Physics - Capacitor Question with Solution | TestHub

Consider the following diagram:$\begin{array}{rl}& Q=CV\\ & E_{\text{capocitor }}=\frac{Q^2}{2C}=\frac{1}{2}C{V}^2\\ & W=\text{ Work done by battery }=QV=C{V}^2\\ & \therefore E_{\text{loss }}=(W)-\left(E_{\text{capocitor }}\right)=C{V}^2-\frac{1}{2}C{V}^2=\frac{1}{2}C{V}^2=\frac{W}{2}\end{array}$ This is $50\%$ off the energy supplied by the battery