Concepts

• Vacuum between plates: Electric field E₀ = σ/ε₀, potential difference V₀ = E₀ d
• Capacitance without dielectric: C₀ = Q/V₀ = ε₀ A/d
• Dielectric inserted fully between plates polarizes, creating surface charges σ_P and -σ_P
• Electric field inside dielectric: E = (σ - σ_P)/ε₀
• Potential difference with dielectric: V = E d = (σ - σ_P)/ε₀ d
• For linear dielectrics, σ_P proportional to σ, so σ - σ_P = σ/K, where K > 1
• Potential difference: V = σ d/ε₀ K = Q d/A ε₀ K
• Capacitance with dielectric: C = Q/V = ε₀ K A/d
• Permittivity of medium: ε = ε₀ K
• Dielectric constant: K = ε/ε₀ = C/C₀

Effect of Conducting Slab

• Conducting slab thickness t < d inserted between plates
• Electric field inside conductor is zero (E = 0)
• Potential difference reduces: V = E₀ (d - t) = Q/A ε₀ (d - t)
• Capacitance: C = ε₀ A/(d - t) = C₀/1 - t/d

Capacitance of Partially Filled Capacitor

• Dielectric thickness f partially fills the gap
• Capacitance: C = K ε₀ A/K(d - t) + t

Solved Examples

• Ex 1: Calculate capacitance with dielectric K=5, plate area A=0.01 m², plate separation d=1 mm, and ε₀ = 8.85 × 10^-12 F/m.
  • Use C = ε₀ K A/d
  • C = 8.85 × 10^-12 × 5 × 0.01/1 × 10^-3 = 4.425 × 10^-10 F
• Ex 2: Conducting slab thickness t=0.2 mm inserted in capacitor with d=1 mm, A=0.01 m², find new capacitance.
  • C = ε₀ A/d - t = 8.85 × 10^-12 × 0.01/1 × 10^-3 - 0.2 × 10^-3 = 1.106 × 10^-10 F

Practice Set

• Q1: What is the dielectric constant K of vacuum?
  • A. 0
  • B. 1
  • C. ε₀
  • D. Infinity
• Q2: Capacitance of a capacitor with dielectric K=4, plate area 2 m², plate separation 0.5 mm, and ε₀ = 8.85 × 10^-12 F/m is?
  • A. 1.42 × 10^-7 F
  • B. 7.08 × 10^-9 F
  • C. 7.08 × 10^-7 F
  • D. 1.42 × 10^-8 F
• Q3: A conducting slab of thickness 0.1 mm is inserted in a capacitor with plate separation 1 mm and capacitance without slab 10 pF. New capacitance is?
  • A. 11.1 pF
  • B. 9.0 pF
  • C. 10.1 pF
  • D. 8.9 pF

Answer Key

• Q1: B. 1 — Vacuum has dielectric constant 1 by definition.
• Q2: C. 7.08 × 10^-7 F — Use C = ε₀ K A/d.
• Q3: A. 11.1 pF — New capacitance C = C₀/1 - t/d = 10/1 - 0.1 = 11.1 pF.

Memory Aids

• Dielectric constant K = C/C₀
• Capacitance with dielectric C = ε₀ K A/d
• Conducting slab reduces effective distance: d - t

Basic Concepts

• Vacuum between plates: Electric field E₀ = σ/ε₀, potential difference V₀ = E₀ d
• Capacitance without dielectric: C₀ = Q/V₀ = ε₀ A/d
• Dielectric inserted fully between plates polarizes, creating surface charges σ_P and -σ_P
• Electric field inside dielectric: E = (σ - σ_P)/ε₀
• Potential difference with dielectric: V = E d = (σ - σ_P)/ε₀ d
• For linear dielectrics, σ_P proportional to σ, so σ - σ_P = σ/K, where K > 1
• Potential difference: V = σ d/ε₀ K = Q d/A ε₀ K
• Capacitance with dielectric: C = Q/V = ε₀ K A/d
• Permittivity of medium: ε = ε₀ K
• Dielectric constant: K = ε/ε₀ = C/C₀

Effect of Conducting Slab

• Conducting slab thickness t < d inserted between plates
• Electric field inside conductor is zero (E = 0)
• Potential difference reduces: V = E₀ (d - t) = Q/A ε₀ (d - t)
• Capacitance: C = ε₀ A/(d - t) = C₀/1 - t/d

Capacitance of Partially Filled Capacitor

• Dielectric thickness f partially fills the gap
• Capacitance: C = K ε₀ A/K(d - t) + t

Solved Examples

• Ex 1: Calculate capacitance with dielectric K=5, plate area A=0.01 m², plate separation d=1 mm, and ε₀ = 8.85 × 10^-12 F/m.
  • Use C = ε₀ K A/d
  • C = 8.85 × 10^-12 × 5 × 0.01/1 × 10^-3 = 4.425 × 10^-10 F
• Ex 2: Conducting slab thickness t=0.2 mm inserted in capacitor with d=1 mm, A=0.01 m², find new capacitance.
  • C = ε₀ A/d - t = 8.85 × 10^-12 × 0.01/1 × 10^-3 - 0.2 × 10^-3 = 1.106 × 10^-10 F

Practice Set

• Q1: What is the dielectric constant K of vacuum?
  • A. 0
  • B. 1
  • C. ε₀
  • D. Infinity
• Q2: Capacitance of a capacitor with dielectric K=4, plate area 2 m², plate separation 0.5 mm, and ε₀ = 8.85 × 10^-12 F/m is?
  • A. 1.42 × 10^-7 F
  • B. 7.08 × 10^-9 F
  • C. 7.08 × 10^-7 F
  • D. 1.42 × 10^-8 F
• Q3: A conducting slab of thickness 0.1 mm is inserted in a capacitor with plate separation 1 mm and capacitance without slab 10 pF. New capacitance is?
  • A. 11.1 pF
  • B. 9.0 pF
  • C. 10.1 pF
  • D. 8.9 pF

Answer Key

• Q1: B. 1 — Vacuum has dielectric constant 1 by definition.
• Q2: C. 7.08 × 10^-7 F — Use C = ε₀ K A/d.
• Q3: A. 11.1 pF — New capacitance C = C₀/1 - t/d = 10/1 - 0.1 = 11.1 pF.

Memory Aids

• Dielectric constant K = C/C₀
• Capacitance with dielectric C = ε₀ K A/d
• Conducting slab reduces effective distance: d - t

Basic Concepts

• Vacuum between plates: Electric field E₀ = σ/ε₀, potential difference V₀ = E₀ d
• Capacitance without dielectric: C₀ = Q/V₀ = ε₀ A/d
• Dielectric inserted fully between plates polarizes, creating surface charges σ_P and -σ_P
• Electric field inside dielectric: E = (σ - σ_P)/ε₀
• Potential difference with dielectric: V = E d = (σ - σ_P)/ε₀ d
• For linear dielectrics, σ_P proportional to σ, so σ - σ_P = σ/K, where K > 1
• Potential difference: V = σ d/ε₀ K = Q d/A ε₀ K
• Capacitance with dielectric: C = Q/V = ε₀ K A/d
• Permittivity of medium: ε = ε₀ K
• Dielectric constant: K = ε/ε₀ = C/C₀

Effect of Conducting Slab

• Conducting slab thickness t < d inserted between plates
• Electric field inside conductor is zero (E = 0)
• Potential difference reduces: V = E₀ (d - t) = Q/A ε₀ (d - t)
• Capacitance: C = ε₀ A/(d - t) = C₀/1 - t/d

Capacitance of Partially Filled Capacitor

• Dielectric thickness f partially fills the gap
• Capacitance: C = K ε₀ A/K(d - t) + t

Solved Examples

• Ex 1: Calculate capacitance with dielectric K=5, plate area A=0.01 m², plate separation d=1 mm, and ε₀ = 8.85 × 10^-12 F/m.
  • Use C = ε₀ K A/d
  • C = 8.85 × 10^-12 × 5 × 0.01/1 × 10^-3 = 4.425 × 10^-10 F
• Ex 2: Conducting slab thickness t=0.2 mm inserted in capacitor with d=1 mm, A=0.01 m², find new capacitance.
  • C = ε₀ A/d - t = 8.85 × 10^-12 × 0.01/1 × 10^-3 - 0.2 × 10^-3 = 1.106 × 10^-10 F

Practice Set

• Q1: What is the dielectric constant K of vacuum?
  • A. 0
  • B. 1
  • C. ε₀
  • D. Infinity
• Q2: Capacitance of a capacitor with dielectric K=4, plate area 2 m², plate separation 0.5 mm, and ε₀ = 8.85 × 10^-12 F/m is?
  • A. 1.42 × 10^-7 F
  • B. 7.08 × 10^-9 F
  • C. 7.08 × 10^-7 F
  • D. 1.42 × 10^-8 F
• Q3: A conducting slab of thickness 0.1 mm is inserted in a capacitor with plate separation 1 mm and capacitance without slab 10 pF. New capacitance is?
  • A. 11.1 pF
  • B. 9.0 pF
  • C. 10.1 pF
  • D. 8.9 pF

Answer Key

• Q1: B. 1 — Vacuum has dielectric constant 1 by definition.
• Q2: C. 7.08 × 10^-7 F — Use C = ε₀ K A/d.
• Q3: A. 11.1 pF — New capacitance C = C₀/1 - t/d = 10/1 - 0.1 = 11.1 pF.

Memory Aids

• Dielectric constant K = C/C₀
• Capacitance with dielectric C = ε₀ K A/d
• Conducting slab reduces effective distance: d - t

Basic Concepts

• Vacuum between plates: Electric field \( E_0 = \frac{\sigma}{\varepsilon_0} \), potential difference \( V_0 = E_0 d \)
• Capacitance without dielectric: \( C_0 = \frac{Q}{V_0} = \frac{\varepsilon_0 A}{d} \)
• Dielectric inserted fully between plates polarizes, creating surface charges \( \sigma_P \) and \( -\sigma_P \)
• Electric field inside dielectric: \( E = \frac{\sigma - \sigma_P}{\varepsilon_0} \)
• Potential difference with dielectric: \( V = E d = \frac{\sigma - \sigma_P}{\varepsilon_0} d \)
• For linear dielectrics, \( \sigma_P \) proportional to \( \sigma \), so \( \sigma - \sigma_P = \frac{\sigma}{K} \), where \( K > 1 \)
• Potential difference: \( V = \frac{\sigma d}{\varepsilon_0 K} = \frac{Q d}{A \varepsilon_0 K} \)
• Capacitance with dielectric: \( C = \frac{Q}{V} = \frac{\varepsilon_0 K A}{d} \)
• Permittivity of medium: \( \varepsilon = \varepsilon_0 K \)
• Dielectric constant: \( K = \frac{\varepsilon}{\varepsilon_0} = \frac{C}{C_0} \)

Effect of Conducting Slab

• Conducting slab thickness \( t < d \) inserted between plates
• Electric field inside conductor is zero \( (E = 0) \)
• Potential difference reduces: \( V = E_0 (d - t) = \frac{Q}{A \varepsilon_0} (d - t) \)
• Capacitance: \( C = \frac{\varepsilon_0 A}{d - t} = \frac{C_0}{1 - \frac{t}{d}} \)

Capacitance of Partially Filled Capacitor

• Dielectric thickness \( f \) partially fills the gap
• Capacitance: \( C = \frac{K \varepsilon_0 A}{K(d - t) + t} \)

Solved Examples

• Ex 1: Calculate capacitance with dielectric \( K=5 \), plate area \( A=0.01 \ \mathrm{m}^2 \), plate separation \( d=1 \ \mathrm{mm} \), and \( \varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{F/m} \).
  • Use \( C = \frac{\varepsilon_0 K A}{d} \)
  • \( C = \frac{8.85 \times 10^{-12} \times 5 \times 0.01}{1 \times 10^{-3}} = 4.425 \times 10^{-10} \ \mathrm{F} \)
• Ex 2: Conducting slab thickness \( t=0.2 \ \mathrm{mm} \) inserted in capacitor with \( d=1 \ \mathrm{mm} \), \( A=0.01 \ \mathrm{m}^2 \), find new capacitance.
  • \( C = \frac{\varepsilon_0 A}{d - t} = \frac{8.85 \times 10^{-12} \times 0.01}{1 \times 10^{-3} - 0.2 \times 10^{-3}} = 1.106 \times 10^{-10} \ \mathrm{F} \)

Practice Set

• Q1: What is the dielectric constant \( K \) of vacuum?
  • A. 0
  • B. 1
  • C. \( \varepsilon_0 \)
  • D. Infinity
• Q2: Capacitance of a capacitor with dielectric \( K=4 \), plate area \( 2 \ \mathrm{m}^2 \), plate separation \( 0.5 \ \mathrm{mm} \), and \( \varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{F/m} \) is?
  • A. \( 1.42 \times 10^{-7} \ \mathrm{F} \)
  • B. \( 7.08 \times 10^{-9} \ \mathrm{F} \)
  • C. \( 7.08 \times 10^{-7} \ \mathrm{F} \)
  • D. \( 1.42 \times 10^{-8} \ \mathrm{F} \)
• Q3: A conducting slab of thickness \( 0.1 \ \mathrm{mm} \) is inserted in a capacitor with plate separation \( 1 \ \mathrm{mm} \) and capacitance without slab \( 10 \ \mathrm{pF} \). New capacitance is?
  • A. 11.1 pF
  • B. 9.0 pF
  • C. 10.1 pF
  • D. 8.9 pF

Answer Key

• Q1: B. 1 — Vacuum has dielectric constant 1 by definition.
• Q2: C. \( 7.08 \times 10^{-7} \ \mathrm{F} \) — Use \( C = \frac{\varepsilon_0 K A}{d} \).
• Q3: A. 11.1 pF — New capacitance \( C = \frac{C_0}{1 - \frac{t}{d}} = \frac{10}{1 - 0.1} = 11.1 \ \mathrm{pF} \).

Memory Aids

• Dielectric constant \( K = \frac{C}{C_0} \)
• Capacitance with dielectric \( C = \varepsilon_0 K \frac{A}{d} \)
• Conducting slab reduces effective distance: \( d - t \)

Combination of Capacitors

Capacitors in Series

• Connected end-to-end; charge path is single.
• Charge Q on each capacitor is the same.
• Total voltage V is sum of voltages across capacitors: V = V₁ + V₂ + V₃ + ...
• Equivalent capacitance C_s given by:
  [ 1/C_s = 1/C₁ + 1/C₂ + 1/C₃ + ... ]
• Voltage divides inversely with capacitance: smaller capacitance gets higher voltage.
• Equivalent capacitance is always less than the smallest individual capacitance.
• For n identical capacitors each of capacitance C, C_s = C/n.

Capacitors in Parallel

• Connected side-by-side; multiple charge paths.
• Voltage V across each capacitor is the same.
• Total charge Q is sum of individual charges: Q = Q₁ + Q₂ + Q₃ + ...
• Equivalent capacitance C_p is sum of capacitances:
  [ C_p = C₁ + C₂ + C₃ + ... ]
• Charge divides proportional to capacitance values.

Solved Examples

Ex 1

• Two capacitors C₁ = 4 μF and C₂ = 6 μF in series. Find C_s.
• Use formula: 1/C_s = 1/4 + 1/6 = 5/12 ⇒ C_s = 12/5 = 2.4 μF.

Ex 2

• Two capacitors C₁ = 3 μF and C₂ = 5 μF in parallel. Find C_p.
• Use formula: C_p = 3 + 5 = 8 μF.

Practice Set

Q1

• Two capacitors 2 μF and 3 μF are connected in series. What is the equivalent capacitance?
• A) 5 μF
• B) 1.2 μF
• C) 6 μF
• D) 0.6 μF

Q2

• Three capacitors 1 μF, 2 μF, and 3 μF are connected in parallel. Find the total capacitance.
• A) 6 μF
• B) 0.6 μF
• C) 1.8 μF
• D) 3 μF

Q3

• Four identical capacitors each of 4 μF are connected in series. What is the equivalent capacitance?
• A) 16 μF
• B) 1 μF
• C) 4 μF
• D) 8 μF

Answer Key

• Q1: B) 1.2 μF - Series formula 1/C_s = 1/2 + 1/3 = 5/6 ⇒ C_s = 1.2 μF.
• Q2: A) 6 μF - Parallel sum C_p = 1 + 2 + 3 = 6 μF.
• Q3: B) 1 μF - Series identical capacitors C_s = C/n = 4/4 = 1 μF.

Memory Aids

• Series: 1/C_s = Σ 1/C_i, less than smallest capacitor.
• Parallel: C_p = Σ C_i, sum of all capacitances.
• Charge same in series, voltage same in parallel.

Capacitors in Series

• Connected end-to-end; charge path is single.
• Charge Q on each capacitor is the same.
• Total voltage V is sum of voltages across capacitors: V = V₁ + V₂ + V₃ + ...
• Equivalent capacitance C_s given by:
  [ 1/C_s = 1/C₁ + 1/C₂ + 1/C₃ + ... ]
• Voltage divides inversely with capacitance: smaller capacitance gets higher voltage.
• Equivalent capacitance is always less than the smallest individual capacitance.
• For n identical capacitors each of capacitance C, C_s = C/n.

Capacitors in Parallel

• Connected side-by-side; multiple charge paths.
• Voltage V across each capacitor is the same.
• Total charge Q is sum of individual charges: Q = Q₁ + Q₂ + Q₃ + ...
• Equivalent capacitance C_p is sum of capacitances:
  [ C_p = C₁ + C₂ + C₃ + ... ]
• Charge divides proportional to capacitance values.

Solved Examples

Ex 1

• Two capacitors C₁ = 4 μF and C₂ = 6 μF in series. Find C_s.
• Use formula: 1/C_s = 1/4 + 1/6 = 5/12 ⇒ C_s = 12/5 = 2.4 μF.

Ex 2

• Two capacitors C₁ = 3 μF and C₂ = 5 μF in parallel. Find C_p.
• Use formula: C_p = 3 + 5 = 8 μF.

Practice Set

Q1

• Two capacitors 2 μF and 3 μF are connected in series. What is the equivalent capacitance?
• A) 5 μF
• B) 1.2 μF
• C) 6 μF
• D) 0.6 μF

Q2

• Three capacitors 1 μF, 2 μF, and 3 μF are connected in parallel. Find the total capacitance.
• A) 6 μF
• B) 0.6 μF
• C) 1.8 μF
• D) 3 μF

Q3

• Four identical capacitors each of 4 μF are connected in series. What is the equivalent capacitance?
• A) 16 μF
• B) 1 μF
• C) 4 μF
• D) 8 μF

Answer Key

• Q1: B) 1.2 μF - Series formula 1/C_s = 1/2 + 1/3 = 5/6 ⇒ C_s = 1.2 μF.
• Q2: A) 6 μF - Parallel sum C_p = 1 + 2 + 3 = 6 μF.
• Q3: B) 1 μF - Series identical capacitors C_s = C/n = 4/4 = 1 μF.

Memory Aids

• Series: 1/C_s = Σ 1/C_i, less than smallest capacitor.
• Parallel: C_p = Σ C_i, sum of all capacitances.
• Charge same in series, voltage same in parallel.

Capacitors in Series

• Connected end-to-end; charge path is single.
• Charge Q on each capacitor is the same.
• Total voltage V is sum of voltages across capacitors: V = V₁ + V₂ + V₃ + ...
• Equivalent capacitance C_s given by:
  [ 1/C_s = 1/C₁ + 1/C₂ + 1/C₃ + ... ]
• Voltage divides inversely with capacitance: smaller capacitance gets higher voltage.
• Equivalent capacitance is always less than the smallest individual capacitance.
• For n identical capacitors each of capacitance C, C_s = C/n.

Capacitors in Parallel

• Connected side-by-side; multiple charge paths.
• Voltage V across each capacitor is the same.
• Total charge Q is sum of individual charges: Q = Q₁ + Q₂ + Q₃ + ...
• Equivalent capacitance C_p is sum of capacitances:
  [ C_p = C₁ + C₂ + C₃ + ... ]
• Charge divides proportional to capacitance values.

Solved Examples

Ex 1

• Two capacitors C₁ = 4 μF and C₂ = 6 μF in series. Find C_s.
• Use formula: 1/C_s = 1/4 + 1/6 = 5/12 ⇒ C_s = 12/5 = 2.4 μF.

Ex 2

• Two capacitors C₁ = 3 μF and C₂ = 5 μF in parallel. Find C_p.
• Use formula: C_p = 3 + 5 = 8 μF.

Practice Set

Q1

• Two capacitors 2 μF and 3 μF are connected in series. What is the equivalent capacitance?
• A) 5 μF
• B) 1.2 μF
• C) 6 μF
• D) 0.6 μF

Q2

• Three capacitors 1 μF, 2 μF, and 3 μF are connected in parallel. Find the total capacitance.
• A) 6 μF
• B) 0.6 μF
• C) 1.8 μF
• D) 3 μF

Q3

• Four identical capacitors each of 4 μF are connected in series. What is the equivalent capacitance?
• A) 16 μF
• B) 1 μF
• C) 4 μF
• D) 8 μF

Answer Key

• Q1: B) 1.2 μF - Series formula 1/C_s = 1/2 + 1/3 = 5/6 ⇒ C_s = 1.2 μF.
• Q2: A) 6 μF - Parallel sum C_p = 1 + 2 + 3 = 6 μF.
• Q3: B) 1 μF - Series identical capacitors C_s = C/n = 4/4 = 1 μF.

Memory Aids

• Series: 1/C_s = Σ 1/C_i, less than smallest capacitor.
• Parallel: C_p = Σ C_i, sum of all capacitances.
• Charge same in series, voltage same in parallel.

Capacitors in Series

• Connected end-to-end; charge path is single.
• Charge Q on each capacitor is the same.
• Total voltage V is sum of voltages across capacitors: V = V₁ + V₂ + V₃ + ...
• Equivalent capacitance C_s given by:
  [ 1/C_s = 1/C₁ + 1/C₂ + 1/C₃ + ... ]
• Voltage divides inversely with capacitance: smaller capacitance gets higher voltage.
• Equivalent capacitance is always less than the smallest individual capacitance.
• For n identical capacitors each of capacitance C, C_s = C/n.

Capacitors in Parallel

• Connected side-by-side; multiple charge paths.
• Voltage V across each capacitor is the same.
• Total charge Q is sum of individual charges: Q = Q₁ + Q₂ + Q₃ + ...
• Equivalent capacitance C_p is sum of capacitances:
  [ C_p = C₁ + C₂ + C₃ + ... ]
• Charge divides proportional to capacitance values.

Solved Examples

Ex 1

• Two capacitors C₁ = 4 μF and C₂ = 6 μF in series. Find C_s.
• Use formula: 1/C_s = 1/4 + 1/6 = 5/12 ⇒ C_s = 12/5 = 2.4 μF.

Ex 2

• Two capacitors C₁ = 3 μF and C₂ = 5 μF in parallel. Find C_p.
• Use formula: C_p = 3 + 5 = 8 μF.

Practice Set

Q1

• Two capacitors 2 μF and 3 μF are connected in series. What is the equivalent capacitance?
• A) 5 μF
• B) 1.2 μF
• C) 6 μF
• D) 0.6 μF

Q2

• Three capacitors 1 μF, 2 μF, and 3 μF are connected in parallel. Find the total capacitance.
• A) 6 μF
• B) 0.6 μF
• C) 1.8 μF
• D) 3 μF

Q3

• Four identical capacitors each of 4 μF are connected in series. What is the equivalent capacitance?
• A) 16 μF
• B) 1 μF
• C) 4 μF
• D) 8 μF

Answer Key

• Q1: B) 1.2 μF - Series formula 1/C_s = 1/2 + 1/3 = 5/6 ⇒ C_s = 1.2 μF.
• Q2: A) 6 μF - Parallel sum C_p = 1 + 2 + 3 = 6 μF.
• Q3: B) 1 μF - Series identical capacitors C_s = C/n = 4/4 = 1 μF.

Memory Aids

• Series: 1/C_s = Σ 1/C_i, less than smallest capacitor.
• Parallel: C_p = Σ C_i, sum of all capacitances.
• Charge same in series, voltage same in parallel.

Capacitors in Series

• Connected end-to-end; charge path is single.
• Charge Q on each capacitor is the same.
• Total voltage V is sum of voltages across capacitors: V = V₁ + V₂ + V₃ + ...
• Equivalent capacitance C_s given by:
  [ 1/C_s = 1/C₁ + 1/C₂ + 1/C₃ + ... ]
• Voltage divides inversely with capacitance: smaller capacitance gets higher voltage.
• Equivalent capacitance is always less than the smallest individual capacitance.
• For n identical capacitors each of capacitance C, C_s = C/n.

Capacitors in Parallel

• Connected side-by-side; multiple charge paths.
• Voltage V across each capacitor is the same.
• Total charge Q is sum of individual charges: Q = Q₁ + Q₂ + Q₃ + ...
• Equivalent capacitance C_p is sum of capacitances:
  [ C_p = C₁ + C₂ + C₃ + ... ]
• Charge divides proportional to capacitance values.

Solved Examples

Ex 1

• Two capacitors C₁ = 4 μF and C₂ = 6 μF in series. Find C_s.
• Use formula: 1/C_s = 1/4 + 1/6 = 5/12 ⇒ C_s = 12/5 = 2.4 μF.

Ex 2

• Two capacitors C₁ = 3 μF and C₂ = 5 μF in parallel. Find C_p.
• Use formula: C_p = 3 + 5 = 8 μF.

Practice Set

Q1

• Two capacitors 2 μF and 3 μF are connected in series. What is the equivalent capacitance?
• A) 5 μF
• B) 1.2 μF
• C) 6 μF
• D) 0.6 μF

Q2

• Three capacitors 1 μF, 2 μF, and 3 μF are connected in parallel. Find the total capacitance.
• A) 6 μF
• B) 0.6 μF
• C) 1.8 μF
• D) 3 μF

Q3

• Four identical capacitors each of 4 μF are connected in series. What is the equivalent capacitance?
• A) 16 μF
• B) 1 μF
• C) 4 μF
• D) 8 μF

Answer Key

• Q1: B) 1.2 μF - Series formula 1/C_s = 1/2 + 1/3 = 5/6 ⇒ C_s = 1.2 μF.
• Q2: A) 6 μF - Parallel sum C_p = 1 + 2 + 3 = 6 μF.
• Q3: B) 1 μF - Series identical capacitors C_s = C/n = 4/4 = 1 μF.

Memory Aids

• Series: 1/C_s = Σ 1/C_i, less than smallest capacitor.
• Parallel: C_p = Σ C_i, sum of all capacitances.
• Charge same in series, voltage same in parallel.

Capacitors in Series

• Connected end-to-end; charge path is single.
• Charge \( Q \) on each capacitor is the same.
• Total voltage \( V \) is sum of voltages across capacitors: \( V = V_1 + V_2 + V_3 + \dots \)
• Equivalent capacitance \( C_s \) given by:
  \[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots \]
• Voltage divides inversely with capacitance: smaller capacitance gets higher voltage.
• Equivalent capacitance is always less than the smallest individual capacitance.
• For \( n \) identical capacitors each of capacitance \( C \), \( C_s = \frac{C}{n} \).

Capacitors in Parallel

• Connected side-by-side; multiple charge paths.
• Voltage \( V \) across each capacitor is the same.
• Total charge \( Q \) is sum of individual charges: \( Q = Q_1 + Q_2 + Q_3 + \dots \)
• Equivalent capacitance \( C_p \) is sum of capacitances:
  \[ C_p = C_1 + C_2 + C_3 + \dots \]
• Charge divides proportional to capacitance values.

Solved Examples

Ex 1

• Two capacitors \( C_1 = 4 \mu F \) and \( C_2 = 6 \mu F \) in series. Find \( C_s \).
• Use formula: \( \frac{1}{C_s} = \frac{1}{4} + \frac{1}{6} = \frac{5}{12} \Rightarrow C_s = \frac{12}{5} = 2.4 \mu F \).

Ex 2

• Two capacitors \( C_1 = 3 \mu F \) and \( C_2 = 5 \mu F \) in parallel. Find \( C_p \).
• Use formula: \( C_p = 3 + 5 = 8 \mu F \).

Practice Set

Q1

• Two capacitors \( 2 \mu F \) and \( 3 \mu F \) are connected in series. What is the equivalent capacitance?
• A) 5 \( \mu F \)
• B) 1.2 \( \mu F \)
• C) 6 \( \mu F \)
• D) 0.6 \( \mu F \)

Q2

• Three capacitors \( 1 \mu F \), \( 2 \mu F \), and \( 3 \mu F \) are connected in parallel. Find the total capacitance.
• A) 6 \( \mu F \)
• B) 0.6 \( \mu F \)
• C) 1.8 \( \mu F \)
• D) 3 \( \mu F \)

Q3

• Four identical capacitors each of \( 4 \mu F \) are connected in series. What is the equivalent capacitance?
• A) 16 \( \mu F \)
• B) 1 \( \mu F \)
• C) 4 \( \mu F \)
• D) 8 \( \mu F \)

Answer Key

• Q1: B) 1.2 \( \mu F \) - Series formula \( \frac{1}{C_s} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \Rightarrow C_s = 1.2 \mu F \).
• Q2: A) 6 \( \mu F \) - Parallel sum \( C_p = 1 + 2 + 3 = 6 \mu F \).
• Q3: B) 1 \( \mu F \) - Series identical capacitors \( C_s = \frac{C}{n} = \frac{4}{4} = 1 \mu F \).

Memory Aids

• Series: \( \frac{1}{C_s} = \sum \frac{1}{C_i} \), less than smallest capacitor.
• Parallel: \( C_p = \sum C_i \), sum of all capacitances.
• Charge same in series, voltage same in parallel.

Series Combination

Key Points

• Capacitors connected one after another.
• Charge Q on each capacitor is the same.
• Total voltage V = V₁ + V₂ + ...
• Voltage across each capacitor: V_i = Q/C_i
• Equivalent capacitance for two capacitors:
  [ 1/C_eq = 1/C₁ + 1/C₂ ]
• General formula for n capacitors:
  [ 1/C_eq = Σ_i=1ⁿ 1/C_i ]
• For identical capacitors: C_eq = C/n
• Equivalent capacitance is always less than the smallest capacitor.

Solved Examples

Ex 1

• Two capacitors 5 μF and 10 μF in series. Find C_eq.
• 1/C_eq = 1/5 + 1/10 = 3/10 ⇒ C_eq = 10/3 = 3.33 μF.

Practice Set

Q1

• Two capacitors 3 μF and 6 μF in series. Find C_eq.
• A) 9 μF
• B) 2 μF
• C) 1.5 μF
• D) 18 μF

Q2

• Three capacitors 2 μF, 4 μF, and 6 μF in series. Find C_eq.
• A) 1 μF
• B) 12 μF
• C) 3 μF
• D) 0.5 μF

Q3

• Four identical capacitors each 8 μF in series. Find C_eq.
• A) 32 μF
• B) 2 μF
• C) 8 μF
• D) 4 μF

Answer Key

• Q1: B) 2 μF - 1/C_eq = 1/3 + 1/6 = 1/2 ⇒ C_eq = 2 μF.
• Q2: A) 1 μF - 1/C_eq = 1/2 + 1/4 + 1/6 = 6+3+2/12 = 11/12 ⇒ C_eq ≈ 1.09 μF (closest 1).
• Q3: B) 2 μF - C_eq = 8/4 = 2 μF.

Memory Aids

• Series: 1/C_eq = Σ 1/C_i, less than smallest capacitor.
• Charge same on all capacitors.
• Voltage divides inversely with capacitance.

Key Points

• Capacitors connected one after another.
• Charge Q on each capacitor is the same.
• Total voltage V = V₁ + V₂ + ...
• Voltage across each capacitor: V_i = Q/C_i
• Equivalent capacitance for two capacitors:
  [ 1/C_eq = 1/C₁ + 1/C₂ ]
• General formula for n capacitors:
  [ 1/C_eq = Σ_i=1ⁿ 1/C_i ]
• For identical capacitors: C_eq = C/n
• Equivalent capacitance is always less than the smallest capacitor.

Solved Examples

Ex 1

• Two capacitors 5 μF and 10 μF in series. Find C_eq.
• 1/C_eq = 1/5 + 1/10 = 3/10 ⇒ C_eq = 10/3 = 3.33 μF.

Practice Set

Q1

• Two capacitors 3 μF and 6 μF in series. Find C_eq.
• A) 9 μF
• B) 2 μF
• C) 1.5 μF
• D) 18 μF

Q2

• Three capacitors 2 μF, 4 μF, and 6 μF in series. Find C_eq.
• A) 1 μF
• B) 12 μF
• C) 3 μF
• D) 0.5 μF

Q3

• Four identical capacitors each 8 μF in series. Find C_eq.
• A) 32 μF
• B) 2 μF
• C) 8 μF
• D) 4 μF

Answer Key

• Q1: B) 2 μF - 1/C_eq = 1/3 + 1/6 = 1/2 ⇒ C_eq = 2 μF.
• Q2: A) 1 μF - 1/C_eq = 1/2 + 1/4 + 1/6 = 6+3+2/12 = 11/12 ⇒ C_eq ≈ 1.09 μF (closest 1).
• Q3: B) 2 μF - C_eq = 8/4 = 2 μF.

Memory Aids

• Series: 1/C_eq = Σ 1/C_i, less than smallest capacitor.
• Charge same on all capacitors.
• Voltage divides inversely with capacitance.

Key Points

• Capacitors connected one after another.
• Charge Q on each capacitor is the same.
• Total voltage V = V₁ + V₂ + ...
• Voltage across each capacitor: V_i = Q/C_i
• Equivalent capacitance for two capacitors:
  [ 1/C_eq = 1/C₁ + 1/C₂ ]
• General formula for n capacitors:
  [ 1/C_eq = Σ_i=1ⁿ 1/C_i ]
• For identical capacitors: C_eq = C/n
• Equivalent capacitance is always less than the smallest capacitor.

Solved Examples

Ex 1

• Two capacitors 5 μF and 10 μF in series. Find C_eq.
• 1/C_eq = 1/5 + 1/10 = 3/10 ⇒ C_eq = 10/3 = 3.33 μF.

Practice Set

Q1

• Two capacitors 3 μF and 6 μF in series. Find C_eq.
• A) 9 μF
• B) 2 μF
• C) 1.5 μF
• D) 18 μF

Q2

• Three capacitors 2 μF, 4 μF, and 6 μF in series. Find C_eq.
• A) 1 μF
• B) 12 μF
• C) 3 μF
• D) 0.5 μF

Q3

• Four identical capacitors each 8 μF in series. Find C_eq.
• A) 32 μF
• B) 2 μF
• C) 8 μF
• D) 4 μF

Answer Key

• Q1: B) 2 μF - 1/C_eq = 1/3 + 1/6 = 1/2 ⇒ C_eq = 2 μF.
• Q2: A) 1 μF - 1/C_eq = 1/2 + 1/4 + 1/6 = 6+3+2/12 = 11/12 ⇒ C_eq ≈ 1.09 μF (closest 1).
• Q3: B) 2 μF - C_eq = 8/4 = 2 μF.

Memory Aids

• Series: 1/C_eq = Σ 1/C_i, less than smallest capacitor.
• Charge same on all capacitors.
• Voltage divides inversely with capacitance.

Key Points

• Capacitors connected one after another.
• Charge Q on each capacitor is the same.
• Total voltage V = V₁ + V₂ + ...
• Voltage across each capacitor: V_i = Q/C_i
• Equivalent capacitance for two capacitors:
  [ 1/C_eq = 1/C₁ + 1/C₂ ]
• General formula for n capacitors:
  [ 1/C_eq = Σ_i=1ⁿ 1/C_i ]
• For identical capacitors: C_eq = C/n
• Equivalent capacitance is always less than the smallest capacitor.

Solved Examples

Ex 1

• Two capacitors 5 μF and 10 μF in series. Find C_eq.
• 1/C_eq = 1/5 + 1/10 = 3/10 ⇒ C_eq = 10/3 = 3.33 μF.

Practice Set

Q1

• Two capacitors 3 μF and 6 μF in series. Find C_eq.
• A) 9 μF
• B) 2 μF
• C) 1.5 μF
• D) 18 μF

Q2

• Three capacitors 2 μF, 4 μF, and 6 μF in series. Find C_eq.
• A) 1 μF
• B) 12 μF
• C) 3 μF
• D) 0.5 μF

Q3

• Four identical capacitors each 8 μF in series. Find C_eq.
• A) 32 μF
• B) 2 μF
• C) 8 μF
• D) 4 μF

Answer Key

• Q1: B) 2 μF - 1/C_eq = 1/3 + 1/6 = 1/2 ⇒ C_eq = 2 μF.
• Q2: A) 1 μF - 1/C_eq = 1/2 + 1/4 + 1/6 = 6+3+2/12 = 11/12 ⇒ C_eq ≈ 1.09 μF (closest 1).
• Q3: B) 2 μF - C_eq = 8/4 = 2 μF.

Memory Aids

• Series: 1/C_eq = Σ 1/C_i, less than smallest capacitor.
• Charge same on all capacitors.
• Voltage divides inversely with capacitance.

Key Points

• Capacitors connected side-by-side.
• Voltage V across each capacitor is the same.
• Charge on each capacitor: Q_i = C_iV.
• Total charge Q = ΣQ_i = V ΣC_i.
• Equivalent capacitance:
  [ C_eq = ΣC_i ]
• Charge divides according to capacitance values.

Solved Examples

Ex 1

• Two capacitors 2 μF and 5 μF in parallel. Find C_eq.
• C_eq = 2 + 5 = 7 μF.

Practice Set

Q1

• Two capacitors 1 μF and 4 μF in parallel. Find C_eq.
• A) 5 μF
• B) 0.8 μF
• C) 3 μF
• D) 4 μF

Q2

• Three capacitors 3 μF, 6 μF, and 9 μF in parallel. Find C_eq.
• A) 18 μF
• B) 6 μF
• C) 9 μF
• D) 3 μF

Q3

• Four capacitors 2 μF, 2 μF, 2 μF, and 2 μF in parallel. Find C_eq.
• A) 8 μF
• B) 2 μF
• C) 4 μF
• D) 1 μF

Answer Key

• Q1: A) 5 μF - Sum of capacitances in parallel.
• Q2: A) 18 μF - 3 + 6 + 9 = 18 μF.
• Q3: A) 8 μF - Sum of four identical capacitors.

Memory Aids

• Parallel: C_eq = ΣC_i, sum of capacitances.
• Voltage same across all capacitors.
• Charge divides by capacitance.

Key Points

• Capacitors connected side-by-side.
• Voltage V across each capacitor is the same.
• Charge on each capacitor: Q_i = C_iV.
• Total charge Q = ΣQ_i = V ΣC_i.
• Equivalent capacitance:
  [ C_eq = ΣC_i ]
• Charge divides according to capacitance values.

Solved Examples

Ex 1

• Two capacitors 2 μF and 5 μF in parallel. Find C_eq.
• C_eq = 2 + 5 = 7 μF.

Practice Set

Q1

• Two capacitors 1 μF and 4 μF in parallel. Find C_eq.
• A) 5 μF
• B) 0.8 μF
• C) 3 μF
• D) 4 μF

Q2

• Three capacitors 3 μF, 6 μF, and 9 μF in parallel. Find C_eq.
• A) 18 μF
• B) 6 μF
• C) 9 μF
• D) 3 μF

Q3

• Four capacitors 2 μF, 2 μF, 2 μF, and 2 μF in parallel. Find C_eq.
• A) 8 μF
• B) 2 μF
• C) 4 μF
• D) 1 μF

Answer Key

• Q1: A) 5 μF - Sum of capacitances in parallel.
• Q2: A) 18 μF - 3 + 6 + 9 = 18 μF.
• Q3: A) 8 μF - Sum of four identical capacitors.

Memory Aids

• Parallel: C_eq = ΣC_i, sum of capacitances.
• Voltage same across all capacitors.
• Charge divides by capacitance.

Key Points

• Capacitors connected side-by-side.
• Voltage V across each capacitor is the same.
• Charge on each capacitor: Q_i = C_iV.
• Total charge Q = ΣQ_i = V ΣC_i.
• Equivalent capacitance:
  [ C_eq = ΣC_i ]
• Charge divides according to capacitance values.

Solved Examples

Ex 1

• Two capacitors 2 μF and 5 μF in parallel. Find C_eq.
• C_eq = 2 + 5 = 7 μF.

Practice Set

Q1

• Two capacitors 1 μF and 4 μF in parallel. Find C_eq.
• A) 5 μF
• B) 0.8 μF
• C) 3 μF
• D) 4 μF

Q2

• Three capacitors 3 μF, 6 μF, and 9 μF in parallel. Find C_eq.
• A) 18 μF
• B) 6 μF
• C) 9 μF
• D) 3 μF

Q3

• Four capacitors 2 μF, 2 μF, 2 μF, and 2 μF in parallel. Find C_eq.
• A) 8 μF
• B) 2 μF
• C) 4 μF
• D) 1 μF

Answer Key

• Q1: A) 5 μF - Sum of capacitances in parallel.
• Q2: A) 18 μF - 3 + 6 + 9 = 18 μF.
• Q3: A) 8 μF - Sum of four identical capacitors.

Memory Aids

• Parallel: C_eq = ΣC_i, sum of capacitances.
• Voltage same across all capacitors.
• Charge divides by capacitance.

Key Points

• Capacitors connected side-by-side.
• Voltage V across each capacitor is the same.
• Charge on each capacitor: Q_i = C_iV.
• Total charge Q = ΣQ_i = V ΣC_i.
• Equivalent capacitance:
  [ C_eq = ΣC_i ]
• Charge divides according to capacitance values.

Solved Examples

Ex 1

• Two capacitors 2 μF and 5 μF in parallel. Find C_eq.
• C_eq = 2 + 5 = 7 μF.

Practice Set

Q1

• Two capacitors 1 μF and 4 μF in parallel. Find C_eq.
• A) 5 μF
• B) 0.8 μF
• C) 3 μF
• D) 4 μF

Q2

• Three capacitors 3 μF, 6 μF, and 9 μF in parallel. Find C_eq.
• A) 18 μF
• B) 6 μF
• C) 9 μF
• D) 3 μF

Q3

• Four capacitors 2 μF, 2 μF, 2 μF, and 2 μF in parallel. Find C_eq.
• A) 8 μF
• B) 2 μF
• C) 4 μF
• D) 1 μF

Answer Key

• Q1: A) 5 μF - Sum of capacitances in parallel.
• Q2: A) 18 μF - 3 + 6 + 9 = 18 μF.
• Q3: A) 8 μF - Sum of four identical capacitors.

Memory Aids

• Parallel: C_eq = ΣC_i, sum of capacitances.
• Voltage same across all capacitors.
• Charge divides by capacitance.

Key Points

• Capacitors connected one after another.
• Charge Q on each capacitor is the same.
• Total voltage V = V₁ + V₂ + ...
• Voltage across each capacitor: V_i = Q/C_i
• Equivalent capacitance for two capacitors:
  [ 1/C_eq = 1/C₁ + 1/C₂ ]
• General formula for n capacitors:
  [ 1/C_eq = Σ_i=1ⁿ 1/C_i ]
• For identical capacitors: C_eq = C/n
• Equivalent capacitance is always less than the smallest capacitor.

Solved Examples

Ex 1

• Two capacitors 5 μF and 10 μF in series. Find C_eq.
• 1/C_eq = 1/5 + 1/10 = 3/10 ⇒ C_eq = 10/3 = 3.33 μF.

Practice Set

Q1

• Two capacitors 3 μF and 6 μF in series. Find C_eq.
• A) 9 μF
• B) 2 μF
• C) 1.5 μF
• D) 18 μF

Q2

• Three capacitors 2 μF, 4 μF, and 6 μF in series. Find C_eq.
• A) 1 μF
• B) 12 μF
• C) 3 μF
• D) 0.5 μF

Q3

• Four identical capacitors each 8 μF in series. Find C_eq.
• A) 32 μF
• B) 2 μF
• C) 8 μF
• D) 4 μF

Answer Key

• Q1: B) 2 μF - 1/C_eq = 1/3 + 1/6 = 1/2 ⇒ C_eq = 2 μF.
• Q2: A) 1 μF - 1/C_eq = 1/2 + 1/4 + 1/6 = 6+3+2/12 = 11/12 ⇒ C_eq ≈ 1.09 μF (closest 1).
• Q3: B) 2 μF - C_eq = 8/4 = 2 μF.

Memory Aids

• Series: 1/C_eq = Σ 1/C_i, less than smallest capacitor.
• Charge same on all capacitors.
• Voltage divides inversely with capacitance.

Key Points

• Capacitors connected one after another.
• Charge Q on each capacitor is the same.
• Total voltage V = V₁ + V₂ + ...
• Voltage across each capacitor: V_i = Q/C_i
• Equivalent capacitance for two capacitors:
  [ 1/C_eq = 1/C₁ + 1/C₂ ]
• General formula for n capacitors:
  [ 1/C_eq = Σ_i=1ⁿ 1/C_i ]
• For identical capacitors: C_eq = C/n
• Equivalent capacitance is always less than the smallest capacitor.

Solved Examples

Ex 1

• Two capacitors 5 μF and 10 μF in series. Find C_eq.
• 1/C_eq = 1/5 + 1/10 = 3/10 ⇒ C_eq = 10/3 = 3.33 μF.

Practice Set

Q1

• Two capacitors 3 μF and 6 μF in series. Find C_eq.
• A) 9 μF
• B) 2 μF
• C) 1.5 μF
• D) 18 μF

Q2

• Three capacitors 2 μF, 4 μF, and 6 μF in series. Find C_eq.
• A) 1 μF
• B) 12 μF
• C) 3 μF
• D) 0.5 μF

Q3

• Four identical capacitors each 8 μF in series. Find C_eq.
• A) 32 μF
• B) 2 μF
• C) 8 μF
• D) 4 μF

Answer Key

• Q1: B) 2 μF - 1/C_eq = 1/3 + 1/6 = 1/2 ⇒ C_eq = 2 μF.
• Q2: A) 1 μF - 1/C_eq = 1/2 + 1/4 + 1/6 = 6+3+2/12 = 11/12 ⇒ C_eq ≈ 1.09 μF (closest 1).
• Q3: B) 2 μF - C_eq = 8/4 = 2 μF.

Memory Aids

• Series: 1/C_eq = Σ 1/C_i, less than smallest capacitor.
• Charge same on all capacitors.
• Voltage divides inversely with capacitance.

Key Points

• Capacitors connected one after another.
• Charge Q on each capacitor is the same.
• Total voltage V = V₁ + V₂ + ...
• Voltage across each capacitor: V_i = Q/C_i
• Equivalent capacitance for two capacitors:
  [ 1/C_eq = 1/C₁ + 1/C₂ ]
• General formula for n capacitors:
  [ 1/C_eq = Σ_i=1ⁿ 1/C_i ]
• For identical capacitors: C_eq = C/n
• Equivalent capacitance is always less than the smallest capacitor.

Solved Examples

Ex 1

• Two capacitors 5 μF and 10 μF in series. Find C_eq.
• 1/C_eq = 1/5 + 1/10 = 3/10 ⇒ C_eq = 10/3 = 3.33 μF.

Practice Set

Q1

• Two capacitors 3 μF and 6 μF in series. Find C_eq.
• A) 9 μF
• B) 2 μF
• C) 1.5 μF
• D) 18 μF

Q2

• Three capacitors 2 μF, 4 μF, and 6 μF in series. Find C_eq.
• A) 1 μF
• B) 12 μF
• C) 3 μF
• D) 0.5 μF

Q3

• Four identical capacitors each 8 μF in series. Find C_eq.
• A) 32 μF
• B) 2 μF
• C) 8 μF
• D) 4 μF

Answer Key

• Q1: B) 2 μF - 1/C_eq = 1/3 + 1/6 = 1/2 ⇒ C_eq = 2 μF.
• Q2: A) 1 μF - 1/C_eq = 1/2 + 1/4 + 1/6 = 6+3+2/12 = 11/12 ⇒ C_eq ≈ 1.09 μF (closest 1).
• Q3: B) 2 μF - C_eq = 8/4 = 2 μF.

Memory Aids

• Series: 1/C_eq = Σ 1/C_i, less than smallest capacitor.
• Charge same on all capacitors.
• Voltage divides inversely with capacitance.

Key Points

• Capacitors connected one after another.
• Charge \( Q \) on each capacitor is the same.
• Total voltage \( V = V_1 + V_2 + \dots \)
• Voltage across each capacitor: \( V_i = \frac{Q}{C_i} \)
• Equivalent capacitance for two capacitors:
  \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]
• General formula for \( n \) capacitors:
  \[ \frac{1}{C_{eq}} = \sum_{i=1}^n \frac{1}{C_i} \]
• For identical capacitors: \( C_{eq} = \frac{C}{n} \)
• Equivalent capacitance is always less than the smallest capacitor.

Solved Examples

Ex 1

• Two capacitors \( 5 \mu F \) and \( 10 \mu F \) in series. Find \( C_{eq} \).
• \( \frac{1}{C_{eq}} = \frac{1}{5} + \frac{1}{10} = \frac{3}{10} \Rightarrow C_{eq} = \frac{10}{3} = 3.33 \mu F \).

Practice Set

Q1

• Two capacitors \( 3 \mu F \) and \( 6 \mu F \) in series. Find \( C_{eq} \).
• A) 9 \( \mu F \)
• B) 2 \( \mu F \)
• C) 1.5 \( \mu F \)
• D) 18 \( \mu F \)

Q2

• Three capacitors \( 2 \mu F \), \( 4 \mu F \), and \( 6 \mu F \) in series. Find \( C_{eq} \).
• A) 1 \( \mu F \)
• B) 12 \( \mu F \)
• C) 3 \( \mu F \)
• D) 0.5 \( \mu F \)

Q3

• Four identical capacitors each \( 8 \mu F \) in series. Find \( C_{eq} \).
• A) 32 \( \mu F \)
• B) 2 \( \mu F \)
• C) 8 \( \mu F \)
• D) 4 \( \mu F \)

Answer Key

• Q1: B) 2 \( \mu F \) - \( \frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2} \Rightarrow C_{eq} = 2 \mu F \).
• Q2: A) 1 \( \mu F \) - \( \frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} = \frac{6+3+2}{12} = \frac{11}{12} \Rightarrow C_{eq} \approx 1.09 \mu F \) (closest 1).
• Q3: B) 2 \( \mu F \) - \( C_{eq} = \frac{8}{4} = 2 \mu F \).

Memory Aids

• Series: \( \frac{1}{C_{eq}} = \sum \frac{1}{C_i} \), less than smallest capacitor.
• Charge same on all capacitors.
• Voltage divides inversely with capacitance.

Parallel Combination

Key Points

• Capacitors connected side-by-side.
• Voltage V across each capacitor is the same.
• Charge on each capacitor: Q_i = C_iV.
• Total charge Q = ΣQ_i = V ΣC_i.
• Equivalent capacitance:
  [ C_eq = ΣC_i ]
• Charge divides according to capacitance values.

Solved Examples

Ex 1

• Two capacitors 2 μF and 5 μF in parallel. Find C_eq.
• C_eq = 2 + 5 = 7 μF.

Practice Set

Q1

• Two capacitors 1 μF and 4 μF in parallel. Find C_eq.
• A) 5 μF
• B) 0.8 μF
• C) 3 μF
• D) 4 μF

Q2

• Three capacitors 3 μF, 6 μF, and 9 μF in parallel. Find C_eq.
• A) 18 μF
• B) 6 μF
• C) 9 μF
• D) 3 μF

Q3

• Four capacitors 2 μF, 2 μF, 2 μF, and 2 μF in parallel. Find C_eq.
• A) 8 μF
• B) 2 μF
• C) 4 μF
• D) 1 μF

Answer Key

• Q1: A) 5 μF - Sum of capacitances in parallel.
• Q2: A) 18 μF - 3 + 6 + 9 = 18 μF.
• Q3: A) 8 μF - Sum of four identical capacitors.

Memory Aids

• Parallel: C_eq = ΣC_i, sum of capacitances.
• Voltage same across all capacitors.
• Charge divides by capacitance.

Key Points

• Capacitors connected side-by-side.
• Voltage V across each capacitor is the same.
• Charge on each capacitor: Q_i = C_iV.
• Total charge Q = ΣQ_i = V ΣC_i.
• Equivalent capacitance:
  [ C_eq = ΣC_i ]
• Charge divides according to capacitance values.

Solved Examples

Ex 1

• Two capacitors 2 μF and 5 μF in parallel. Find C_eq.
• C_eq = 2 + 5 = 7 μF.

Practice Set

Q1

• Two capacitors 1 μF and 4 μF in parallel. Find C_eq.
• A) 5 μF
• B) 0.8 μF
• C) 3 μF
• D) 4 μF

Q2

• Three capacitors 3 μF, 6 μF, and 9 μF in parallel. Find C_eq.
• A) 18 μF
• B) 6 μF
• C) 9 μF
• D) 3 μF

Q3

• Four capacitors 2 μF, 2 μF, 2 μF, and 2 μF in parallel. Find C_eq.
• A) 8 μF
• B) 2 μF
• C) 4 μF
• D) 1 μF

Answer Key

• Q1: A) 5 μF - Sum of capacitances in parallel.
• Q2: A) 18 μF - 3 + 6 + 9 = 18 μF.
• Q3: A) 8 μF - Sum of four identical capacitors.

Memory Aids

• Parallel: C_eq = ΣC_i, sum of capacitances.
• Voltage same across all capacitors.
• Charge divides by capacitance.

Key Points

• Capacitors connected side-by-side.
• Voltage V across each capacitor is the same.
• Charge on each capacitor: Q_i = C_iV.
• Total charge Q = ΣQ_i = V ΣC_i.
• Equivalent capacitance:
  [ C_eq = ΣC_i ]
• Charge divides according to capacitance values.

Solved Examples

Ex 1

• Two capacitors 2 μF and 5 μF in parallel. Find C_eq.
• C_eq = 2 + 5 = 7 μF.

Practice Set

Q1

• Two capacitors 1 μF and 4 μF in parallel. Find C_eq.
• A) 5 μF
• B) 0.8 μF
• C) 3 μF
• D) 4 μF

Q2

• Three capacitors 3 μF, 6 μF, and 9 μF in parallel. Find C_eq.
• A) 18 μF
• B) 6 μF
• C) 9 μF
• D) 3 μF

Q3

• Four capacitors 2 μF, 2 μF, 2 μF, and 2 μF in parallel. Find C_eq.
• A) 8 μF
• B) 2 μF
• C) 4 μF
• D) 1 μF

Answer Key

• Q1: A) 5 μF - Sum of capacitances in parallel.
• Q2: A) 18 μF - 3 + 6 + 9 = 18 μF.
• Q3: A) 8 μF - Sum of four identical capacitors.

Memory Aids

• Parallel: C_eq = ΣC_i, sum of capacitances.
• Voltage same across all capacitors.
• Charge divides by capacitance.

Key Points

• Capacitors connected side-by-side.
• Voltage \( V \) across each capacitor is the same.
• Charge on each capacitor: \( Q_i = C_i V \).
• Total charge \( Q = \sum Q_i = V \sum C_i \).
• Equivalent capacitance:
  \[ C_{eq} = \sum C_i \]
• Charge divides according to capacitance values.

Solved Examples

Ex 1

• Two capacitors \( 2 \mu F \) and \( 5 \mu F \) in parallel. Find \( C_{eq} \).
• \( C_{eq} = 2 + 5 = 7 \mu F \).

Practice Set

Q1

• Two capacitors \( 1 \mu F \) and \( 4 \mu F \) in parallel. Find \( C_{eq} \).
• A) 5 \( \mu F \)
• B) 0.8 \( \mu F \)
• C) 3 \( \mu F \)
• D) 4 \( \mu F \)

Q2

• Three capacitors \( 3 \mu F \), \( 6 \mu F \), and \( 9 \mu F \) in parallel. Find \( C_{eq} \).
• A) 18 \( \mu F \)
• B) 6 \( \mu F \)
• C) 9 \( \mu F \)
• D) 3 \( \mu F \)

Q3

• Four capacitors \( 2 \mu F \), \( 2 \mu F \), \( 2 \mu F \), and \( 2 \mu F \) in parallel. Find \( C_{eq} \).
• A) 8 \( \mu F \)
• B) 2 \( \mu F \)
• C) 4 \( \mu F \)
• D) 1 \( \mu F \)

Answer Key

• Q1: A) 5 \( \mu F \) - Sum of capacitances in parallel.
• Q2: A) 18 \( \mu F \) - \( 3 + 6 + 9 = 18 \mu F \).
• Q3: A) 8 \( \mu F \) - Sum of four identical capacitors.

Memory Aids

• Parallel: \( C_{eq} = \sum C_i \), sum of capacitances.
• Voltage same across all capacitors.
• Charge divides by capacitance.

Energy Stored in a Capacitor

Concepts and Formulas

• If q is charge and V is potential difference at any instant, small work done to add charge dq is dW = V · dq = q/C dq.
• Total work done to store maximum charge Q is W = ∫₀^Q q/C dq = 1/2 Q²/C.
• Energy stored in capacitor (electrostatic energy):
  U = 1/2 Q²/C = 1/2 CV² = 1/2 QV.
• Energy density (energy per unit volume) in parallel plate capacitor:
  u = 1/2 ε₀ E², where ε₀ is permittivity of free space and E is electric field.
• Effect of dielectric:
  
  • With dielectric connected to battery: Capacitance C = K C₀, voltage V = V₀, electric field E = E₀, energy stored U = K U₀ (increases).
  • With dielectric removed (isolated capacitor): Charge Q = Q₀ constant, voltage V = V₀/K, electric field E = E₀/K, energy stored U = U₀/K (decreases).

Solved Examples

• Ex 1: Calculate energy stored in a capacitor of capacitance 5 μF charged to 10 V.
  Solution:
  U = 1/2 CV² = 1/2 × 5 × 10^-6 × 10² = 2.5 × 10^-4 J.
• Ex 2: Find energy density if electric field E = 2 × 10⁶ V/m.
  Solution:
  u = 1/2 ε₀ E² = 1/2 × 8.85 × 10^-12 × (2 × 10⁶)² = 1.77 × 10^-5 J/m³.

Practice Set

• Q1: Energy stored in a capacitor of 2 μF charged to 5 V is:
  A) 25 μJ B) 10 μJ B) 5 μJ C) 50 μJ
• Q2: Energy density in a capacitor depends on:
  A) Capacitance only B) Electric field only C) Voltage only D) Charge only
• Q3: A capacitor with dielectric constant 4 is connected to battery. Energy stored becomes:
  A) Same B) 4 times C) Half D) 2 times

Answer Key

• Q1: A) 25 μJ - Using U=1/2CV² = 1/2 × 2 × 10^-6 × 5² = 25 × 10^-6 J.
• Q2: B) Electric field only - Energy density formula u=1/2 ε₀ E² depends on electric field.
• Q3: B) 4 times - Energy stored increases by dielectric constant K when connected to battery.

Memory Aids

• Energy stored: U = 1/2 CV² or 1/2 QV.
• Energy density: u = 1/2 ε₀ E².
• Dielectric effect: Energy U scales by dielectric constant K if battery connected.

Concepts and Formulas

• If q is charge and V is potential difference at any instant, small work done to add charge dq is dW = V · dq = q/C dq.
• Total work done to store maximum charge Q is W = ∫₀^Q q/C dq = 1/2 Q²/C.
• Energy stored in capacitor (electrostatic energy):
  U = 1/2 Q²/C = 1/2 CV² = 1/2 QV.
• Energy density (energy per unit volume) in parallel plate capacitor:
  u = 1/2 ε₀ E², where ε₀ is permittivity of free space and E is electric field.
• Effect of dielectric:
  
  • With dielectric connected to battery: Capacitance C = K C₀, voltage V = V₀, electric field E = E₀, energy stored U = K U₀ (increases).
  • With dielectric removed (isolated capacitor): Charge Q = Q₀ constant, voltage V = V₀/K, electric field E = E₀/K, energy stored U = U₀/K (decreases).

Solved Examples

• Ex 1: Calculate energy stored in a capacitor of capacitance 5 μF charged to 10 V.
  Solution:
  U = 1/2 CV² = 1/2 × 5 × 10^-6 × 10² = 2.5 × 10^-4 J.
• Ex 2: Find energy density if electric field E = 2 × 10⁶ V/m.
  Solution:
  u = 1/2 ε₀ E² = 1/2 × 8.85 × 10^-12 × (2 × 10⁶)² = 1.77 × 10^-5 J/m³.

Practice Set

• Q1: Energy stored in a capacitor of 2 μF charged to 5 V is:
  A) 25 μJ B) 10 μJ B) 5 μJ C) 50 μJ
• Q2: Energy density in a capacitor depends on:
  A) Capacitance only B) Electric field only C) Voltage only D) Charge only
• Q3: A capacitor with dielectric constant 4 is connected to battery. Energy stored becomes:
  A) Same B) 4 times C) Half D) 2 times

Answer Key

• Q1: A) 25 μJ - Using U=1/2CV² = 1/2 × 2 × 10^-6 × 5² = 25 × 10^-6 J.
• Q2: B) Electric field only - Energy density formula u=1/2 ε₀ E² depends on electric field.
• Q3: B) 4 times - Energy stored increases by dielectric constant K when connected to battery.

Memory Aids

• Energy stored: U = 1/2 CV² or 1/2 QV.
• Energy density: u = 1/2 ε₀ E².
• Dielectric effect: Energy U scales by dielectric constant K if battery connected.

Solved Examples

• Ex 1: Calculate energy stored in a capacitor of capacitance 5 μF charged to 10 V.
  Solution:
  U = 1/2 CV² = 1/2 × 5 × 10^-6 × 10² = 2.5 × 10^-4 J.
• Ex 2: Find energy density if electric field E = 2 × 10⁶ V/m.
  Solution:
  u = 1/2 ε₀ E² = 1/2 × 8.85 × 10^-12 × (2 × 10⁶)² = 1.77 × 10^-5 J/m³.

Practice Set

• Q1: Energy stored in a capacitor of 2 μF charged to 5 V is:
  A) 25 μJ B) 10 μJ B) 5 μJ C) 50 μJ
• Q2: Energy density in a capacitor depends on:
  A) Capacitance only B) Electric field only C) Voltage only D) Charge only
• Q3: A capacitor with dielectric constant 4 is connected to battery. Energy stored becomes:
  A) Same B) 4 times C) Half D) 2 times

Answer Key

• Q1: A) 25 μJ - Using U=1/2CV² = 1/2 × 2 × 10^-6 × 5² = 25 × 10^-6 J.
• Q2: B) Electric field only - Energy density formula u=1/2 ε₀ E² depends on electric field.
• Q3: B) 4 times - Energy stored increases by dielectric constant K when connected to battery.

Memory Aids

• Energy stored: U = 1/2 CV² or 1/2 QV.
• Energy density: u = 1/2 ε₀ E².
• Dielectric effect: Energy U scales by dielectric constant K if battery connected.

Concepts and Formulas

• If q is charge and V is potential difference at any instant, small work done to add charge dq is dW = V · dq = q/C dq.
• Total work done to store maximum charge Q is W = ∫₀^Q q/C dq = 1/2 Q²/C.
• Energy stored in capacitor (electrostatic energy):
  U = 1/2 Q²/C = 1/2 CV² = 1/2 QV.
• Energy density (energy per unit volume) in parallel plate capacitor:
  u = 1/2 ε₀ E², where ε₀ is permittivity of free space and E is electric field.
• Effect of dielectric:
  
  • With dielectric connected to battery: Capacitance C = K C₀, voltage V = V₀, electric field E = E₀, energy stored U = K U₀ (increases).
  • With dielectric removed (isolated capacitor): Charge Q = Q₀ constant, voltage V = V₀/K, electric field E = E₀/K, energy stored U = U₀/K (decreases).

Solved Examples

• Ex 1: Calculate energy stored in a capacitor of capacitance 5 μF charged to 10 V.
  Solution:
  U = 1/2 CV² = 1/2 × 5 × 10^-6 × 10² = 2.5 × 10^-4 J.
• Ex 2: Find energy density if electric field E = 2 × 10⁶ V/m.
  Solution:
  u = 1/2 ε₀ E² = 1/2 × 8.85 × 10^-12 × (2 × 10⁶)² = 1.77 × 10^-5 J/m³.

Practice Set

• Q1: Energy stored in a capacitor of 2 μF charged to 5 V is:
  A) 25 μJ B) 10 μJ B) 5 μJ C) 50 μJ
• Q2: Energy density in a capacitor depends on:
  A) Capacitance only B) Electric field only C) Voltage only D) Charge only
• Q3: A capacitor with dielectric constant 4 is connected to battery. Energy stored becomes:
  A) Same B) 4 times C) Half D) 2 times

Answer Key

• Q1: A) 25 μJ - Using U=1/2CV² = 1/2 × 2 × 10^-6 × 5² = 25 × 10^-6 J.
• Q2: B) Electric field only - Energy density formula u=1/2 ε₀ E² depends on electric field.
• Q3: B) 4 times - Energy stored increases by dielectric constant K when connected to battery.

Memory Aids

• Energy stored: U = 1/2 CV² or 1/2 QV.
• Energy density: u = 1/2 ε₀ E².
• Dielectric effect: Energy U scales by dielectric constant K if battery connected.

Concepts and Formulas

• If \( q \) is charge and \( V \) is potential difference at any instant, small work done to add charge \( dq \) is \( dW = V \cdot dq = \frac{q}{C} dq \).
• Total work done to store maximum charge \( Q \) is \( W = \int_0^Q \frac{q}{C} dq = \frac{1}{2} \frac{Q^2}{C} \).
• Energy stored in capacitor (electrostatic energy):
  \( U = \frac{1}{2} \frac{Q^2}{C} = \frac{1}{2} CV^2 = \frac{1}{2} QV \).
• Energy density (energy per unit volume) in parallel plate capacitor:
  \( u = \frac{1}{2} \epsilon_0 E^2 \), where \( \epsilon_0 \) is permittivity of free space and \( E \) is electric field.
• Effect of dielectric:
  
  • With dielectric connected to battery: Capacitance \( C = K C_0 \), voltage \( V = V_0 \), electric field \( E = E_0 \), energy stored \( U = K U_0 \) (increases).
  • With dielectric removed (isolated capacitor): Charge \( Q = Q_0 \) constant, voltage \( V = \frac{V_0}{K} \), electric field \( E = \frac{E_0}{K} \), energy stored \( U = \frac{U_0}{K} \) (decreases).

Solved Examples

• Ex 1: Calculate energy stored in a capacitor of capacitance 5 \( \mu F \) charged to 10 V.
  Solution:
  \( U = \frac{1}{2} CV^2 = \frac{1}{2} \times 5 \times 10^{-6} \times 10^2 = 2.5 \times 10^{-4} \text{ J} \).
• Ex 2: Find energy density if electric field \( E = 2 \times 10^6 \text{ V/m} \).
  Solution:
  \( u = \frac{1}{2} \epsilon_0 E^2 = \frac{1}{2} \times 8.85 \times 10^{-12} \times (2 \times 10^6)^2 = 1.77 \times 10^{-5} \text{ J/m}^3 \).

Practice Set

• Q1: Energy stored in a capacitor of 2 \( \mu F \) charged to 5 V is:
  A) 25 \( \mu J \) B) 10 \( \mu J \) C) 5 \( \mu J \) D) 50 \( \mu J \)
• Q2: Energy density in a capacitor depends on:
  A) Capacitance only B) Electric field only C) Voltage only D) Charge only
• Q3: A capacitor with dielectric constant 4 is connected to battery. Energy stored becomes:
  A) Same B) 4 times C) Half D) 2 times

Answer Key

• Q1: A) 25 \( \mu J \) - Using \( U=\frac{1}{2}CV^2 = \frac{1}{2} \times 2 \times 10^{-6} \times 5^2 = 25 \times 10^{-6} \text{ J} \).
• Q2: B) Electric field only - Energy density formula \( u=\frac{1}{2} \epsilon_0 E^2 \) depends on electric field.
• Q3: B) 4 times - Energy stored increases by dielectric constant \( K \) when connected to battery.

Memory Aids

• Energy stored: \( U = \frac{1}{2} CV^2 \) or \( \frac{1}{2} QV \).
• Energy density: \( u = \frac{1}{2} \epsilon_0 E^2 \).
• Dielectric effect: Energy \( U \) scales by dielectric constant \( K \) if battery connected.