Question

A 900 pF capacitor is charged by 100 V battery.
(a) How much electrostatic energy is stored by the capacitor? The capacitor is disconnected from the battery and connected to another 900 pF capacitor. How much is the electrostatic energy stored in the system?

Answer 1

$$\begin{gathered} Dearstudent, \\ energystored=\frac{1}{2}C{V}^2=\frac{1}{2}\times 900\times {10}^{-12}\times {10}^4=4.5\mu J \\ andchargeonit=CV=900\times {10}^{-12}\times 100=90nC \\ whenconnectedtoacapacitiorinparallelcombination \\ equivalentcapacitonce=\frac{900\times 900}{900+900}=450pF \\ sonowenergystoredinthiscombination=\frac{q^2}{2C_{eq}}=\frac{{\left(90nC\right)}^2}{2\times 450pF} \\ =\frac{8100\times {10}^{-18}}{900\times {10}^{-12}}=9\mu J \\ Regards \end{gathered}$$