Question

Deduce the expression for the electrostatic energy stored in a capacitor of capacitance ${}^{\prime }{C}^{\prime }$ and having charge ${}^{\prime }{Q}^{\prime }$.
How will the (i) energy stored and (ii) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant ${}^{\prime }{K}^{\prime }$?

Answer 1

Energy stored in a charged capacitor. The energy of a charged capacitor is measured by the total work done in charging the capacitor to a given potential we know that capacitance is $C=Vq$ ; where q is the charge on the plates and V is potential difference. When an additional amount of charge $dq$ is transferred from negative to positive plate.

The total work stored as electrostatic potential energy $U$ in the capacitor,

$Q=CV$

When dielectric of dielectric constant $K$ is filled fully in between the plates. Then the charge stored in the capacitor can be written as

$C=\frac{KAϵ}{d}$; d= gap between plates; A= surface area of the plate.

Electric field $E=\frac{KV}{d}$