Question

A 4µF capacitor is charged by a 200V supply. The supply is then disconnected and the charged capacitor is connected to another uncharged 2µF capacitor. How much electrostatic energy of the first capacitor is lost in the process of attaining the steady situation?

Answer 1

$$\begin{gathered} chargingof4\mu fcapacitorconnectedto200Vsupply, \\ q=cvi.e.q=4\times {10}^{-6}\times 200=8\times {10}^{-4}C. \\ whenthischargedcapacitorisconnectedto2\mu funchargedcapacitor,chargesareequallysharedbythetwo. \\ chargingof2\mu fcapacitor=4\times {10}^{-4}C. \\ theenergygainedbythe2\mu fcapacitor=energylostby4\mu fcapacitor=\frac{1}{2}\frac{q^2}{c}=\frac{1\times (4\times {10}^{-4}{)}^2}{2\times 2\times {10}^{-6}}=4\times {10}^{-2}J. \end{gathered}$$