
(i) When the switch is closed Ceq=2C
Charge on each capacitor will be q=CV.
Energy E1=21CeqV2 =212C×V2
E1=CV2
(ii) When the switch is opened charge on the right capacitor remain CV while the potential on the left capacitor remains the same.
Dielectric k=5
C′=kC C′=5C
Now to calculate E2,
E2=21C′V2+2C′q2
⇒E2=21(5C)V2+2(5C)(CV)2
⇒E2=25CV2+10CV2
⇒E2=513CV2
⇒E2E1=135