Question

Two parallel plate capacitors of capacitances $C_1$ and $C_2$ such that $C_1=2C_2$ are connected across a battery of $V$ volts as shown in the figure. Initially the key ($k$) is kept closed to fully charge the capacitors. The key is now thrown open and a dielectric slab of dielectric constant '$K$' is inserted in the two capacitors to completely fill the gap between the plates. The ratio of the energies stored in the combination, before and after the introduction of the dielectric slab:

Answer 1

Before slab inserted, net capacitance is $C=C_1+C_2=2C_2+C_2=3C_2$ (as they are in parallel)
Energy stored $U=\frac{1}{2}C{V}^2=\frac{3}{2}{C}_2{V}^2$
After insert slab $C_1$ becomes $C_1^{\prime }=K{C}_1$ and $C_2$ becomes $C_2^{\prime }=K{C}_2$
now net capacitance $C^{\prime }=C_1^{\prime }+C_2^{\prime }=K{C}_1+K{C}_2=K(C_1+C_2)=3K{C}_2$
When switch is opened ,but the capacitors fully charged by voltage $V$ and the energy stored $U^{\prime }=\frac{1}{2}{C}^{\prime }{V}^2=\frac{3}{2}K{C}_2{V}^2$
$\therefore \frac{U}{U^{\prime }}=\frac{\left(3/2\right)C_2{V}^2}{\left(3/2\right)K{C}_2{V}^2}=\frac{1}{K}$