Question

A parallel plate capacitor of capacitance 20μF,  is connected to a 100 V, supply. After sometime,  the battery is disconnected, and the space,  between the plates of the capacitor is filled with  a dielectric, of dielectric constant 5. Calculate the  energy stored in the capacitor.

(i) Before 

(ii) After the dielectric has been put in between its plates

Answer 1

Charge stored, Q = CV = 20 x 10^-6 x 100 C

= 2000μC 

New value of capacitance

= 5 x 20μF

= 100μF

Energy stored in a capacitor

(1) :.  Energy stored before dielectric