Two parallel plate capacitors of capacitances `C_1` and `C_2` such that `C_2 = 2 C_1` are connected across battery of V volts as shown in the Fig . Initially the key K is kept closed to fully charge the capacitors . The key K is now thrown open and dielectric slabs of dielectric constant `K_0` are inserted in the two capacitors to completely fill the gap between the plates .

Find the ratio of the energies stored in the combination before and after the introduction of dielectric slabs .

Explanation

Total initial energy of the combination `U_i = 1/2 C_i V^(2)`
As the key K is then thrown open , net charge Q remains constant but voltage V changes to `V_(f) = (V)/(K_(0))`
`therefore` Total final energy of the combination `U_(f) = 1/2 C_f V_f^2 = 1/2 (K_(0) . C_(f)) ((V)/(K_(0)))^(2) = (C_(1) * V^(2))/(2 K_(0)) implies (U_(i))/(U_(f)) = K_(0)`