Ncert Solution for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance

Electrostatic Potential and Capacitance ncert solutions: Ncert Class 12 Physics Chapter 2 Exercise Solutions

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Question 2.1: Two charges 5 × 10^-8 C and -3 x 10^-8 C are located 16 cm apart from each other. At what point (s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Solution 2.1:

Given, there are two charges,

q₁ = 5 x 10^-8 C

q₂ = -3 x 10^-8 C

• the two charges are at a distance, d = 16 cm = 0.16 m from each other.
• Let us consider a point “P” over the line joining charges q₁ and q₂.
• Let the distance of the considered point P from q₁ be ‘r’
• Let us consider point P to have zero electric potential (V).
• The electric potential at point P is the summation of potentials due to charges q₁ and q₂.

Therefore, 

\( V = \frac{q_{1}}{4\pi \epsilon_{0}{r}} + \frac{q_{2}}{4\pi \epsilon_{0}{(d – r)}} \) ……….(1)

Here,

\( \epsilon _{o} \) = permittivity of free space.

Putting  V = 0, in equation (1), we get,

0 = \( V = \frac{q_{1}}{4\pi \epsilon_{0}{r}} + \frac{q_{2}}{4\pi \epsilon_{0}{(d – r)}} \)

\( \frac{q_{1}}{4\pi \epsilon_{0}{r}} = – \frac{q_{2}}{4\pi \epsilon_{0}{(d – r)}} \)

\( \frac{q_{1}}{r} = – \frac{q_{2}}{d – r} \)

\( \frac{5 \times 10^{-8}}{r} = – \frac{(- 3 \times 10^{-8})}{0.16 – r} \)

r = 0.1m = 10cm.

Therefore, at a distance of 10 cm from the positive charge, the potential is zero between the two charges.

Let us assume a point P at a distance ‘s’ from the negative charge is outside the system, having a potential zero.

So, for the above condition, the potential is given by

\( V = \frac{q_{1}}{4\pi \epsilon_{0}{s}} + \frac{q_{2}}{4\pi \epsilon_{0}{(s – d)}} \) ……….(2)

For V = 0, equation (2) can be written as :

0 = \( \frac{q_{1}}{4\pi \epsilon_{0}{s}} + \frac{q_{2}}{4\pi \epsilon_{0}{(s – d)}} \)

\( \frac{q_{1}}{4\pi \epsilon_{0}{s}} = – \frac{q_{2}}{4\pi \epsilon_{0}{(s – d)}} \)

\( \frac{q_{1}}{s} = – \frac{-q_{2}}{s – d} \)

\( \frac{5 \times 10^{-8}}{s} = – \frac{(- 3 \times 10^{-8})}{s – 0.16} \)

5(s – 0.16) = 3s

0.8  = 2s

S = 0.4m = 40cm.

Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.

Question 2.2: A regular hexagon of side 10 cm has a charge of 5 \(µC\) at each of its vertices. Calculate the potential at the centre of the hexagon.

Solution 2.2: The figure shows a regular hexagon containing charges q at each of its vertices.

Here,

q =  5 µC = 5 × 10^-6 C.

• Length of each side of the hexagon, AB =BC = CD = DE = EF = FA = 10 cm.
• The distance of the vertices from the centre O, d = 10 cm.
• The electric potential at point O,
• The electric potential, V = \( \frac{6 \times q}{4\pi \epsilon _{o}{d}} \)

Here,

\( \epsilon _{o} \) = permittivity of free space.

\(\frac{ 1 }{ 4 \pi \epsilon _{ 0 }} \) = 9 x 10⁹ NC^-2 m²

V = \( \frac{6 × 9 × 10^{9} × 5 × 10^{-6}}{0.1} \) = 2.7 × 10⁶ V.

Therefore, the potential at the centre of hexagon is 2.7 × 10⁶ V.

Question 2.3: Two charges, 2 \(µC\) and -2 \(µC\), are placed at points A and B, 6 cm apart. (a) Identify the equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface?

Solution 2.3: (a) An equipotential surface is defined as the surface over which the total potential is zero. In the given question, the plane is normal to line AB. The plane is located at the mid-point of line AB as the magnitude of the charges is the same.

(b) At every point on this surface, the direction of the electric field is normal to the plane in the direction of AB.

Question 2.4: A spherical conductor of radius 12 cm has a charge of 1.6 x 10^-7C 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 of the sphere?

Solution 2.4: (a) Radius of spherical conductor, r = 12cm = 0.12m

The charge is distributed uniformly over the surface, q = 1.6 x 10^-7 C.

The electric field inside a spherical conductor is zero. this is because is there is field inside the conductor, then charges will move to neutralize it.

(b) Electric field E, just outside the conductor, is given by the relation,

\( E = \frac{ q }{ 4\pi \epsilon _{ 0 } r ^{ 2 }}\)

Here,

\( \epsilon _{o} \) = permittivity of free space.

\( \frac{ 1 }{ 4\pi \epsilon _{ 0 }}\) = 9 x 10⁹ N m²C^-2

Therefore,

E = \( \frac{9 \times 10^{-9} \times 1.6 \times 10^{-7}}{(0.12)^{2}} \)

=10⁵ N C^-1.

Therefore, just outside the sphere, the electric field is 10⁵ N C^-1.

(C) From the centre of the sphere, the electric field at a point 18m = E₁.

From the centre of the sphere, the distance of point d = 18 cm = 0.18m.

E₁ =\( \frac{ q }{ 4\pi \epsilon _{ 0 } d ^{ 2 }}\)

= \( \frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{(1.8 \times 10^{-2})^{2}} \)

= 4.4 x 10⁴ NC^-1

So, from the centre of the sphere, the electric field at a point 18 cm away is 4.4 x 10⁴ NC^-1.

Question 2.5: A parallel plate capacitor with air between the plates has a capacitance of 8pF  (1pF = 10^-12 F). 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?

Solution 2.5: Given: Capacitance of capacitor when medium between two plates is air, C = 8 pF

initially, distance between the parallel plates was d and it was filled with air. Air has a dielectric constant, k = 1

Capacitance, C = \( \frac{k \epsilon _{o} A}{d} \)

= \( \frac{\epsilon _{o} \times A}{d} \) … eq(1)

Here,

A = area of each plate

\( \epsilon _{o} \) = permittivity of free space.

Now, if the distance between the parallel plates is reduced to half, then d₁ = d/2

Given the dielectric constant of the substance, k₁ = 6

Hence, the capacitance of the capacitor,

C₁ = \( \frac{k_{1} \times \epsilon_{o} \times A}{d_{1}} \) = \( \frac{6 \epsilon _{o} \times A}{d/2} \)

= \( \frac{12 \epsilon _{o} A}{d} \) … (2)

Taking ratios of equations (1) and (2), we get, C₁ = 2 x 6 C = 12 C = 12 x  8 pF = 96pF.

Hence, the capacitance between the plates is 96pF.

Question 2.6: Three capacitors connected in series have a capacitance of 9pF each. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

Solution 2.6: (a)  Given,

The capacitance of the three capacitors, C = 9 pF

Equivalent capacitance (c_eq) is the capacitance of the combination of the capacitors given by

\( \frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} = \frac{3}{9} =\frac{1}{3}\)

\( \frac{1}{C_{eq}} = \frac{1}{3}\)

Therefore, the total capacitance = 3pF.

(B) Given supply voltage, V = 100V

The potential difference (V₁) across the capacitors will be equal to one-third of the supply voltage.

Therefore, V₁ = \( \frac{V}{3} \)

= \( \frac{120}{3} \)

= 40V.

Hence, the potential difference across each capacitor is 40V.

Question 2.7: Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

Solution 2.7: (a) Given, C₁ = 2pF, C₂ = 3pF and C₃ = 4pF.

Equivalent capacitance for the parallel combination is given by C_eq.

Therefore, C_eq = C₁ + C₂ + C₃ = 2 + 3 + 4 = 9pF

Hence, the total capacitance of the combination is 9pF.

(b) Supply voltage, V = 100V

The three capacitors have the same voltage, V = 100v

q = VC

where,

q = charge

C = capacitance of the capacitor

V = potential difference

for capacitance, c = 2pF

q = 100 x 2 = 200pC = 2 x 10^-10C

for capacitance, c = 3pF

q = 100 x 3 = 300pC = 3 x 10^-10C

for capacitance, c = 4pF

q = 100 x 4 = 400pC = 4 x 10^-10 C

Question 2.8: In a parallel plate capacitor with air between the plates, each plate has an area of 6 x 10^-3 m ² and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

Solution 2.8: Given,

Let’s solve the problem step by step to understand how the values are calculated.

Given:

• Area of each plate, A = 6 x 10^-3 m²
• Distance between the plates, d = 3mm = 3 x 10^-3 m
• Permittivity of free space (air), \( \epsilon _{o} \) = 8.854 × 10^-12 F/m
• Voltage across the plates, V = 100V

(a) Capacitance Calculation:

The capacitance ( C ) of a parallel plate capacitor is given by: \(C = \frac{\epsilon_0 A}{d}\)

Substituting the values:
\(C = \frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}}\)

Simplifying the expression:

\(C = \frac{8.854 \times 10^{-12} \times 6}{3} \, \text{F}\)

\(C = \frac{53.124 \times 10^{-12}}{3} \, \text{F}\)

\(C = 17.708 \times 10^{-12} \, \text{F}\)

This can be approximated to:

\(C \approx 18 \times 10^{-12} \, \text{F} = 18 \, \text{pF}\)

(b) Charge on Each Plate:

The charge ( Q ) on each plate is given by: Q = CV

Substituting the values:

\(Q = 18 \times 10^{-12} \times 100 \, \text{C}\)
\(Q = 1.8 \times 10^{-9} \, \text{C}\)

Final Answer:

• Capacitance C = 18 pF
• Charge on each plate \(Q = 1.8 \times 10^{-9} \, \text{C}\)

These match the values given in the problem statement: C = 18 pF and \(Q = 1.8 \times 10^{-9} \, \text{C}\).

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.

Solution 2.9: (a) Dielectric constant of the mica sheet, k = 6

If the voltage supply remains connected, the voltage between the two plates will be constant.

• Supply voltage, V = 100 V
• Initial capacitance, C = 1.771 × 10⁻¹¹ F
• New capacitance, C₁ = kC = 6 × 1.771 × 10⁻¹¹ F = 106 pF
• New charge, q₁ = C₁V = 106 × 100 pC = 1.06 × 10^–8 C
• Potential across the plates remains 100 V.

(b) Dielectric constant, k = 6

• Initial capacitance, C = 1.771 × 10⁻¹¹ F
• New capacitance, C₁ = kC = 6 × 1.771 × 10⁻¹¹ F = 106 pF
• If the supply voltage is removed, then there will be a constant amount of charge in the plates.
• Charge = 1.771 × 10⁻⁹ C
• Potential across the plates is given by,

V₁ = q/C₁ = \(\frac{1.771 \times 10^{-9}}{106 \times 10^{-12}}\)

= 16.7 V

Question 2.10: A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

Solution 2.10: To find the electrostatic energy stored in a capacitor, you can use the formula:

\(E = \frac{1}{2} C V^2 \)

where:

• ( E ) is the energy stored in the capacitor,
• ( C ) is the capacitance,
• ( V ) is the voltage.

Given:

• Capacitance, C = 12pF = 12 x 10^-12 F,
• Voltage, V = 50v.

Plugging in these values:

\( E = \frac{1}{2} \times (12 \times 10^{-12} \text{ F}) \times (50 \text{ V})^2 \)

\( E = \frac{1}{2} \times 12 \times 10^{-12} \text{ F} \times 2500 \text{ V}^2 \)

\( E = \frac{1}{2} \times 30 \times 10^{-9} \text{ J} \)

E = 1.5 x 10^-8 J

So, the electrostatic energy stored in the capacitor is indeed 1.5 x 10^-8 J.

Question 2.11: A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Solution 2.11: To determine the electrostatic energy lost when a charged capacitor is connected to an uncharged capacitor, follow these steps:

1. Calculate the initial energy stored in the charged capacitor:

The energy ( E ) stored in a capacitor is given by:
\(E = \frac{1}{2} C V^2\)

where ( C ) is the capacitance and ( V ) is the voltage.

For the initially charged capacitor:

• Capacitance C = 600pF = 600 x 10^-12 F
• Voltage V = 200v

The initial energy \( E_i \) is:

\(E_i = \frac{1}{2} \times 600 \times 10^{-12} \text{ F} \times (200 \text{ V})^2\)
\(E_i = \frac{1}{2} \times 600 \times 10^{-12} \times 40000\)
\(E_i = 12 \times 10^{-6} \text{ J} = 1.2 \times 10^{-5} \text{ J}\)

1. Determine the final energy after connecting to the second capacitor: When the charged capacitor is connected to an uncharged capacitor, they will share charge and reach the same final voltage \( V_f \).

The total capacitance of the two capacitors in series is:

\(C_{\text{total}} = C_1 + C_2 = 600 \text{ pF} + 600 \text{ pF} = 1200 \text{ pF} = 1200 \times 10^{-12} \text{ F}\)

The total charge ( Q ) before connection is:

\(Q = C_1 \times V = 600 \times 10^{-12} \text{ F} \times 200 \text{ V} = 120 \times 10^{-12} \text{ C}\)

After connection, the charge ( Q ) is distributed equally over both capacitors, so:

\(V_f = \frac{Q}{C_{\text{total}}} = \frac{120 \times 10^{-12}}{1200 \times 10^{-12}} = 0.1 \text{ V}\)

The final energy ( E_f ) in each capacitor is:
\( E_f = \frac{1}{2} \times C_{\text{total}} \times V_f^2\)
\(E_f = \frac{1}{2} \times 1200 \times 10^{-12} \text{ F} \times (0.1 \text{ V})^2\)
\(E_f = \frac{1}{2} \times 1200 \times 10^{-12} \times 0.01\)
\(E_f = 6 \times 10^{-12} \text{ J} = 6 \times 10^{-6} \text{ J}\)

3. Calculate the energy lost: The energy lost is:
\(E_{\text{lost}} = E_i – E_f\)
\(E_{\text{lost}} = 1.2 \times 10^{-5} – 6 \times 10^{-6}\)
\(E_{\text{lost}} = 6 \times 10^{-6} \text{ J}\)

So the electrostatic energy lost in the process is indeed \( 6 \times 10^{-6} \text{ J} \).