Physics - Capacitor Question with Solution | TestHub

Before inserting dielectric capacitance is given ${\mathrm{C}}_0=12.5\mathrm{pF}$ and charge on the capacitor $\mathrm{Q}={\mathrm{C}}_0\mathrm{V}$ After inserting dielectric capacitance will become ${ϵ}_{\mathrm{r}}{\mathrm{C}}_0$. Change in potential energy of the capacitor $\begin{array}{rl}& ={\mathrm{E}}_{\mathrm{i}}-{\mathrm{E}}_{\mathrm{f}}\\ & =\frac{{\mathrm{Q}}^2}{2{\mathrm{C}}_{\mathrm{i}}}-\frac{{\mathrm{Q}}^2}{2{\mathrm{C}}_{\mathrm{f}}}=\frac{{\mathrm{Q}}^2}{2{\mathrm{C}}_0}\left[1-\frac{1}{{ϵ}_{\mathrm{r}}}\right]\\ & =\frac{{\left({\mathrm{C}}_0\mathrm{V}\right)}^2}{2{\mathrm{C}}_0}\left[1-\frac{1}{{ϵ}_{\mathrm{r}}}\right]=\frac{1}{2}{\mathrm{C}}_0{\mathrm{V}}^2\left[1-\frac{1}{{ϵ}_{\mathrm{r}}}\right]\end{array}$
Using ${\mathrm{C}}_0=12.5\mathrm{pF},\mathrm{V}=12\mathrm{V},{ϵ}_{\mathrm{r}}=6$ $\begin{array}{rl}& =\frac{1}{2}(12.5)\times 12^2\left[1-\frac{1}{6}\right]=\frac{1}{2}(12.5)\times 12^2\times \frac{5}{6}\\ & =750\mathrm{pJ}=750\times 10^{-12}\mathrm{J}\end{array}$