Physics - Magnetism Question with Solution | TestHub

The charge on the capacitor when connected to the battery is given by
$\mathrm{Q}=\mathrm{CV}=\left(900\times 10^{-12}\mathrm{F}\right)\times 100\mathrm{V}=9\times 10^{-8}\mathrm{C}$ When the battery is replaced by another capacitor of $900\mathrm{pF}$ capacitance, the charge of $9\times 10^{-8}\mathrm{C}$ is distributed on both. Let ${\mathrm{Q}}_1$ and $Q_2$ be the charge on each of them. $\therefore Q=Q_1+Q_2$ $={\mathrm{C}}_1\mathrm{V}+{\mathrm{C}}_2\mathrm{V}$, where $\mathrm{V}$ is the common potential. or $\mathrm{V}=\frac{\mathrm{Q}}{{\mathrm{C}}_1+{\mathrm{C}}_2}$ As the two capacitors are in parallel, the equivalent capacitance is given by $\mathrm{C}={\mathrm{C}}_1+{\mathrm{C}}_2$ $\therefore$ Total energy of the capacitors $=\frac{1}{2}{\mathrm{CV}}^2$ $\begin{array}{l}=\frac{1}{2}\times \left({\mathrm{C}}_1+{\mathrm{C}}_2\right)\times \frac{{\mathrm{Q}}^2}{{\left({\mathrm{C}}_1+{\mathrm{C}}_2\right)}^2}=\frac{{\mathrm{Q}}^2}{2\left({\mathrm{C}}_1+{\mathrm{C}}_2\right)}\\ =\frac{{\left(9\times 10^{-8}\right)}^2}{2\left(900\times 10^{-12}+900\times 10^{-12}\right)}\\ =\frac{(9{)}^2\times 10^{-16}}{2\times 2\times 9\times 10^{-10}}=2.25\times 10^{-6}\mathrm{J}\end{array}$