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

A, C, D

Explanation

The potential difference across each capacitor is V.
Total Energy = Energy in A+ Energy in B
`=1/2CV^2+1/2CV^2=CV^2`
When the switch opened and a dielectric is inserted between the plants of capacitors, the new capacitance is 3C.
Energy in `A=1/2(3C)V^2=3/2CV^2` (V is the same)
Energy in `B=1/2(q^2)/(KC)=1/2xx((CV)^2)/(3C)` `=(CV^2)/(6)` (charge on capacitor B remains same when switch is opened)
Total Energy = Energy in A+Energy in B
`:.` Total En ergy =3/2CV^2+1/6CV^2=5/3CV^2` ...(ii)
Total Energy initially/Total energy finally`=(CV^2)/(5/3CV^2)=3/5`