Get free **NCERT Solutions** on the Aasoka platform to achieve good grades in the Class 12 board exams. The solutions are formulated by the subject matter experts in a simple and lucid language that can easily be understood by all students. The **NCERT Solutions for Class 12th** makes it easier for students to get a good grasp on the concepts or topics related to the Physics chapter **“Electrostatic Potential and Capacitance”**.

The class 12th chapter of Physics **“Electrostatic Potential and Capacitance”** explains electrostatic potential, problems based on the effective capacitance of capacitors, energy stored in a capacitor, important derivations and formulas, development of electrostatic potential, etc.

##### Question 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.

##### Answer:

Given

q_{1} = 5×10^{–8} C, r = 16 cm= 0.16 m

q_{2} = –3 × 10^{–8} C

Let potential be zero at a distance x metre from

positive charge q_{1}

$\therefore $ r_{1} = x metre

r_{2} = (0.16 – x) metre

##### Question 2:

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

##### Answer:

From the figure TBQ 2.2 we have
OP = OQ = OR = OS = OT = OU = r = 10 cm = 0.1 m
And given q = 5 $\mathrm{\mu}$C = 5 × 10–6 C

$\therefore $ Potential at O due to all the charges

##### Question 3:

Two charges 2 $\mathrm{\mu}$C and –2 $\mathrm{\mu}$C are placed at points A and B 6 cm apart.
Identify an equipotential surface of the system.

What is the direction of the electric field at every point on this surface ?

##### Answer:

The plane normal to AB and passing through its mid-point has zero potential everywhere hence the plane is equipotential.

Normal to the plane is the direction AB.

##### Question 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

- inside the sphere
- just outside the sphere
- at point 18 cm from the centre of the sphere ?

##### Answer:

(a) Zero

##### Question 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 ?

##### Answer:

The capacitance of capacitor with air as dielectric is given by

##### Question 6:

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

What is the total capacitance of the combination ?

What is the potential difference across each capacitor, if the combination is connected to a 120 volt supply ?

##### Answer:

Given C _{1}= C _{2}= C _{3} = 9 pF = 9 × 10 ^{–12} F; V = 120 volt.

Total capacitance of the series combination is given by

Let q be charge on each capacitor. Then, sum of the potential differences across their plates must be equal to 120 V

##### Question 7:

Three capacitors of capacitance 2pF, 3pF and 4pF are connected in parallel.

What is the total capacitance of the combination ?

Determine the charge on each capacitor, if the combination is connected to a 100 V supply.

##### Answer:

Here C_{1} = 2pF; C_{2} = 3pF; C_{3} = 4pF ; V = 100 volt

Total capacitance of the parallel combination is given by

C = C_{1} + C_{2} + _{C3} = 2 + 3 + 4 = 9 pF.

Let q_{1}, q_{2} and q _{3} be the charges on C_{1}, C_{2} and
C_{3} respectively. In the parallel combination, the potential difference across each capacitor will be equal to the supply voltage i.e. 100 V.

Therefore, q_{1} = C_{1} V = 2 × 10^{–12 × 100
= 2 × 10–10 C,
q2 = C2 V=3×10–12×100 =3×10–10 C
and q 3 = C 3 V = 4×10–12×100=4×10 –12 C.}

##### Question 8:

In a parallel plate capacitor with air
between the plates, each plate has an area of
6 × 10^{–3} 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 of the capacitor ?

##### Answer:

Given

A = 6 × 10^{–3} m^{2}, d = 3 mm = 3 × 10^{–3} m
C_{0} = ?

V_{0} = 100 V, q = ?

##### Question 9:

Explain what would happen if in the capacitor given in exercise 2.8, a 3mm thick mica sheet (of dielectric constant = 6) were inserted between the plates.

- While the voltage supply remained connected.
- After the supply was disconnected.

##### Answer:

(a) When supply voltage remains connected

V = 100 volt (voltage remains constant)
Capacitance C = KC _{0} = 6 × 18 = 108 pF
Charge q =1.8 × 10 ^{–9} C (remains constant)

(b) After the supply is disconnected

##### Question 10:

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

##### Answer:

Given C = 12 pF = 12 × 10^{–12} C

V = 50 volt

##### Question 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 ?

##### Answer:

Given C_{1} = 600 pF = 600 × 10^{–12} F

V_{1} = 200 volt