Question

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’?

Answer 1

 Potential difference between the plates of capacitor V =q/C

Work done to add additional charge dq on the capacitor

dW = V x dq=q/C xdq

 ∵ Total energy stored in the capacitor

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

Alternatively, if a student attempts to answer by keeping the battery connected, then 

(i) energy stored will increase or become K times the initial energy. 

(ii) electric field will not change.