The figure shows two identical parallel plate capacitors connected to a battery with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant K (or relative permittivity) 3. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.

Explanation

Initially, when the switch is closed, both the capacitors `A and B` are in parallel and, therefore, the energuy stored in the capacirors is
`U_(i)=2xx1/2CV^(2)` (i)
When switch S is opened, (B)gets disconnected from the battery. The capacitor `B` is now isolated, and the charge on an isolated capacitor remains constant, often referred to as bound charge. On the other hand, A remains connected to the battery. Hence, potential V remains constant on it.
When the capacitors are filled with dieletric, their capacitance increases to `KC`. Therefore, energy stored in B changes to `Q^(2)//2KC`, where `Q=CV` is the charge on `B`, which remains constant, and energy stored in A changes to `1/2 KCV^(2)`
where V is the potential on A, which remains constant. Thus, the final total energy stored in the capacitors is
`U_(f)=1/2(CV)^(2)/(KC)+1/2KCV^(2)=1/2CV^(2)(K+1/K)` (ii)
from Eqs. (i) and (ii), we find
`U_(i)/U_(f)=(2K)/(K^(2)+1)`
It is given that `K=3`. Therefore, we have
`U_(i)/U_(f)=3/5`.