# NCERT Solutions for Class 12 Physics Chapter 2 - Electrostatic Potential and Capacitance

## NCERT Solutions for Class 12 Physics Chapter 2 Free PDF Download

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1. Two charges 5 ×10–8 C and –3 × 10–8 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.

Sol. Let the potential be zero at point P at a distance x from point B. Let us consider two charges q1 and q2. According to the question.

q1 = 5 × 10–8 C, q2 = – 3 × 10–8 C, AB = 16 cm = 16 × 10–2m

The potential at point P due to charge q1

$$\text{V}_{\text{A}}=\frac{1}{4\pi\epsilon_{0}}.\frac{q_1}{\text{AP}}\\=\frac{9×10^{9}×5×10^{\normalsize-8}}{(16-x)×10^{\normalsize-2}}\qquad\text{...(i)}$$

The potential at point P due to charge q2

$$\text{V}_\text{B}=\frac{1}{4\pi\epsilon_{0}}.\frac{\text{q}_{2}}{\text{BP}}\\=\frac{9×10^{9}×(-3×10^{\normalsize-8})}{x×10^{\normalsize-2}}\qquad...(\text{ii})$$

Now the net potential at point P is zero i.e., VA + VB = 0

Putting the values from Eqs. (i) and (ii), we get

$$\frac{9×10^{9}×5×10^{\normalsize-8}}{(16-x)×10^{-2}} + \bigg(\frac{-9×10^{9}×3×10^{\normalsize-8}}{x×10^{\normalsize-2}}\bigg)=0\\\frac{5}{16-x}- \frac{3}{x}=0$$

or  5x – 3 (16 – x) = 0

or  5x – 48 + 3x = 0

or 8x = 48

or x = 6 cm

Thus, the electric potential is zero at point P i.e. at the distance of 6 cm from q2 (= – 3 × 10–8 C).

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

Sol. ABCDEF is a regular hexagon of side 10 cm each.

i.e., AB = BC = CD = DE = EF = FA = 10 cm

At each corner the charge q = 5µC is placed.

As the hexagon has six equilateral triangles.

∴ OA = OB = OC = OD = OE = OF = 10 cm.
Potential at point O =
Sum of potential at centre O due to individual point charges.

∴ VO = VA + VB + VC + VD + VE + VF

$$\text{V}_{o}=\frac{1}{4\pi\epsilon_{0}}\bigg[\frac{q}{\text{OA}}+ \frac{q}{\text{OB}}+ \frac{q}{\text{OC}} + \frac{q}{\text{OD}} + \frac{q}{\text{OE}} + \frac{q}{\text{OF}}\bigg]\\\bigg(\because\text{V}=\frac{1}{4\pi\epsilon_{0}}.\frac{q}{r}\bigg)\\\text{V}_{0}=9×10^{9}\bigg[\frac{5×10^{\normalsize-6}}{10×10^{\normalsize-2}} + \frac{5×10^{\normalsize-6}}{10×10^{\normalsize-2}} + \frac{5×10^{-6}}{10×10^{\normalsize-2}} + \frac{5×10^{\normalsize-6}}{10×10^{\normalsize-2}} + \frac{5×10^{\normalsize-6}}{10×10^{-2}} + \frac{5×10^{\normalsize-6}}{10×10^{-2}}\bigg]\\\text{or\space V}_{o}=9×10^{9}×\frac{6×10^{\normalsize-6}×5}{10×10^{\normalsize-2}}$$

or VO = 27 × 104

VO = 2.7 × 106 V

3. Two charges 2mC and – 2mC 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?

Sol. (a) As we know that equipotential surface means the surface where potential remains same at each point.

∴ The potential at P

$$\text{V}=\frac{1}{4\pi\epsilon_{0}}\bigg[\frac{2×10^{6}}{0.03} + \frac{(-2×10^{\normalsize-6})}{0.03}\bigg]=0$$

So, the potential is zero at each point on the line passes through the mid-point of AB and perpendicular to it. So, a plane passing through the mid-poind P of AB is an equipotential surface.

(b) We know that Edr = dV, the value of dV = 0 at each point of equipotential surface.

∴ E dr = 0

So, the angle between electric field vector and distance vector will be 90°. Thus, the electric field is always normal to the plane passing through AB.

4. A spherical conductor of radius 12 cm has a charge of 1.6 × 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?

Sol. (a) As we know that electric field, inside the conductor is zero. So, the electric field inside the spherical sphere is zero.

(b) For a point just outside the sphere i.e., lying on the surface of the sphere, the charge may be supposed to be concentrated on the centre of the sphere by using the formula of electric field.

$$\text{E}=\frac{1}{4\pi\epsilon_{0}}.\frac{q}{r^{2}}=\frac{9×10^{9}×1.6×10^{\normalsize-7}}{0.12×0.12}$$

= 1×105 N/C

(c)

Using the formula of electric field

$$\text{E}=\frac{1}{4\pi\epsilon_{0}}.\frac{q}{(\text{OP})^{2}}=\frac{9×10^{9}×1.6×10^{\normalsize-7}}{0.18×0.18}\\=4.4×10^{4}\text{N/C}$$

5. A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1 pF = 10–12F). 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 6F.

Sol. Let the distance between the plates be d and air the filled between the plates. Now, the capacitance is C0.

C0 = 8pF = 8 × 10–12F

Using the formula of capacitance

$$\text{C}_{0}=\frac{\epsilon_{0}\text{A}}{d}$$

(where A is the area of plates and d is the distance between them)

$$8×10^{\normalsize-12}=\frac{\epsilon_{0}\text{A}}{d}\qquad...(\text{i})$$

Now, the distance between the plates is reduced to half i.e.,

$$\text{d'}=\frac{d}{2}$$

and the space between the plates is filled with a dielectric of dielectric constant 6. Now, let the new capacitance be C. In this condition area remains constant.

$$\therefore\space\text{C}=\frac{\text{K}\epsilon_{0}\text{A}}{d'}$$

(where K is the dielectric constant i.e., 6)

$$\text{C}=6.\frac{\epsilon_{0}\text{A}}{d}.2$$

or  C = 12 × 8 × 10–12 [Using Eq. (i)]

or  C = 96 × 10–12 F = 96 pF

The new capacitance becomes 96 pF.

6. Three capacitors each of capacitance 9 pF are connected in series.

(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 120V supply?

Sol. There are three capacitors each of capacitance 9 pF

i.e. C1 = C2 = C3 = 9pF

and Voltage, V= 120V

(a) The total capacitance in series combination

$$\frac{1}{\text{C}}=\frac{1}{\text{C}_{1}}+\frac{1}{\text{C}_{2}}+\frac{1}{\text{C}_{3}}=\frac{1}{9} + \frac{1}{9} +\frac{1}{9}\\\frac{1}{\text{C}}=\frac{3}{9}\Rarr\text{C}_{s}=3\text{pF}$$

(b) Let the charge across the system be q and potentials across C1, C2 and C3 be V1, V2 and V3 respectively. Charge q = C.V = 3 × 120 = 360 pC

Potential difference across C1,

$$\text{(V}_{1})=\frac{q}{\text{C}_{1}}=\frac{360}{9}=40\text{V}$$

Potential difference across C2,

$$\text{(V}_{2})=\frac{q}{\text{C}_{2}}=\frac{360}{9}=40\text{V}$$

Potential difference across C3,

$$\text{(V}_{3})=\frac{q}{\text{C}_{3}}=\frac{360}{9}=40\text{V}$$

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

7. Three capacitors of capacitances 2 pF, 3pF and 4pF 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.

Sol. Given: C1 = 2pF, C2 = 3pF and C3 = 4pF

(a) The total capacitance of the parallel combination is given by

CP = C1 + C2 + C3 = 2 + 3 + 4 = 9pF

(b) Let the charges on the capacitors C1, C2 and C3 be q1, q2 and q3 respectively.

Charge q1 at C1= C1V = 2 × 100 = 200pC

Charge q2 at C2 = C2V = 3 × 100 = 300 pC

Charge q3 at C3 = C3V = 4 × 100 = 400 pC

8. In a parallel plate capacitor with air between the 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 this capacitor is connected to a 100V supply, what is the charge on each plate of the capacitor?

Sol. Capacitance of a parallel plate capacitor

$$\text{C}=\frac{\epsilon_{0}\text{A}}{d}\\=\frac{8.854×10^{\normalsize-12}×6×10^{\normalsize-3}}{3×10^{\normalsize-3}}$$

C = 1.77 × 10–11F

When the capacitor
is connected to a 100V supply, charge on each plate of the capacitor

q = CV = 1.77 × 10–11 × 100

q = 1.77 × 10–9C = 1.77 nC

9. Explain what would happen, if in the capacitor given in Q.8, a 3 mm thick mica sheet (of dielectric constant = 6) where inserted between the plates.

(a) While the voltage supply remained connected?

(b) After the supply was disconnected?

Sol. (a) As we introduce a mica sheet of dielectric constant K = 6 between the plates and the voltage supply remained connected.

We know that the capacitance is directly proportional to the dielectric constant.

C ∝ K

By inserting the dielectric the new capacitance becomes

C’= CK = 6 × 1.77 × 10–11

= 1.062 × 10–10F

Now, we see that the capacitance increases, so the charge will also increase.

Charge q’ on the capacitor = C’V

= 1.062 × 10–10 ×100

or

q’ = 1.062 × 10–8C

(b) When the voltage supply was disconnected, charge of the capacitor remains constant.

New capacitance

C’ = 1.062 × 10–10F

The potential difference across the plates of the capacitor

$$\text{V'}=\frac{q}{C'}\frac{1.77×10^{\normalsize-9}}{1.062×10^{\normalsize-10}}=16.67\text{V}$$

Note : When the battery remains connected, the capacitance and charge are is changed.

On disconnecting the battery charge remains same but capacitance and potential are changed.

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

Sol. Energy stored in the capacitor

$$\text{E}=\frac{1}{2}\text{CV}^{2}\\=\frac{1}{2}×12×10^{\normalsize-12}×50×50$$

= 1.5 × 10-8 J

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

Sol. Capacitance of capacitor C1 = 600 pF and supply voltage V1 = 200V

C2 = 600pF = 600 × 10–12F and V2 = 0

$$\text{Loss in energy (E) =}\frac{\text{C}_1\text{C}_2(\text{V}_{1}-\text{V}_{2})^{2}}{2(\text{C}_{1} + \text{C}_{2})}\\\text{E}=\frac{600×10^{\normalsize-12}×600×10^{\normalsize-12}(200-0)^{2}}{2(600+600)×10^{\normalsize-12}}$$

= 6 × 10-6 J

Thus, the 6 × 10-6J amount of electrostatic energy is lost in the sharing of charges.