Physics - Capacitor Question with Solution | TestHub

Energy stored in a charged capacitor is $$\mathrm{U}=\frac{{\mathrm{Q}}^2}{2\mathrm{C}}\Rightarrow \mathrm{U}\propto {\mathrm{Q}}^2$$ As per given condition, when, ${\mathrm{Q}}_2=\left({\mathrm{Q}}_1+3\right)$, $$\begin{array}{ll}\therefore & {\mathrm{U}}_2=44\%\text{ of }{\mathrm{U}}_1=\frac{144}{100}{\mathrm{U}}_1\\ \therefore & \frac{{\mathrm{Q}}_2^2}{2\mathrm{C}}=\frac{144}{100}\times \frac{{\mathrm{Q}}_1^2}{2\mathrm{C}}\\ \therefore & {\mathrm{Q}}_1=\frac{10}{12}{\mathrm{Q}}_2\\ \therefore & {\mathrm{Q}}_1=\frac{5}{6}\left({\mathrm{Q}}_1+3\right)\\ \therefore & 6{\mathrm{Q}}_1=5{\mathrm{Q}}_1+15\\ \therefore & {\mathrm{Q}}_1=15\mathrm{C}\end{array}$$