NCERT CLASS 12 PHYSICS

CHAPTER 2 – ELECTROSTATIC POTENTIAL AND CAPACITANCE

IN-TEXT / CONCEPT-BASED QUESTIONS AND ANSWERS

1. What is electric potential?

Electric potential at a point is defined as the work done per unit positive test charge by an external agent in bringing the test charge from infinity to that point without changing its kinetic energy.

Mathematically,

V = W/q

Where:

V = electric potential
W = work done
q = test charge

SI unit: volt (V).

2. Define one volt.

One volt is the electric potential at a point when one joule of work is required to bring a positive charge of one coulomb from infinity to that point.

1 V = 1 J/C

3. What is potential difference?

The potential difference between two points is the work done per unit positive charge in moving the charge from one point to another.

V_B − V_A = W_ext/q

SI unit: volt (V).

4. Is electric potential a scalar or vector quantity?

Electric potential is a scalar quantity because it has magnitude but no direction.

5. What is the electric potential due to a point charge?

The electric potential at a distance r from a point charge q is:

V = (1/4πε₀) × q/r

Where:

q = point charge
r = distance from the charge
ε₀ = permittivity of free space

6. What is the electric potential due to a positive point charge?

The electric potential due to a positive point charge is positive when the potential at infinity is taken as zero.

V = (1/4πε₀) × q/r

7. What is the electric potential due to a negative point charge?

The electric potential due to a negative point charge is negative when the potential at infinity is taken as zero.

V = (1/4πε₀) × q/r

Since q is negative, V is also negative.

8. State the principle of superposition of electric potential.

According to the principle of superposition, the total electric potential at a point due to several charges is equal to the algebraic sum of the potentials produced by each charge separately.

V = V₁ + V₂ + V₃ + ...

Electric potential is a scalar quantity, so ordinary algebraic addition is used.

9. What is an equipotential surface?

An equipotential surface is a surface on which the electric potential is the same at every point.

The potential difference between any two points on an equipotential surface is zero.

10. Write any four properties of equipotential surfaces.

The electric potential is the same at every point on an equipotential surface.
No work is done in moving a charge along an equipotential surface.
The electric field is always perpendicular to an equipotential surface.
Two equipotential surfaces never intersect each other.
Equipotential surfaces are closer together in regions where the electric field is stronger.

11. Why is no work done in moving a charge along an equipotential surface?

The potential difference between any two points on an equipotential surface is zero.

Work done:

W = qΔV

Since ΔV = 0,

W = 0

Therefore, no work is done in moving a charge along an equipotential surface.

12. What is the relation between electric field and electric potential?

The electric field is equal to the negative rate of change of electric potential with distance in the direction of the field.

For one-dimensional motion:

E = −dV/dr

The negative sign indicates that electric potential decreases in the direction of the electric field.

13. What is electrostatic potential energy?

Electrostatic potential energy is the energy possessed by a system of charges due to their relative positions and electrostatic interactions.

For two point charges:

U = (1/4πε₀) × q₁q₂/r

SI unit: joule (J).

14. What is the electrostatic potential energy of two like charges?

The electrostatic potential energy of two like charges is positive when the potential energy is taken as zero at infinite separation.

This is because external work must be done to bring like charges closer together against their repulsive force.

15. What is the electrostatic potential energy of two unlike charges?

The electrostatic potential energy of two unlike charges is negative when the potential energy is taken as zero at infinite separation.

This is because the charges attract each other.

16. What is an electric dipole?

An electric dipole is a system of two equal and opposite charges separated by a small distance.

17. Define electric dipole moment.

Electric dipole moment is the product of the magnitude of either charge and the separation vector directed from the negative charge to the positive charge.

p⃗ = q × 2a⃗

SI unit: C m.

18. What is the electric potential due to an electric dipole?

The electric potential at a point due to an electric dipole is:

V = (1/4πε₀) × (p cos θ/r²)

Where:

p = dipole moment
r = distance from the centre of the dipole
θ = angle between the dipole moment and the position vector

This expression is valid for a short dipole when the observation distance is much greater than its size.

19. What is the electric potential on the axial line of a short electric dipole?

On the axial line, θ = 0° or 180°.

V = ±(1/4πε₀) × p/r²

The sign depends on the side of the dipole.

20. What is the electric potential on the equatorial line of an electric dipole?

On the equatorial line, θ = 90°.

V = (1/4πε₀) × p cos 90°/r²

Therefore,

V = 0

21. What is a conductor in electrostatic equilibrium?

A conductor is said to be in electrostatic equilibrium when its free charges are at rest and there is no net movement of charge within it.

22. Write the properties of a conductor in electrostatic equilibrium.

The electric field inside the conductor is zero.
The electric potential is constant throughout the conductor.
Any excess charge resides on the outer surface of an isolated conductor.
The electric field just outside the conductor is perpendicular to its surface.
The electric field just outside the surface is E = σ/ε₀ for a conductor in vacuum.

23. What is dielectric material?

A dielectric is an insulating material that does not allow electric charges to move freely through it but can become polarised when placed in an electric field.

Examples: glass, mica, paper and air.

24. What is dielectric polarisation?

Dielectric polarisation is the process in which the positive and negative charges within a dielectric material undergo small relative displacements when an external electric field is applied.

This produces an induced electric field that generally opposes the applied field inside the dielectric.

25. Define capacitance.

Capacitance is the ability of a conductor or capacitor to store electric charge.

It is defined as the ratio of the charge stored on a conductor to its electric potential.

C = Q/V

SI unit: farad (F).

26. Define one farad.

A capacitor has a capacitance of one farad if it stores a charge of one coulomb when the potential difference between its plates is one volt.

1 F = 1 C/V

27. What is a capacitor?

A capacitor is an electrical device consisting of two conductors separated by an insulating material or air. It is used to store electric charge and electrical energy.

28. What is a parallel plate capacitor?

A parallel plate capacitor consists of two large, parallel conducting plates separated by a small distance. The space between the plates may contain air, vacuum or a dielectric material.

29. What is the capacitance of a parallel plate capacitor?

For two large parallel plates separated by a distance d and having area A:

C = ε₀A/d

If a dielectric of relative permittivity K completely fills the space:

C = Kε₀A/d

30. What happens to the capacitance when the distance between the plates is reduced?

The capacitance of a parallel plate capacitor is inversely proportional to the separation between its plates.

C ∝ 1/d

Therefore, reducing the distance between the plates increases the capacitance.

31. What happens to the capacitance when a dielectric is inserted between the plates?

When a dielectric of relative permittivity K completely fills the space between the plates, the capacitance becomes K times its original value.

C' = KC

32. What is the difference between series and parallel combinations of capacitors?

Series combination:

Capacitors are connected one after another.
Each capacitor carries the same magnitude of charge.
The potential difference is divided among the capacitors.
Equivalent capacitance is less than the smallest individual capacitance.

Parallel combination:

All positive plates are connected together and all negative plates are connected together.
Each capacitor has the same potential difference.
The total charge is the sum of the individual charges.
Equivalent capacitance is the sum of individual capacitances.

33. Write the formula for equivalent capacitance in series.

For capacitors C₁, C₂ and C₃ connected in series:

1/C = 1/C₁ + 1/C₂ + 1/C₃

34. Write the formula for equivalent capacitance in parallel.

For capacitors C₁, C₂ and C₃ connected in parallel:

C = C₁ + C₂ + C₃

35. What is the energy stored in a capacitor?

The energy stored in a capacitor is the work done in charging it. It is stored in the electric field between the plates.

U = ½ CV²

Other forms:

U = ½ QV

U = Q²/2C

SI unit: joule (J).

NCERT EXERCISE QUESTIONS AND ANSWERS

Q2.1. Two point charges of 5 × 10⁻⁸ C and −3 × 10⁻⁸ C are located 16 cm apart. At what point or points on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Given:

q₁ = +5 × 10⁻⁸ C

q₂ = −3 × 10⁻⁸ C

Distance between charges = 16 cm

Let the positive charge be at A and the negative charge at B.

Let P be the point where the potential is zero.

Formula:

V = (1/4πε₀) × q/r

For zero potential:

V₁ + V₂ = 0

Case 1: Point between the charges

Let the distance of P from the positive charge be x cm.

Distance from negative charge = (16 − x) cm.

Therefore,

(1/4πε₀) × [5 × 10⁻⁸/x − 3 × 10⁻⁸/(16 − x)] = 0

Cancelling common terms:

5/x = 3/(16 − x)

5(16 − x) = 3x

80 − 5x = 3x

80 = 8x

x = 10 cm

Case 2: Point beyond the negative charge

Let the distance of P from the positive charge be x cm.

Distance from negative charge = (x − 16) cm.

For zero potential:

5/x = 3/(x − 16)

5(x − 16) = 3x

5x − 80 = 3x

2x = 80

x = 40 cm

There is no zero-potential point on the other side of the positive charge.

Answer:

The electric potential is zero at two points:

Between the charges, 10 cm from the positive charge.
Beyond the negative charge, 40 cm from the positive charge.

Q2.2. A regular hexagon of side 10 cm has a charge of 5 μC at each of its six vertices. Calculate the electric potential at the centre of the hexagon.

Given:

Side of hexagon = 10 cm = 0.1 m

Charge at each vertex:

q = 5 μC = 5 × 10⁻⁶ C

Number of charges = 6

k = 9 × 10⁹ N m²/C²

Solution:

In a regular hexagon, the distance from the centre to each vertex is equal to the side of the hexagon.

Therefore,

r = 0.1 m

Potential due to one charge:

V₁ = kq/r

V₁ = (9 × 10⁹ × 5 × 10⁻⁶)/0.1

V₁ = 4.5 × 10⁵ V

Since electric potential is a scalar quantity, the potentials due to all six charges are added.

V = 6V₁

V = 6 × 4.5 × 10⁵

V = 2.7 × 10⁶ V

Answer: Electric potential at the centre = 2.7 × 10⁶ V.

Q2.3. Two charges of 2 μC and −2 μC are placed at points A and B, respectively, separated by a distance of 6 cm. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface?

Given:

qA = +2 μC

qB = −2 μC

AB = 6 cm

(a) Equipotential surface

Consider the plane passing through the midpoint of AB and perpendicular to the line joining the two charges.

Every point on this plane is at equal distances from the two charges.

Potential due to positive charge:

V₁ = kq/r

Potential due to negative charge:

V₂ = −kq/r

Total potential:

V = V₁ + V₂

V = kq/r − kq/r

V = 0

Therefore, the perpendicular bisector plane is an equipotential surface of zero potential.

(b) Direction of electric field

The electric field is always perpendicular to an equipotential surface.

On this plane, the electric field is perpendicular to the plane and directed from the positive charge A towards the negative charge B.

Answer:
(a) The plane passing through the midpoint of AB and perpendicular to AB is an equipotential surface.
(b) The electric field is perpendicular to this plane and directed from A towards B.

Q2.4. A spherical conductor of radius 12 cm has a charge of 1.6 × 10⁻⁷ C distributed uniformly over its surface. Calculate the electric field at (a) a point inside the sphere, (b) a point just outside the sphere and (c) a point 18 cm from the centre of the sphere.

Given:

Radius R = 12 cm = 0.12 m

Charge Q = 1.6 × 10⁻⁷ C

k = 9 × 10⁹ N m²/C²

(a) Electric field inside the sphere

The electric field inside a conductor in electrostatic equilibrium is zero.

Therefore,

E = 0 N/C

(b) Electric field just outside the sphere

At the surface, the field is:

E = kQ/R²

E = (9 × 10⁹ × 1.6 × 10⁻⁷)/(0.12)²

E = 1440/0.0144

E = 1 × 10⁵ N/C

(c) Electric field at 18 cm from the centre

r = 18 cm = 0.18 m

E = kQ/r²

E = 1440/(0.18)²

E = 1440/0.0324

E = 4.44 × 10⁴ N/C

Answer:

(a) Inside the sphere: E = 0 N/C.

(b) Just outside the sphere: E = 1 × 10⁵ N/C.

The field outside is directed radially outward because the charge is positive.

Q2.5. A parallel plate capacitor with air between its plates has a capacitance of 8 pF. What will be its capacitance if the distance between the plates is reduced to half and the space between the plates is filled with a substance having a dielectric constant of 6?

Given:

Initial capacitance C₀ = 8 pF

Dielectric constant K = 6

New separation d' = d/2

Formula:

Initial capacitance:

C₀ = ε₀A/d

New capacitance:

C' = Kε₀A/(d/2)

C' = 2Kε₀A/d

Since ε₀A/d = C₀,

C' = 2KC₀

Substitution:

C' = 2 × 6 × 8

C' = 96 pF

Answer: New capacitance = 96 pF.

Q2.6. Three capacitors, each having a capacitance of 9 pF, are connected in series. (a) Calculate the total capacitance of the combination. (b) Calculate the potential difference across each capacitor if the combination is connected to a 120 V supply.

Given:

C₁ = C₂ = C₃ = 9 pF

V = 120 V

(a) Equivalent capacitance in series

For capacitors connected in series:

1/C = 1/C₁ + 1/C₂ + 1/C₃

1/C = 1/9 + 1/9 + 1/9

1/C = 3/9

1/C = 1/3

C = 3 pF

(b) Potential difference across each capacitor

In a series combination, the charge on each capacitor is the same.

Since all three capacitors have equal capacitance, the applied voltage is divided equally.

V₁ = V₂ = V₃ = 120/3

V₁ = V₂ = V₃ = 40 V

Answer:

(a) Equivalent capacitance = 3 pF.

(b) Potential difference across each capacitor = 40 V.

Q2.7. Three capacitors having capacitances of 2 pF, 3 pF and 4 pF are connected in parallel. (a) Calculate the total capacitance of the combination. (b) Determine the charge on each capacitor when the combination is connected to a 100 V supply.

Given:

C₁ = 2 pF

C₂ = 3 pF

C₃ = 4 pF

V = 100 V

(a) Equivalent capacitance in parallel

C = C₁ + C₂ + C₃

C = 2 + 3 + 4

C = 9 pF

(b) Charge on each capacitor

In parallel, the potential difference across each capacitor is the same.

V = 100 V

Using:

Q = CV

For the first capacitor:

Q₁ = 2 × 10⁻¹² × 100

Q₁ = 2 × 10⁻¹⁰ C

For the second capacitor:

Q₂ = 3 × 10⁻¹² × 100

Q₂ = 3 × 10⁻¹⁰ C

For the third capacitor:

Q₃ = 4 × 10⁻¹² × 100

Q₃ = 4 × 10⁻¹⁰ C

Answer:

(a) Equivalent capacitance = 9 pF.

(b) Charges:

Q₁ = 2 × 10⁻¹⁰ C

Q₂ = 3 × 10⁻¹⁰ C

Q₃ = 4 × 10⁻¹⁰ C

Q2.8. A parallel plate capacitor has air between its plates. Each plate has an area of 6 × 10⁻³ m², and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If it is connected to a 100 V supply, calculate the charge on each plate.

Given:

Area A = 6 × 10⁻³ m²

Distance d = 3 mm = 3 × 10⁻³ m

Potential difference V = 100 V

ε₀ = 8.854 × 10⁻¹² F/m

(a) Capacitance

For a parallel plate capacitor:

C = ε₀A/d

Substituting:

C = (8.854 × 10⁻¹² × 6 × 10⁻³)/(3 × 10⁻³)

C = 17.708 × 10⁻¹² F

C ≈ 1.77 × 10⁻¹¹ F

C ≈ 17.7 pF

(b) Charge on each plate

Q = CV

Q = 1.77 × 10⁻¹¹ × 100

Q = 1.77 × 10⁻⁹ C

One plate carries positive charge and the other carries an equal negative charge.

Answer:

(a) Capacitance = 1.77 × 10⁻¹¹ F = 17.7 pF.

(b) Magnitude of charge on each plate = 1.77 × 10⁻⁹ C.

Q2.9. Explain what happens to the capacitance, charge and potential difference of the capacitor described in Exercise 2.8 when a 3 mm thick mica sheet having a dielectric constant of 6 is inserted between its plates: (a) while the voltage supply remains connected, and (b) after the voltage supply has been disconnected.

Given:

Initial capacitance:

C = 1.77 × 10⁻¹¹ F

Initial potential difference:

V = 100 V

Initial charge:

Q = 1.77 × 10⁻⁹ C

Dielectric constant:

K = 6

New capacitance

When mica completely fills the space between the plates:

C' = KC

C' = 6 × 1.77 × 10⁻¹¹

C' = 1.062 × 10⁻¹⁰ F

C' ≈ 106.2 pF

(a) When the voltage supply remains connected

The battery remains connected, so the potential difference remains constant.

V' = 100 V

New charge:

Q' = C'V'

Q' = 1.062 × 10⁻¹⁰ × 100

Q' = 1.062 × 10⁻⁸ C

Therefore:

Capacitance becomes six times the original value.
Potential difference remains 100 V.
Charge becomes six times the original value.

(b) When the voltage supply is disconnected

After disconnecting the battery, the charge on the capacitor remains constant because there is no conducting path for charge to leave the plates.

Therefore,

Q' = Q = 1.77 × 10⁻⁹ C

New potential difference:

V' = Q'/C'

V' = (1.77 × 10⁻⁹)/(1.062 × 10⁻¹⁰)

V' ≈ 16.7 V

Therefore:

Capacitance becomes six times the original value.
Charge remains constant.
Potential difference becomes one-sixth of its original value.

Answer:

(a) Battery connected:

C' ≈ 106.2 pF
V' = 100 V
Q' ≈ 1.062 × 10⁻⁸ C

(b) Battery disconnected:

C' ≈ 106.2 pF
Q' = 1.77 × 10⁻⁹ C
V' ≈ 16.7 V

Q2.10. A capacitor of capacitance 12 pF is connected to a 50 V battery. Calculate the electrostatic energy stored in the capacitor.

Given:

C = 12 pF = 12 × 10⁻¹² F

V = 50 V

Formula:

U = ½ CV²

Substitution:

U = ½ × 12 × 10⁻¹² × (50)²

U = 6 × 10⁻¹² × 2500

U = 15000 × 10⁻¹²

U = 1.5 × 10⁻⁸ J

Answer: Energy stored = 1.5 × 10⁻⁸ J.

Q2.11. A capacitor of capacitance 600 pF is charged by a 200 V supply. It is then disconnected from the supply and connected to another uncharged capacitor of the same capacitance, 600 pF. Calculate the amount of electrostatic energy lost in the process.

Given:

C₁ = 600 pF = 600 × 10⁻¹² F

C₂ = 600 pF = 600 × 10⁻¹² F

Initial potential difference:

V₁ = 200 V

Step 1: Calculate the initial charge

Q = C₁V₁

Q = 600 × 10⁻¹² × 200

Q = 1.2 × 10⁻⁷ C

Step 2: Calculate the initial energy

U₁ = ½ C₁V₁²

U₁ = ½ × 600 × 10⁻¹² × (200)²

U₁ = 300 × 10⁻¹² × 40000

U₁ = 1.2 × 10⁻⁵ J

Step 3: Calculate the final capacitance

The two capacitors are connected in parallel.

C' = C₁ + C₂

C' = 600 + 600

C' = 1200 pF

C' = 1.2 × 10⁻⁹ F

Step 4: Calculate the final potential difference

The total charge is conserved.

V' = Q/C'

V' = (1.2 × 10⁻⁷)/(1.2 × 10⁻⁹)

V' = 100 V

Step 5: Calculate the final energy

U₂ = ½ C'V'²

U₂ = ½ × 1.2 × 10⁻⁹ × (100)²

U₂ = 0.6 × 10⁻⁹ × 10000

U₂ = 6 × 10⁻⁶ J

Step 6: Calculate energy lost

Energy lost = Initial energy − Final energy

Energy lost = U₁ − U₂

Energy lost = 1.2 × 10⁻⁵ − 6 × 10⁻⁶

Energy lost = 6 × 10⁻⁶ J

Answer: Energy lost = 6 × 10⁻⁶ J.

The energy difference is dissipated during charge redistribution, mainly as heat in the connecting circuit and electromagnetic radiation.

IMPORTANT QUESTIONS AND ANSWERS

1. Derive the expression for electric potential due to a point charge.

Consider a point charge Q placed at point O.

Let P be a point at a distance r from the charge.

The electric field at distance x from the charge is:

E = (1/4πε₀) × Q/x²

The work done by an external agent in bringing a unit positive test charge from infinity to P is equal to the potential at P.

For a small displacement dx towards the charge, the work done by the external agent is:

dW = −E dx

Therefore,

V = ∫∞ʳ −E dx

Substituting the electric field:

V = −∫∞ʳ (1/4πε₀)(Q/x²) dx

Taking constants outside:

V = −Q/(4πε₀) ∫∞ʳ x⁻² dx

Integrating:

V = −Q/(4πε₀) [−1/x]∞ʳ

V = Q/(4πε₀r)

Therefore,

V = (1/4πε₀) × Q/r

This is the expression for electric potential due to a point charge.

2. Derive the expression for electric potential due to an electric dipole at a general point.

Consider an electric dipole consisting of charges +q and −q separated by a distance 2a.

Let P be a point at a distance r from the centre of the dipole, making an angle θ with the dipole axis.

Distance of P from positive charge:

r₊ = √(r² + a² − 2ar cos θ)

Distance of P from negative charge:

r₋ = √(r² + a² + 2ar cos θ)

The potential at P is the algebraic sum of the potentials due to the two charges.

V = (1/4πε₀) [q/r₊ − q/r₋]

For a short dipole, where r is much greater than a, using the approximation:

1/r₊ − 1/r₋ ≈ 2a cos θ/r²

Therefore,

V = (1/4πε₀) × q × 2a cos θ/r²

Since dipole moment:

p = 2aq

Therefore,

V = (1/4πε₀) × p cos θ/r²

This is the electric potential due to a short electric dipole at a general point.

3. Derive the expression for electrostatic potential energy of two point charges.

Consider two point charges q₁ and q₂ separated by a distance r.

Initially, charge q₁ is placed at a point. Its potential at the position of q₂ is:

V₁ = (1/4πε₀) × q₁/r

The work done in bringing charge q₂ from infinity to this point is:

W = q₂V₁

Therefore, the electrostatic potential energy is:

U = q₂ × (1/4πε₀) × q₁/r

Hence,

U = (1/4πε₀) × q₁q₂/r

If both charges are like charges, U is positive.

If the charges are unlike, U is negative.

4. Derive the expression for electrostatic potential energy of a system of three point charges.

Consider three charges q₁, q₂ and q₃ separated by distances r₁₂, r₂₃ and r₃₁.

First, bring q₁ from infinity. No work is required because no other charge is present.

Next, bring q₂ from infinity.

Work done:

W₂ = (1/4πε₀) × q₁q₂/r₁₂

Finally, bring q₃ from infinity.

The potential at its position due to q₁ and q₂ is:

V₃ = (1/4πε₀) × [q₁/r₁₃ + q₂/r₂₃]

Work done:

W₃ = q₃V₃

W₃ = (1/4πε₀) × [q₁q₃/r₁₃ + q₂q₃/r₂₃]

Total potential energy:

U = W₂ + W₃

Therefore,

U = (1/4πε₀) [q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₃q₁/r₃₁]

This is the electrostatic potential energy of a system of three point charges.

5. Derive the relation between electric field and electric potential.

Consider a positive test charge q moving through a small displacement dr in an electric field E.

The work done by the electric field is:

dW = qE dr

The decrease in potential energy is equal to the work done by the electric field.

dU = −dW

dU = −qE dr

Electric potential is:

V = U/q

Therefore,

dV = dU/q

dV = −E dr

Hence,

E = −dV/dr

In vector form:

E⃗ = −∇V

The negative sign indicates that the electric field points in the direction of decreasing potential.

6. Explain equipotential surfaces and their properties.

An equipotential surface is a surface on which the electric potential is the same at every point.

Properties:

The potential is constant at every point on the surface.
No work is done in moving a charge along the surface.
The electric field is perpendicular to the surface.
Two equipotential surfaces cannot intersect.
Closely spaced equipotential surfaces indicate a stronger electric field.

Examples:

Around a point charge, equipotential surfaces are concentric spherical surfaces.
In a uniform electric field, equipotential surfaces are parallel planes perpendicular to the field.
Around an isolated spherical conductor, equipotential surfaces outside it are concentric spherical surfaces.

7. Explain the electrostatic properties of conductors.

A conductor in electrostatic equilibrium has the following properties:

1. Electric field inside a conductor is zero.

If an electric field existed inside a conductor, free charges would move. In electrostatic equilibrium, charges are at rest. Hence, the internal electric field must be zero.

2. Charge resides on the outer surface.

Any excess charge given to an isolated conductor resides on its outer surface in electrostatic equilibrium.

3. Potential is constant throughout the conductor.

Since the electric field inside the conductor is zero, there is no potential difference between any two points inside it.

4. Electric field is perpendicular to the surface.

The electric field just outside a conductor is normal to its surface. A tangential component would cause charges to move along the surface.

5. Electric field just outside the conductor.

In vacuum:

E = σ/ε₀

Where σ is the surface charge density.

8. Derive the expression for capacitance of a parallel plate capacitor.

Consider two large parallel conducting plates, each having area A and separated by a distance d.

Let the plates carry charges +Q and −Q.

Surface charge density:

σ = Q/A

Electric field between the plates, neglecting edge effects:

E = σ/ε₀

E = Q/(Aε₀)

Potential difference between the plates:

V = Ed

V = Qd/(Aε₀)

Capacitance is defined as:

C = Q/V

Substituting:

C = Q/[Qd/(Aε₀)]

Therefore,

C = ε₀A/d

This is the capacitance of a parallel plate capacitor with vacuum or approximately air between the plates.

If a dielectric of relative permittivity K completely fills the space:

C = Kε₀A/d

9. Derive the expression for equivalent capacitance when capacitors are connected in series.

Consider three capacitors C₁, C₂ and C₃ connected in series across a potential difference V.

In a series combination, each capacitor carries the same magnitude of charge Q.

Let the potential differences across the capacitors be V₁, V₂ and V₃.

Total potential difference:

V = V₁ + V₂ + V₃

Using V = Q/C:

V = Q/C₁ + Q/C₂ + Q/C₃

For the equivalent capacitance C:

V = Q/C

Therefore,

Q/C = Q/C₁ + Q/C₂ + Q/C₃

Cancelling Q:

1/C = 1/C₁ + 1/C₂ + 1/C₃

For n capacitors:

1/C = 1/C₁ + 1/C₂ + ... + 1/Cₙ

10. Derive the expression for equivalent capacitance when capacitors are connected in parallel.

Consider three capacitors C₁, C₂ and C₃ connected in parallel across a potential difference V.

In a parallel combination, the potential difference across each capacitor is the same.

Therefore,

V₁ = V₂ = V₃ = V

Total charge:

Q = Q₁ + Q₂ + Q₃

Using Q = CV:

Q = C₁V + C₂V + C₃V

Q = V(C₁ + C₂ + C₃)

For the equivalent capacitance C:

Q = CV

Therefore,

CV = V(C₁ + C₂ + C₃)

Cancelling V:

C = C₁ + C₂ + C₃

For n capacitors:

C = C₁ + C₂ + ... + Cₙ

11. Derive the expression for energy stored in a capacitor.

Consider a capacitor being charged gradually.

At any instant, let the charge on the capacitor be q and its potential difference be:

V = q/C

The small amount of work done in transferring an additional charge dq is:

dW = V dq

dW = (q/C)dq

Total work done in charging the capacitor from zero charge to Q:

U = ∫₀Q (q/C)dq

Taking 1/C outside:

U = (1/C) ∫₀Q q dq

U = (1/C) [q²/2]₀Q

U = Q²/(2C)

Since Q = CV:

U = ½ CV²

Also,

U = ½ QV

Therefore, the three expressions for energy stored in a capacitor are:

U = ½ CV²

U = ½ QV

U = Q²/(2C)

12. Explain the effect of inserting a dielectric between the plates of a capacitor.

When a dielectric material is inserted between the plates of a capacitor, it becomes polarised.

The induced electric field inside the dielectric opposes the original electric field. Consequently, for a fixed charge, the potential difference decreases and the capacitance increases.

If the dielectric completely fills the space between the plates:

C' = KC

Where K is the dielectric constant or relative permittivity.

Case 1: Battery remains connected

Potential difference remains constant.
Capacitance increases K times.
Charge increases K times.
Stored energy increases K times.

Case 2: Battery is disconnected

Charge remains constant.
Capacitance increases K times.
Potential difference decreases to 1/K of its original value.
Stored energy decreases to 1/K of its original value.

IMPORTANT FORMULAS – CHAPTER 2

1. Electric potential

V = W/q

2. Potential difference

ΔV = W/q

3. Potential due to a point charge

V = (1/4πε₀) × q/r

4. Potential due to several point charges

V = V₁ + V₂ + V₃ + ...

5. Potential due to an electric dipole

V = (1/4πε₀) × p cos θ/r²

This is for a short dipole at a distant point.

6. Potential on the axial line of a short dipole

V = ±(1/4πε₀) × p/r²

7. Potential on the equatorial line of a dipole

V = 0

8. Electrostatic potential energy of two charges

U = (1/4πε₀) × q₁q₂/r

9. Potential energy of a system of charges

U = (1/4πε₀) Σ(qᵢqⱼ/rᵢⱼ)

Here, each pair of charges is counted once.

10. Relation between electric field and potential

E = −dV/dr

11. Capacitance

C = Q/V

12. Parallel plate capacitor

C = ε₀A/d

13. Parallel plate capacitor with dielectric

C = Kε₀A/d

14. Capacitors in series

1/C = 1/C₁ + 1/C₂ + 1/C₃

15. Capacitors in parallel

C = C₁ + C₂ + C₃

16. Charge stored in a capacitor

Q = CV

17. Energy stored in a capacitor

U = ½ CV²

U = ½ QV

U = Q²/2C

18. Energy density of an electric field in vacuum

u = ½ ε₀E²

19. Energy density in a linear dielectric

u = ½ εE²

Where ε is the permittivity of the dielectric.

IMPORTANT SI UNITS

Physical quantity	SI unit
Electric potential	Volt (V)
Potential difference	Volt (V)
Work	Joule (J)
Potential energy	Joule (J)
Capacitance	Farad (F)
Electric charge	Coulomb (C)
Electric field	N/C or V/m
Dipole moment	C m
Permittivity	F/m
Area	m²
Distance	Metre (m)
Energy density	J/m³

QUICK REVISION POINTS

Electric potential is a scalar quantity.
The SI unit of electric potential is volt.
One volt is equal to one joule per coulomb.
Electric potential due to a positive point charge is positive.
Electric potential due to a negative point charge is negative.
The total potential due to several charges is the algebraic sum of individual potentials.
An equipotential surface has the same potential at every point.
No work is done in moving a charge along an equipotential surface.
Electric field is always perpendicular to an equipotential surface.
Electric potential energy of two like charges is positive.
Electric potential energy of two unlike charges is negative.
The electric field points in the direction of decreasing potential.
The electric field inside a conductor in electrostatic equilibrium is zero.
The potential throughout a conductor in electrostatic equilibrium is constant.
Excess charge on an isolated conductor resides on its outer surface.
Capacitance is the ability of a system to store charge.
The SI unit of capacitance is farad.
Capacitance of a parallel plate capacitor increases with plate area.
Capacitance decreases when the distance between the plates increases.
Inserting a dielectric increases the capacitance.
In a series combination, the charge on each capacitor is the same.
In a parallel combination, the potential difference across each capacitor is the same.
The energy stored in a capacitor is U = ½ CV².
When a battery is connected, the potential difference remains constant.
When a capacitor is disconnected from the battery, its charge remains constant if it is isolated.