NCERT Questions for Class 12 Physics Chapter 2 – Electrostatic Potential and Capacitance

Important Questions with Solutions of Class 12 Physics Chapter 2 – Electrostatic Potential and Capacitance

1) What is responsible for reducing the electric field inside a dielectric material kept under an external electric field?

Ans – During the dielectric polarization, the internal electric field will reduce during an external electric field. This process builds an inner electric field opposing the outer electric field within a dielectric by lowering the net electric field.

2) The force of an attraction over the 2 poles of the electric charges kept at a distance d in a medium F. At what distance, do you place these electric charges in the same medium for the force between them to turn F/3?

Ans – Two pole charges are named q₁ & q₂ separated by the distance d. It can be denoted as,

F = (K*q₁*q₂)/(d²)

Now, when the force turns F/3 then assume distance d = x

3) The plates of a completely charged capacitor are merged using a voltmeter. What will happen to the reading on the voltmeter when the capacitor plates are moved far from their original position?

Ans – The connection between area, distance capacitance & dielectric constant is

C = Aϵ0/d

C ∝ 1/d

Therefore, once the distance increases then the capacitance will decrease. Since V= Q/C & the capacitor charge remains constant. Then the final voltmeter reading will be increased.

4) A parallel plate capacitor with air over the plates consists of a capacitance named 8pF (1pF = 10^-12F). Calculate the capacitance when the distance between these 2 plates gets lowered to half & the free space is filled with an element of dielectric constant 6.

Ans – For air, the capacitance can be denoted as C₀ = Aϵ₀/d

C₀ = 8pF = 8 * 10^-12F

Now d’ = d/2 & K = 6

So, C’ = (A*C₀)/(d’) * 2 * K

C’ = 8 * 10^-12 * 2 * 6

C’ = 96 * 10^-12pF

5) Draw the equipotential surfaces related to

a) Constant electric field in the z-direction

Ans – Equidistant planes parallel to the x-y plane are denoted as equipotential surfaces.

b) A field that uniformly increases in magnitude but stays constant. For instance, z-direction.

Ans – Equipotential surfaces are planes parallel to the x-y plane with the exception that once the planes get closer to each other, the field starts to increase.

c) A single positive charge placed at the origin point

Ans – Equipotential surfaces will be concentric spherical surfaces placed at the origin.

d) A uniform grid with long & evenly spaced parallel charged wires in a plane.

Ans – An equipotential surface is a shape near the desired grid whose size changes constantly. The shape will slowly gain its form of planes parallel to the grid at a distance that is far away.

6) 2 dielectric slabs with dielectric constant K1 & K2 are placed between the 2 plates with an area A of parallel plate capacitor as depicted below. Calculate the capacitor’s net capacitance where the area of each plate is denoted as A/2.

Ans – Since 2 capacitors are in parallel the net capacitance will be, C = C₁ + C₂

is the net capacitance of the capacitor.

7) Give evidence that the energy kept in a parallel plate capacitor is 1/2CV².

Ans – Assume that a capacitor is linked with a battery & delivers a small amount of charge dq with a constant potential V. Then, a small amount of work done by the battery will be dw = Vdq

⇒ dW = qc/dq  Since, q = CV

Now, the total work done in which the capacitor is completely charged to q

Work done is stored in the capacitor as electrostatic potential energy.

So, W = U = 1/2 CV²

Hence proved.

8) A normal hexagon of side 10cm contains 5µC of charge at its vertices. Figure out the potential at the centre of the hexagon.

Ans – The diagram below shows six equal charges q, taken at the vertices of a normal hexagon.

Here, charge (q) = 5µC = 5*10^-5 C

The side of the hexagon are AB = BC = CD = DE = EF = FA = 10cm

The distance of each vertex from center O is d = 10cm The electric potential at point O is,

Where ϵ₀ is the permittivity in free space

So, the potential at the centre of the hexagon is denoted as 2.7*10⁶V

9) A 12 pF capacitor is charged by a 50 V battery. Find the total electrostatic energy stored in the capacitor.

Ans – Capacitor of the capacitance will be C = 12pF = 12 * 10^-12 F

Potential difference is mentioned as V = 50V

Now, the electrostatic energy in the capacitor can be depicted as,

θ = cos^-1 0.5556 = 56.25

⇒ 2θ = 112.5

⇒ cos 2θ = – 0.38

Now,

E(2πdL) = 21.73 x 10^-20 J – 27.2eV

E(2πdL) = 13.58eV – 27.2eV

E(2πdL) = 13.58eV

E = 1/2 CV²

⇒ E = 1/2 x 12 x 10⁻¹² x (50)²

⇒ E = 1.5 x 10^-8 J

Thus, electrostatic energy in a capacitor is 1.5*10^-8 J

10) An electrical expert needs a 2µF potential difference across a circuit with a 1 kV existing potential difference. He possesses massive 2µF capacitors that can hold a max range of upto 400V. Provide a reliable setup that needs fewer capacitors.

Ans – Total required capacitance will be C = 2µF

Potential difference is V = 1kV = 1000V

The capacitance of each capacitor C1 = 1µF

Now we know each capacitor can withstand a potential difference V1 = 400V

1000/400 = 2.5

Now, assume several capacitors are connected in series & parallel. The potential difference via each row must be 1000V & potential difference via each capacitor must be 400V. So, the number of capacitors in rows can be defined as,

When there are 3 capacitors in every row

Total capacitance of the each row = 1/(1+1+1) = 1/3 µF

1/3 + 1/3 + 1/3 +…n terms =  n/3

But the circuit capacitance is 2µF

Therefore, n/3 = 2

Now, N = 6

Assume there are n number of rows & each row has 3 capacitors connected in parallel. Therefore, the equivalent capacitance of the circuit will be N = 6. Therefore, 6 rows of 3 capacitors are connected in the circuit. At least 6 * 3 i.e. 18 capacitors are needed for this setup.