Question

a 80 microfarad capacitor is charged by 50V battery . d capacitor is disconnected from it and connected across uncharged 320 microfarad capacitor calculate the charge on second capacitor

Answer 1

$$\begin{gathered} \mathrm{Given}, \\ {\mathrm{C}}_1=80\mathrm{\mu F}=80\times {10}^{-6}\mathrm{F}=8\times {10}^{-5}\mathrm{F} \\ {\mathrm{V}}_1=50\mathrm{V} \\ {\mathrm{Q}}_2=0 \\ {\mathrm{C}}_2=320\mathrm{\mu F}=320\times {10}^{-6}\mathrm{F}=3.2\times {10}^{-4}\mathrm{F} \\ \mathrm{Charge}\mathrm{on}\mathrm{first}\mathrm{capacitor}={\mathrm{Q}}_1={\mathrm{C}}_1{\mathrm{V}}_1 \\ =8\times {10}^{-5}\times 50=4\times {10}^{-3}\mathrm{C}=4\mathrm{mC} \\ \mathrm{Now}, \\ \mathrm{The}\mathrm{common}\mathrm{potential}\mathrm{after}\mathrm{connecting}\mathrm{them}\mathrm{together}\mathrm{is}\mathrm{given}\mathrm{by} \\ \mathrm{V}=\frac{4\times {10}^{-3}}{(8\times {10}^{-5})+(3.2\times {10}^{-4})} \\ =10\mathrm{V} \\ \mathrm{Now},\mathrm{charge}\mathrm{on}\mathrm{second}\mathrm{capacitor}\mathrm{will}\mathrm{be} \\ {\mathrm{C}}_2\mathrm{V}=3.2\times {10}^{-4}\times 10=3.2\times {10}^{-3}\mathrm{F}\mathrm{or}3.2\mathrm{mF}\mathrm{or}3200\mathrm{\mu F} \end{gathered}$$