Question

Derivation of energy stored in a parallel plate capacitor?

Answer 1

A charged Capacitor is a store of electrical potential energy.

To find the energy stored in a capacitor, let us consider a capacitor of capacitance C, with a potential difference V between the plates.

There is a charge +q on one plate and –q on the other.

Potential of capacitor =q/C

Work done in giving additional charge dq to capacitor is

dW=q/C *dq

total work done in giving a charge Q to the capacitor is q.

Q=Q

W=1/C Q²/2

Energy stored in the capacitor

U=W

=1/2 Q²/C

WE know that Q=CV

U=1/2 (CV)²/C

=1/2 CV²

Now CV=Q

U=1/2 QV

Thus,

U=1/2 Q2/C

=1/2 CV2

=1/2 QV

Where Q=Coloumb,V=Volt Energy U is in joule