Question

A parallel plate capacitor is charged by a battery.

The battery is disconnected and a dielectric slab is inserted to completely fill the space between the plates.

How will (i) its capacitance, (ii) electric field between the plates and (iii) energy stored in the capacitor be affected ? Justify your answer giving necessary mathematical expressions for each case.

Answer 1

The capacitance without dielectric is $C=\frac{A{ϵ}_0}{d}$

When dielectric slab is inserted, the capacitance becomes, $C^{\prime }=\frac{AK{ϵ}_0}{d}=KC$ where K be the dielectric constant.

i)Thus, the capacitance will increase K times of the initial.

ii) As the battery is disconnected so the charge on capacitor remains constant.

Since, $Q=CV$ so potential V will decrease and also $E=V/d$ so the field E will also decrease.

iii) Stored energy, $U=\frac{Q^2}{2C}$ . As charge Q is constant and $C$ is increasing so energy will decrease.