Physics - Capacitor Question with Solution | TestHub

Capacitance of capacitor (C) $=6\mu \mathrm{F}$ $=6\times 10^{-6}\mathrm{F}$; Initial potential $\left({\mathrm{V}}_1\right)=10\mathrm{V}$ and final potential $\left({\mathrm{V}}_2\right)=20\mathrm{V}$. The increase in energy $(\Delta \mathrm{U})$ $$\begin{array}{l}=\frac{1}{2}\mathrm{C}\left({\mathrm{V}}_2^2-{\mathrm{V}}_1^2\right)\\ =\frac{1}{2}\times \left(6\times 10^{-6}\right)\times \left[(20{)}^2-(10{)}^2\right]\\ =\left(3\times 10^{-6}\right)\times 300=9\times 10^{-4}\mathrm{J}.\end{array}$$