Two identical parallel plate capacitors A and B are
connected to a battery of V volts with the switch S
closed . The switch is now opend and the free
space between the plantes of the capacitors is filled
with a dielectric of dielectric constant K. Find the
rario 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`.