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(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

Let C is the capacitance of both the capacitor .
(i) Initially, when switch S is closed
Net capacitance `C_("inj") = C_(1) + C_(2) = C + C = 2C `
Total initial energy = `((1)/(2)) CV^(2) + ((1)/(2)) CV^(2) = CV^(2)`
After filling the gap , `C_("inj") = KC + KC = 2KC `
(ii) When switch is open, the capacitance in each of the capacitions varies, hence , the energy
also varies i.e., in case of 'B' , the change remains same,.i.e., CV and the capacity of each
capacitor becomes `C_(1) = KC` potenial of conneted, capacitor, remains same `V_(1) V` so CV
= KCV' here `V_(1)` is potential of disconnected capacitor
`V_(1) = (V)/(K)`
`C_(f) = (1)/(2) CV_(1)^(2) + (1)/(2) C'V'^(2) = (1)/(2) KCV^(2) + (1)/(2) KC((V)/(K))^(2)`
`= (1)/(2) KCV^(2) + (1)/(2) (CV^(2))/(K) = (K^(2) CV^(2) + CV^(2))/(2k) = (CV^(2) (K +1))/(2K)`
Ratio`=(U_(i))/(U_(f)) = (CV^(2) xx2K)/(CV^(2) (K^(2) +1)) = (2K)/((K^(2) +1))` .