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.

Answer

0.6

Explanation

Before opening the switch potential difference across both the capacitors is V. as they are in parallel.Hence energy
`U_A=U_B=1/2CV^(2) therefore U_(Total)=VC^(2)=U_1`
After opening the switchpotential difference across it is V and its capacity is 3C
`therefore U_A=1/2(3C)V^(2)=3/2CV^(2)`
In case of capacitor B, charges stored in it is q=CV and its capacity is also 3C
Therefore,`U_B=(q^(2))/(2(3C)) (CV^(2))/6 therefore U_(Total)=(3Cv^(2))/2+(CV^(2))/6=10/6CV^(2)=(5CV^(2))/3U_f` .....(ii)
From Eqs (i) and (ii), we get `(U_f)/(Uf)=3/5`