Answer to Question #232573 in Electric Circuits for Gabu

Question

Suppose, the two sides of a parallel plate capacitor are connected with a battery of V electromagnetic force. The capacitance is C, plate area is A and the distance between the plates is d.

Then,

(i) How much energy is used by the battery to fully charge the capacitor?

(ii) what is the Energy stored by the capacitor?

(ii) If the battery is removed after the capacitor is charged, and the d(distance between the plates) is doubled, what is the current energy stored in the capacitor?

Answer 1

i) work done = $\frac{1}{2}QV$

where Q= charge and V is voltage.

ii) $C=\frac{Q}{V}$

$Q=CV$

Energy stored$=\frac{1}{2QV}=\frac{1}{2}(CV)(V)=\frac{1}{2}CV^2$

iii) doubling the distance d, current energy stored =$(\frac{1}{2})(\frac{1}{2}CV^2)=\frac{1}{4}CV^2$