Electric Charge

Definition and Types

Electric charge is the amount of energy or electrons that pass from one body to another by different modes like conduction, induction, etc. There are two types of electric charges: positive charges and negative charges.

Properties of Electric Charge

Addition of Charges: If a system contains n point charges q1, q2, q3, A0A0A0A0q9 then the total charge of the system is

q = q1 + q2 + q3 + A0A0A0A0A0A0A0A0q9

Conservation of Charges: The sum of positive and negative charges present in an isolated system always remains constant. Charge can neither be created nor destroyed but only exists in positive-negative pairs and can only be transferred from one body to another.

Quantisation of Charges: Electric charge is always quantised. The net charge q#8327#8326, of an object having N#8327 electrons, N#8324 protons and N#8325 neutrons is:

q#8327#8326 = -eN#8327 + eN#8324 + 0 D7 N#8325 = e (N#8324 - N#8327) = B1ne, where n is an integer.

Solved Examples

Example 1: Time to Accumulate Charge

If 10B9 electrons move out of a body to another body every second, how much time is required to get a total charge of 1C on the other body?

Solution:

10B9 electrons will take 1 second.

So, 1 electron will take 1/10B9 seconds.

Since, 1C = 6.25 D7 10B9 electrons,

Time taken = 6.25 D7 10B9 / 10B9 = 6.25 D7 10B9 seconds.

Converting to years:

= 6.25 D7 10B9 / (60 D7 60 D7 24 D7 365) = 198.18 years  198 years.

Practice Set

• Level 1: What are the two types of electric charges?

• Level 2: Explain the principle of conservation of charge.

• Level 3: Calculate the net charge of an object having 10B9 electrons and 10#8324 protons.

Answer Key

• Level 1: Positive and negative charges.

• Level 2: In an isolated system, the total charge remains constant; charge can neither be created nor destroyed but only transferred.

• Level 3: Net charge q = e (N#8324 - N#8327) = 1.6 D7 10B9 D7 (10#8324 - 10B9) = 1.6 D7 10B9 D7 (10#8324 - 10B9) C.

Coulomb's Law

Statement and Formula

The force of attraction or repulsion between two point charges q1 and q2, separated by a distance r, is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between charges. Mathematically,

F = k D7 |q1 q2| / rB2

where,

• F = Force of attraction/repulsion between charges q1 and q2.

• q1, q2 = Magnitudes of charges.

• r = Distance between charges q1 and q2.

• k = Constant whose value depends on medium where charges are kept (8.99 D7 10B9 N.mB2 /CB2 for vacuum/air).

Force in Vacuum and Medium

In vacuum, Coulomb's law is given by:

F = D7 D7 |q1 q2| / (4 D7 C0 D7 B5B0 rB2)

where B5B0 = Permittivity of vacuum = 8.854 D7 10B9B2 F/m.

In a medium with permittivity B5,

F = D7 D7 |q1 q2| / (4 D7 C0 D7 B5 rB2)

where B5 = K D7 B5B0, K is the dielectric constant of the medium.

Vector Form

The force on charge q1 due to q2 is given by:

9F12 = k D7 q1 q2 / rB2 D7 C112

and force on q2 due to q1 is:

9F21 = -9F12

This shows the forces are equal in magnitude and opposite in direction, consistent with Newton's third law.

Solved Examples

Example 2: Force Between Charged Spheres

A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by an insulating handle is brought close to A such that the distance between their centres is 10 cm.

After touching spheres A and C, and B and D (uncharged spheres), and then removing C and D, B is brought closer to A to a distance of 5 cm.

What is the expected repulsion of A on the basis of Coulomb's law?

Solution:

Let the original repulsive force between A and B be F = k q1 q2 / rB2.

As A and C have the same size, their charges are shared equally.

Charges on A and B are halved and distance between them is also halved from 10 cm to 5 cm.

New force F' = k (q1/2)(q2/2) / (r/2)B2 = k q1 q2 / rB2 = F.

Therefore, the repulsive force remains the same.

Example 3: Force on Charge at Centroid

Three charges q1, q2, q3 each equal to q are placed at the vertices of an equilateral triangle of side l. Find the force on a charge Q (with the same sign as q) placed at the centroid of the triangle.

Solution:

Distance of centroid from vertex AO = BO = CO = l / A3 3.

Force due to one charge on Q is F = k Qq / (l / A3 3)B2 = 3k Qq / lB2.

Using the parallelogram law, the resultant force of F2 and F3 is F23 = 3k Qq / lB2 along OA.

Total force on Q is F1 + F23 = 3k Qq / lB2 rC + 3k Qq / lB2 (-rC) = 0.

Hence, the net force on Q is zero.

Practice Set

• Level 1: State Coulomb's law.

• Level 2: How does the force between two charges change if the distance between them is doubled?

• Level 3: Calculate the force between two charges of 2 D7 10B6 C and 3 D7 10B6 C placed 0.5 m apart in air.

Answer Key

• Level 1: The force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

• Level 2: The force becomes one-fourth of the original force.

• Level 3: Using Coulomb's law, F = 8.99 D7 10B9 D7 (2 D7 10B6) D7 (3 D7 10B6) / (0.5)B2 = 0.21576 N.

Electric Field

Definition

The space around a charge in which a test charge experiences an electric force is called the electric field. It is a vector quantity.

Electric Field Due to a Point Charge

The electric field strength E due to a point charge q at a distance r is given by:

E = 9q / (4 D7 C0 D7 B5B0 rB2)

where B5B0 is the permittivity of free space.

Electric Field Lines

Electric field lines are imaginary lines that originate from positive charges and terminate at negative charges. They never intersect and are perpendicular to the surface of conductors.

Solved Examples

Example 4: Electric Field at Points

Two point charges q1 = +10B8 C and q2 = -10B8 C are placed 0.1 m apart. Calculate the electric fields at points A, B, and C.

Solution:

Calculate the electric field due to each charge at points A, B, and C using Coulomb's law and vector addition.

Practice Set

• Level 1: What is the direction of electric field lines around a positive charge?

• Level 2: Why do electric field lines never intersect?

• Level 3: Calculate the electric field at a point 0.2 m away from a charge of 5 D7 10B6 C.

Answer Key

• Level 1: Electric field lines originate from positive charges and terminate at negative charges.

• Level 2: Electric field lines never intersect because the electric field at any point has a unique direction.

• Level 3: E = q / (4 D7 C0 D7 B5B0 rB2) = 5 D7 10B6 / (4 D7 3.1416 D7 8.854 D7 10B9 D7 0.2B2) = calculated value.

Electric Dipole

Definition and Dipole Moment

An electric dipole consists of two equal and opposite charges separated by a small distance 2a. The dipole moment C0 is a vector quantity given by:

C0 = q D7 2a

Electric Field Due to a Dipole

At a point P at distance r from the centre of the dipole (r BB a), the electric field is:

On the axial line:

E = 2p / (4 D7 C0 D7 B5B0 rB3)

On the equatorial line:

E = p / (4 D7 C0 D7 B5B0 rB3)

Torque on a Dipole

When placed in a uniform electric field E, a dipole experiences a torque C4 given by:

C4 = p E D7 sin B8

where B8 is the angle between the dipole moment and the electric field.

The dipole is in stable equilibrium when B8 = 0B0 or 360B0, and unstable equilibrium when B8 = 180B0.

Solved Examples

Example 5: Electric Field of a Dipole

Calculate the electric field at 15 cm from a dipole of charges B110 BCC separated by 5 mm on axial and equatorial lines.

Solution:

Dipole moment p = q D7 2a = 10B5 D7 5 D7 10B3 = 5 D7 10B8 C m.

Electric field on axial line E5 = 2p / (4 D7 C0 D7 B5B0 rB3) = 2.67 D7 10B5 N/C.

Electric field on equatorial line E4 = p / (4 D7 C0 D7 B5B0 rB3) = 1.33 D7 10B5 N/C.

Practice Set

• Level 1: Define electric dipole moment.

• Level 2: What is the direction of torque on a dipole in an electric field?

• Level 3: Calculate the torque on a dipole with moment 5 D7 10B8 C m in a field of 10B3 N/C at 30B0.

Answer Key

• Level 1: Electric dipole moment is a vector quantity equal to the product of charge and separation distance.

• Level 2: Torque acts to align the dipole moment with the electric field.

• Level 3: C4 = p E sin B8 = 5 D7 10B8 D7 10B3 D7 sin 30B0 = 2.5 D7 10B5 N m.

Gauss's Theorem

Electric Flux

Electric flux is proportional to the algebraic number of electric field lines passing through a surface. Outgoing lines have positive sign, incoming lines have negative sign.

For a uniform electric field E making an angle B8 with surface area A, flux is:

A6 = E A cos B8

For a non-uniform electric field, flux is given by the surface integral:

A6 = 2 E7 B7 d2

Gauss's Law

The net outward electric flux through any closed surface is equal to the net charge enclosed divided by the permittivity of free space B5B0:

A6 = q / B5B0

Applications

• Electric field due to uniformly charged shell:
• Outside the shell: E = q / (4 D7 C0 D7 B5B0 rB2)
• Inside the shell: E = 0
• On the surface: E = q / (4 D7 C0 D7 B5B0 RB2)
• Electric field due to infinitely long charged wire:
• E = 2 D7 BB / (4 D7 C0 D7 B5B0 r)

Solved Examples

Example 6: Electric Flux and Charge

Given electric field components EB8 = B1 xB9BD, EB9 = EB2 = 0, with B1 = 800 N/C mB9BD, calculate (a) the flux through the cube of side a = 0.1 m, and (b) the charge within the cube.

Solution:

Calculate flux on each face of the cube and sum them.

Use Gauss's law A6 = q / B5B0 to find the charge enclosed.

Practice Set

• Level 1: What is electric flux?

• Level 2: State Gauss's law.

• Level 3: Calculate the electric field due to a uniformly charged shell at a point outside the shell.

Answer Key

• Level 1: Electric flux is the measure of electric field lines passing through a surface.

• Level 2: The net electric flux through a closed surface equals the charge enclosed divided by permittivity of free space.

• Level 3: E = q / (4 D7 C0 D7 B5B0 rB2) for a point outside the shell.

Quick Reference Table

Coulomb's Law: F = k |q1 q2| / rB2, k = 8.99 D7 10B9 N mB2 / CB2

Electric Field: E = q / (4 D7 C0 D7 B5B0 rB2)

Electric Dipole Moment: p = q D7 2a

Torque on Dipole: C4 = p E sin B8

Electric Flux: A6 = E A cos B8

Gauss's Law: A6 = q / B5B0

Electric Field of Charged Shell: Outside: E = q / (4 D7 C0 D7 B5B0 rB2), Inside: E = 0

Electric Field of Infinite Wire: E = BB / (2 D7 C0 D7 B5B0 r)

Common Mistakes and Misconceptions

• Confusing electric field and electric potential; they are related but represent different physical quantities.
• Incorrect application of Coulomb's law without considering vector nature of forces.
• Neglecting units, especially distance in meters and charge in coulombs.
• Misunderstanding Gauss's law and not choosing symmetric Gaussian surfaces.
• Assuming electric field inside conductors is non-zero in electrostatic equilibrium; it is zero.
• Incorrect sign conventions for charges and forces leading to wrong force directions.

Glossary

• Electric Charge: A fundamental property of matter causing it to experience force in an electric field.
• Coulomb's Law: Law describing the force between two point charges.
• Electric Field: The force per unit positive charge at a point in space.
• Electric Dipole: A pair of equal and opposite charges separated by a distance.
• Electric Dipole Moment: A vector quantity representing the strength and orientation of a dipole.
• Electric Flux: The number of electric field lines passing through a surface.
• Gauss's Law: The net electric flux through a closed surface equals the charge enclosed divided by permittivity.
• Torque: A measure of the turning force on an object.