Deduce the expression for the electrostatic energy stored in a capacitor of capacitance ‘C’ and having charge ‘Q’.

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 ‘K’? 

When a capacitor is charged by a battery, work is done by the charging battery at the expense of its chemical energy. This work is stored in the capacitor in the form of electrostatic potential energy.

Consider a capacitor of capacitance C. Initial charge on capacitor is zero. Initial potential difference between capacitor plates =0. Let a charge Q be given to it in small steps. When charge is given to capacitor, the potential difference between its plates increases.

Potential difference between its plates, V= q/C

Work done to give an infinitesimal charge dq to the capacitor is given by, 
 

If V is the final potential difference between capacitor plates, then Q = CV

Work is stored in the form of electrostatic potential energy.
Electrostatic potential energy, U = 

When battery is disconnected,

i) Energy stored will decrease.

Energy becomes, U = 

So, energy is reduced to 1/K times its initial energy.

i) In the presence of dielectric, electric field becomes, E =