A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity C and is charged to a potential V volts. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is
( Capacitor is connected to the battery throughout)

Question

Toppr Admin · Accepted Answer

Initial potential energy of the capacitor is $U_{i} = \frac{1}{2} C V^{2}$ .
When dielectric is removed potential energy becomes $U^{'} = \frac{1}{2} (C / K) V^{2}$ .
When again reinsert the dielectric slab potential energy becomes $U_{f} = \frac{1}{2} C V^{2}$ .
So net work done $W = U_{f} - U_{i} = 0$

Initial potential energy of the capacitor is Ui=12CV2.When dielectric is removed potential energy becomes U′=12(C/K)V2.When again reinsert the dielectric slab potential energy becomes Uf=12CV2.So net work done W=Uf−Ui=0