Physics - Capacitor Question with Solution | TestHub

When a dielectric of dielectric constant $K$ is inserted between the plates of a capacitor $C_0$, the capacitance becomes $K{C}_0$.

Energy stored in a capacitor is given by, $U=\frac{1}{2}C{V}^2$. Now,

$U_i=\frac{1}{2}\left(K_1{C}_0\right)V^2$ and $U_f=\frac{1}{2}\left(K_2{C}_0\right)V^2$

$\Rightarrow \Delta U=U_f-U_i=\frac{1}{2}\left(K_2-K_1\right)C_0{V}^2$

Percentage change in the energy will be,

$\frac{\Delta U}{U_i}\times 100=\frac{\frac{1}{2}\times 5\times {C_0}^2}{\frac{1}{2}\times 10\times {C_0}^2}\times 100=50\%$