NCERT Solutions for class-12 Physics Chapter 2 Electrostatic Potential and Capacitance is prepared by our senior and renowned teachers of Param Himalaya .
Question 2.1 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.
Solution :
Let the Potential be zero at O, then VA + VB = 0 ,
$$9 \times10^{9}\frac{q_{A}}{x}+9\times10^{9}\frac{q_{B}}{r-x} = 0$$
$$9 \times10^{9}[\frac{5\times10^{-8}}{x}+\frac{(-3\times10^{-8})}{(0.16-x)}] = 0$$
$$\frac{5\times10^{-8}}{x}=\frac{(3\times10^{-8})}{(0.16-x)}$$
$$5(0.16-x) = 3x $$
$$0.8 = 3x+5x = 8x$$
$$ x = 0.1 m = 10 cm$$
Question 2.2 A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the centre of the hexagon.
Solution :
Total potential at O is given by :
$$V = 6\times(\frac{1}{4\pi \varepsilon _{0}}\frac{q}{r})$$
$$V = 6\times(9\times10^{9})\times\frac{5\times10^{-6}}{.1}$$
$$V=2.7\times10^{6}V$$
Question 2.3 Two charges 2 μC and –2 μC 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?
Solution :
$$q_{1} = 2\mu C =2\times10^{-6}C$$
$$q_{2} = -2\mu C =-2\times10^{-6}C$$
$$r=6cm = 0.06m$$
(a) The equipotential surface is the plane perpendicular to the line AB joining the two charges and passing through the mid-point . on this plane potential zero everywhere.
$$\frac{1}{4\pi \varepsilon _{0}}[\frac{q_{1}}{x}+\frac{q_{2}}{(0.06-x)}] = 0$$
$$\frac{q_{1}}{x}=-\frac{q_{2}}{(0.06-x)}$$
$$\frac{2\times10^{-6}}{x}=-\frac{(-2\times10^{-6})}{(0.06-x)}$$
$$x=0.06-x $$
$$x=\frac{0.06}{2}=0.03m$$
(b) The direction of electric field is from positive to negative charge i.e A to B , Which is infect perpendicular to the equipotential plane.
Question 2.4 A spherical conductor of radius 12 cm has a charge of 1.6 × $10^{7}$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 of the sphere?
Solution:
(a) Inside a conductor , the electric field is zero because the charge resides on the surface of a conductor.
(b) Electric field just outside the sphere is given by :
$$E = \frac{1}{4\pi \varepsilon _{0}}\frac{q}{R^{2}}$$
$$E = 9 \times 10^{9} \times \frac{(1.6 \times 10^{-7})}{(12\times 10^{-2})^{2}}$$
$$E= 10^{5}NC^{-1}$$
(c) Electric field at a distant point is given by :
$$E = \frac{1}{4\pi \varepsilon _{0}}\frac{q}{r^{2}}$$
$$E = 9 \times 10^{9} \times \frac{(1.6 \times 10^{-7})}{(18\times 10^{-2})^{2}}$$
$$E= 4.44 \times 10^{4}NC^{-1}$$
Question 2.5 A parallel plate capacitor with air between the plates has a capacitance of 8 pF (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 :
$$C^{'} = \frac{\varepsilon_{0}\varepsilon_{r}A}{d}$$
$$C^{'} = \varepsilon _{r}C = 12(8\times10^{-12})$$
$$C^{'} = 96 \times 10^{-12}F = 96 pF$$
Question 2.6 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 120 V supply?
Solution :
(a) $$\frac{1}{C_{eq}} = \frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}$$
$$= \frac{1}{C}+\frac{1}{C}+\frac{1}{C} = 3 \times \frac{1}{C}$$
i.e $$\frac{1}{C_{eq}} = 3 \times \frac{1}{9 \times 10^{-12}}$$
$$\frac{1}{C_{eq}}=\frac{1}{3 \times 10^{-12}}$$
or $$C_{eq} = 3 \times 10^{-12}F=3pF$$
(b) Here
$$V_{1}+V_{2}+V_{3} = 120$$
i.e $$\frac{q}{C_{1}}+\frac{q}{C_{2}}+\frac{q}{C_{3}}= 120$$
$$\because C=\frac{q}{V} \ or \ V =\frac{q}{C}$$
$$\frac{3q}{C} = 120$$
$$\frac{3\times q}{9\times 10^{-12} } = 120$$
or $$q= 360\times10^{-12}F$$
$\therefore$ P.D across a capacitor
$$V= \frac{q}{C} = \frac{360\times 10^{-12}}{9\times 10^{-12}}$$
$$V=40 V.$$
Question 2.7 Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Solution :
(a) Total Capacitance =
$$ C_{1} + C_{2} + C_{3}$$
$$C= 2+3+4 = 9pF$$
(b) Using
$$C=\frac{q}{C}$$
we get q=CV
$$q_{1} = C_{1}V=2\times 10^{-12} \times 100$$
$$q_{1} = 2 \times 10^{-10}C=200pC$$
$$q_{2} = C_{2}V$$
$$q_{2} = 3 \times 10^{-12}\times100$$
$$q_{2}= 3 \times 10^{-10} = 300pC$$
$$q_{3} = C_{3}V$$
$$q_{3}= 4 \times 10^{-12}\times100$$
$$q_{3}= 4 \times 10^{-10} = 400pC$$
Question 2.8 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 100 V supply, what is the charge on each plate of the capacitor?
Solution :
$$C = \frac{\varepsilon _{0}A}{d}$$
we get
$$C = \frac{(8.854\times 10^{-12})(6\times10^{-3})}{(3\times 10^{-3})}$$
$$C = 17.708 \times 10^{-12} \cong 18 pF$$
Using
$$C=\frac{q}{V}$$
we get q = CV
$$q = 18 \times 100 = 1.8 \times 10^{-9}C$$
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 :
(a) (i)$${C}'= \varepsilon _{r}C$$
$${C}'= 6 \times 18 = 108 pF$$
(ii) $$q ={C}'V = 108 \times 100$$
$$q= $1.08 \times 10^{-8}C$$
(b) q remains $1.8\times 10^{-9}C$
Capacitance = 108 pF
$$V = \frac{q}{C} = \frac{1.8 \times 10^{-9}}{108 \times 10^{-12}}=16.6 V$$
Question 2.10 A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
Solution :
$$E = \frac{1}{2}CV^{2}$$
$$E= \frac{1}{2} \times 12 \times 10^{-12} \times 50 \times 50$$
$$E = 1.5 \times 10^{-8}J$$
Questions 2.11 A 600pF capacitor is charged by a 200V 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?
Solution :
$$C_{1} = 600pF = 6 \times 10^{-10}F$$
$$C_{2} = 6 \times 10^{-10}F $$
$$V_{1} = 200V , V_{2} = 0$$
$$Energy loss = \frac{C_{1}C_{2}(V_{1}-V_{2})^{2}}{2(C_{1}+C_{2})}$$
$$= \frac{36 \times10^{-20}\times 4 \times 10^{4}}{2\times 12\times 10^{-10}}$$
$$= 6 \times 10^{-6}J$$