In the circuit shown in figure, capacitor `A` has capacitance `C_(1)=2muF` when filled with a di-electric slab of dielectric constant `k=2`. Capacitors `B` and `C` are air capacitors and have capacitances `C_(2)=3muF` and `C_(3)=6muF`, respectively. A is charged by closing switch `S_(1)` alone

(a) Calculate energy supplied by battery during process of charging. Switch `S_(1)` is now opened and `S_(2)` is closed.
(b) Calculate charge on `B` and energy stored in the system when electrical equilibrium is attained. Now switch `S_(2)` is also opened, slab of `A` is removed. Another di-electric slab of `K=2`, which can just fill the space in `B`, is inserted into it and then switch `S_(2)` alone is closed.
(c) Calculate by how many times electric field in `B` is increased. Calculate also, loss. of energy during redistribution of charge.

Answer

(a) `0.0648J`, Energy stored in capacitor is `1/2CV^(2)`, but energy supplied by battery is `CV^(2)`; (b) `180muC, 0.0162J`; © `0.75, 0.0054J`