Question

A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electrical capacity of 3400.0 μF and is charged to a potential difference of 72.0 V.
1. Calculate the amount of energy stored in the capacitor.
2. Calculate the charge on this capacitor when the electrical energy stored in the capacitor is 11.15 J.
3. If the two plates of the capacitor have their separation increased by a factor of 8 while the charge on the plates remains constant, by what factor is the energy stored in the capacitor increased?

Answer 1

1. The amount of energy stored in the capacitor is given by

U = (1/2) C V² -----------------------(1)

here C = capacitance =3400.0 μF = 3400.0 x 10^-6 F .

V = Potential difference between the metal conductors =72.0 V.

from (1) on substituting the values we get

U = (1/2) ( 3400.0 x 10^-6 F) (72.0 V)² = 8.813 J.

Therefore the energy stored in the capacitor is  8.813 J.

2. Here the energy stored in the capacitor is given as U =11.15 J

from (1) , U = (1/2) C V²   but Q = C V and V = Q /C then

U = Q² / (2C) and

Q = sqrt ( 2 C U) { U =11.15 J , C = 3400.0 μF = 3400.0 x 10^-6 F }

Q = sqrt ( 2 x 3400.0 x 10^-6 F x 11.15 J )

Q = 275.35 x 10^-3 C = 0.275 C

Therefore the charge on the capacitor is  0.275 C.

3. From the equation (1)

U =  (1/2) C V² here V = E d and C = (₀ A ) / d

{ E = electrical field between the conductors , d =separation between the conductors , A= area of the conductor }

U = (1/2) ((₀ A ) / d ) ( E d )²

U =(1/2) (₀ A E²) d here   (1/2) (₀ A E²) =constant.

from the above equation it is clear that the energy stored in the capacitor is directly proportional to the separation between the conductors.

Therefore if the separation increased by a factor of 8 the energy is also increased by the factor 8.