A parallel plate capacitor filled with a medium of dielectric...

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, $U = \dfrac{1}{2}C{V}^{2}$. Now,

${U}_{i} = \dfrac{1}{2}\left({K}_{1}{C}_{0}\right){V}^{2} and${U}_{f} = \dfrac{1}{2}\left({K}_{2}{C}_{0}\right){V}^{2}$</p><p>$⇒ \DeltaU = {U}_{f} - {U}_{i} = \dfrac{1}{2}\left({K}_{2} - {K}_{1}\right){C}_{0}{V}^{2}$</p><p>Percentage change in the energy will be,</p><p>$\dfrac{\DeltaU}{{U}_{i}} \times 100 = \dfrac{\dfrac{1}{2} \times 5 \times {{C}_{0}}^{2}}{\dfrac{1}{2} \times 10 \times {{C}_{0}}^{2}} \times 100 = 50 %$