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' ?

Explanation

Potential difference between the plates of capacitor
V = q/C
Work done to add additional charge dg on the capacitor
` dw = V xx dq`
` therefore ` Total energy stored in the capacitor
` U = int d w = int_0^Q dq = 1/2 (Q^2)/(C )`
When battery is disconnected,
(i) Energy stored will be decreased or energy stored = 1/K times the initial energy
(ii) Electric field would decrease or E. = E/K