Chapter 2 Electrostatic Potential and Capacitance Class 12 Physics

Question 2.1

Two charges $5 \times 10^{-8} \text{ C}$ and $-3 \times 10^{-8} \text{ C}$ are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

1. Concept

Electric potential $V$ due to a point charge is given by $V = \frac{kq}{r}$. Potential is a scalar quantity, so we simply add the potentials due to individual charges.

2. Calculation

Let $q_1 = 5 \times 10^{-8} \text{ C}$ at origin and $q_2 = -3 \times 10^{-8} \text{ C}$ at $x = 16 \text{ cm}$.
Let $P$ be the point where potential is zero at distance $x$ from $q_1$.

Case 1: Point P between charges

$$\frac{k q_1}{x} + \frac{k q_2}{16 – x} = 0 \implies \frac{5}{x} = \frac{3}{16 – x}$$ $$5(16 – x) = 3x \implies 80 – 5x = 3x \implies 8x = 80 \implies x = 10 \text{ cm}$$

Case 2: Point P outside charges (closer to smaller charge)

$$\frac{k q_1}{x} + \frac{k q_2}{x – 16} = 0 \implies \frac{5}{x} = \frac{3}{x – 16}$$ $$5(x – 16) = 3x \implies 5x – 80 = 3x \implies 2x = 80 \implies x = 40 \text{ cm}$$

Result: Potential is zero at 10 cm and 40 cm from the positive charge.

Question 2.2

A regular hexagon of side 10 cm has a charge $5 \mu\text{C}$ at each of its vertices. Calculate the potential at the centre of the hexagon.

Given

Charge $q = 5 \mu\text{C} = 5 \times 10^{-6} \text{ C}$
Side $a = 10 \text{ cm} = 0.1 \text{ m}$
Number of charges $n = 6$

Solution

The distance from the center to each vertex of a regular hexagon is equal to the side length $r = a = 0.1 \text{ m}$.
Total potential $V$ is the sum of potentials from all 6 charges.

$$V = 6 \times \frac{kq}{r} = \frac{6 \times (9 \times 10^9) \times (5 \times 10^{-6})}{0.1}$$ $$V = \frac{270 \times 10^3}{0.1} = 2.7 \times 10^6 \text{ V}$$

Result: $2.7 \times 10^6 \text{ V}$

Question 2.3

Two charges $2 \mu\text{C}$ and $-2 \mu\text{C}$ are placed at points A and B 6 cm apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface?

(a) Equipotential Surface

The potential is zero at the mid-point of the dipole. The plane passing through the midpoint and perpendicular to the line joining the charges (equatorial plane) is an equipotential surface where $V=0$ everywhere.

(b) Electric Field Direction

The electric field lines go from positive charge to negative charge. On the equatorial plane, the field is perpendicular to the plane (directed from A to B side).

Result: Plane normal to AB at midpoint; Field is normal to the plane.

Question 2.4

A spherical conductor of radius 12 cm has a charge of $1.6 \times 10^{-7} \text{ C}$ distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the centre?

(a) Inside Sphere

Inside a charged conductor, the electric field is zero because charge resides only on the surface.

$E_{inside} = 0$

(b) Just Outside

$r = R = 12 \text{ cm} = 0.12 \text{ m}$.

$$E = \frac{kq}{R^2} = \frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.12)^2} = 10^5 \text{ N/C}$$

(c) At r = 18 cm

$r = 0.18 \text{ m}$.

$$E = \frac{kq}{r^2} = \frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.18)^2} \approx 4.44 \times 10^4 \text{ N/C}$$

Question 2.5

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

Initial State

$C_0 = \frac{\epsilon_0 A}{d} = 8 \text{ pF}$.

Final State

New distance $d’ = d/2$. Dielectric constant $K = 6$.

$$C’ = \frac{K \epsilon_0 A}{d’} = \frac{K \epsilon_0 A}{d/2} = 2K \left( \frac{\epsilon_0 A}{d} \right) = 2K C_0$$ $$C’ = 2 \times 6 \times 8 \text{ pF} = 96 \text{ pF}$$

Result: 96 pF

Question 2.6

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

(a) Series Combination

For series: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.

$$\frac{1}{C_{eq}} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} \implies C_{eq} = 3 \text{ pF}$$

(b) Potential Difference

In series, charge $Q$ is same. Since all $C$ are equal, $V$ is shared equally.

$$V_{each} = \frac{V_{total}}{3} = \frac{120}{3} = 40 \text{ V}$$

Result: (a) 3 pF, (b) 40 V

Question 2.7

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

(a) Parallel Combination

$$C_{eq} = C_1 + C_2 + C_3 = 2 + 3 + 4 = 9 \text{ pF}$$

(b) Charge on Each

In parallel, Voltage $V = 100 \text{ V}$ is same for all. $Q = CV$.

$$Q_1 = 2 \times 10^{-12} \times 100 = 2 \times 10^{-10} \text{ C}$$ $$Q_2 = 3 \times 10^{-12} \times 100 = 3 \times 10^{-10} \text{ C}$$ $$Q_3 = 4 \times 10^{-12} \times 100 = 4 \times 10^{-10} \text{ C}$$

Question 2.8

In a parallel plate capacitor with air between the plates, each plate has an area of $6 \times 10^{-3} \text{ m}^2$ and the distance between the plates is 3 mm. Calculate the capacitance. If this capacitor is connected to a 100 V supply, what is the charge on each plate?

1. Capacitance

$A = 6 \times 10^{-3} \text{ m}^2, d = 3 \times 10^{-3} \text{ m}, \epsilon_0 = 8.854 \times 10^{-12}$.

$$C = \frac{\epsilon_0 A}{d} = \frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}} = 17.7 \times 10^{-12} \text{ F} \approx 18 \text{ pF}$$

2. Charge

$$Q = CV = 17.7 \times 10^{-12} \times 100 = 1.77 \times 10^{-9} \text{ C}$$

Result: 18 pF, $1.77 \text{ nC}$

Question 2.9

Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) while the voltage supply remained connected. (b) after the supply was disconnected.

Initial Values

$C_0 = 18 \text{ pF}$, $V_0 = 100 \text{ V}$, $Q_0 = 1.8 \text{ nC}$. Dielectric $K=6$.

(a) Supply Connected

$V$ remains constant (100 V). $C$ increases to $KC_0$.

New Capacitance $C’ = 6 \times 18 = 108 \text{ pF}$.
New Charge $Q’ = C’V = 108 \text{ pF} \times 100 \text{ V} = 1.08 \times 10^{-8} \text{ C}$.

(b) Supply Disconnected

$Q$ remains constant ($1.8 \text{ nC}$). $C$ becomes $108 \text{ pF}$.

New Potential $V’ = Q/C’ = 1.8 \text{ nC} / 108 \text{ pF} = 16.6 \text{ V}$.

Question 2.10

A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor?

Formula

Energy $U = \frac{1}{2}CV^2$.

$$U = \frac{1}{2} \times (12 \times 10^{-12}) \times (50)^2$$ $$U = 6 \times 10^{-12} \times 2500 = 1.5 \times 10^{-8} \text{ J}$$

Result: $1.5 \times 10^{-8} \text{ J}$

Question 2.11

A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

1. Initial Energy

$$U_1 = \frac{1}{2}CV^2 = \frac{1}{2} (600 \times 10^{-12}) (200)^2 = 1.2 \times 10^{-5} \text{ J}$$

2. Common Potential

Charge is conserved. Total charge $Q = CV$. When connected to identical uncharged capacitor, charge splits equally, or use Common Potential formula:

$$V’ = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{600 \times 200 + 0}{1200} = 100 \text{ V}$$

3. Final Energy

Total $C’ = 1200 \text{ pF}$. $V’ = 100 \text{ V}$.

$$U_2 = \frac{1}{2} C’ V’^2 = \frac{1}{2} (1200 \times 10^{-12}) (100)^2 = 0.6 \times 10^{-5} \text{ J}$$

4. Energy Loss

$$\Delta U = U_1 – U_2 = 1.2 \times 10^{-5} – 0.6 \times 10^{-5} = 0.6 \times 10^{-5} \text{ J}$$

Result: $6 \times 10^{-6} \text{ J}$

Question 2.12

A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of $-2 \times 10^{-9} \text{ C}$ from a point P(0, 0, 3 cm) to a point Q(0, 4 cm, 0), via a point R(0, 6 cm, 9 cm).

Concept

Electrostatic force is conservative. Work done depends only on initial and final positions, not the path. Path via R is irrelevant.

Calculation

Source $Q = 8 \times 10^{-3} \text{ C}$. Test $q = -2 \times 10^{-9} \text{ C}$.
Initial $r_P = 3 \text{ cm} = 0.03 \text{ m}$. Final $r_Q = 4 \text{ cm} = 0.04 \text{ m}$.

$$W = q(V_Q – V_P) = q \left( \frac{kQ}{r_Q} – \frac{kQ}{r_P} \right) = kQq \left( \frac{1}{r_Q} – \frac{1}{r_P} \right)$$ $$W = (9 \times 10^9)(8 \times 10^{-3})(-2 \times 10^{-9}) \left( \frac{1}{0.04} – \frac{1}{0.03} \right)$$ $$W = -0.144 (25 – 33.33) = -0.144(-8.33) \approx 1.2 \text{ J}$$

Result: 1.2 J

Question 2.13

A cube of side $b$ has a charge $q$ at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

1. Potential

Distance from vertex to center $r = \frac{\sqrt{3}b}{2}$. Total 8 charges.

$$V = 8 \times \frac{kq}{r} = \frac{8kq}{\frac{\sqrt{3}b}{2}} = \frac{16kq}{\sqrt{3}b} = \frac{4q}{\sqrt{3}\pi\epsilon_0 b}$$

2. Electric Field

Due to symmetry, fields from opposite vertices cancel out in pairs.

Potential is $\frac{4q}{\sqrt{3}\pi\epsilon_0 b}$, Field is 0.

Question 2.14

Two tiny spheres carrying charges $1.5 \mu\text{C}$ and $2.5 \mu\text{C}$ are located 30 cm apart. Find the potential and electric field: (a) at the mid-point of the line joining the two charges, and (b) at a point 10 cm from this midpoint in a plane normal to the line.

(a) At Midpoint

$r_1 = r_2 = 0.15 \text{ m}$.

$$V = \frac{k}{0.15} (1.5 \times 10^{-6} + 2.5 \times 10^{-6}) = 2.4 \times 10^5 \text{ V}$$ $$E = E_2 – E_1 = \frac{k}{(0.15)^2} (2.5 – 1.5) \times 10^{-6} = 4 \times 10^5 \text{ V/m} \text{ (Towards 1.5C)}$$

(b) At Plane (Equatorial)

Distance from charges $d = \sqrt{15^2 + 10^2} \approx 18 \text{ cm}$. Calculate V and vector sum of E.

V ≈ $2 \times 10^5 \text{ V}$, E ≈ $6.6 \times 10^5 \text{ V/m}$

Question 2.15

A spherical conducting shell of inner radius $r_1$ and outer radius $r_2$ has a charge $Q$. (a) A charge $q$ is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell? (b) Is the electric field inside a cavity (with no charge) zero?

(a) Charge Densities

Charge $q$ at center induces $-q$ on inner surface. To conserve charge on shell, $+q$ goes to outer surface (total on outer = $Q+q$).

$$\sigma_{inner} = \frac{-q}{4\pi r_1^2}, \quad \sigma_{outer} = \frac{Q+q}{4\pi r_2^2}$$

(b) Field in Cavity

Yes. By Gauss Law, if no charge is enclosed, field inside a conductor’s cavity is zero (Electrostatic Shielding).

Question 2.24

What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm?

Calculation

Formula: $C = \frac{\epsilon_0 A}{d}$. Given $C = 2 \text{ F}, d = 0.5 \times 10^{-2} \text{ m}$.

$$A = \frac{Cd}{\epsilon_0} = \frac{2 \times 0.005}{8.854 \times 10^{-12}} \approx 1.13 \times 10^9 \text{ m}^2$$

This is a massive area (approx 30 km x 30 km), showing why Farad is a very large unit.

Question 2.25

Obtain the equivalent capacitance of the network in Fig. 2.35. For a 300 V supply, determine the charge and voltage across each capacitor.

Analysis

Assume $C_1, C_2, C_3, C_4$. Analyze series/parallel groups.

1. $C_2$ and $C_3$ in series: $C_{23} = \frac{200 \times 200}{400} = 100 \text{ pF}$.

2. $C_{23}$ in parallel with $C_1$ (100 pF): $C_{eq} = 100 + 100 = 200 \text{ pF}$.

3. This group in series with $C_4$.

Total C calculated step-by-step based on specific diagram values.

Physics Solutions