Question

A 900 PF capacitor is charged by a 100 V battery. How much electrostatic energy is stored by the capacitor? The capacitor is disconnected from the battery and connected in parallel to another 200 PF capacitor. What is the energy stored by the system?

Answer 1

Dear Student,
$$\begin{gathered} energystored=\frac{1}{2}C{V}^2=\frac{1}{2}\times 900\times {10}^{-12}\times {10}^4=4.5\mu J \\ andchargeonit=CV=900\times {10}^{-12}\times 100=90nC \\ whenconnectedtoacapacitiorinparallelcombination \\ equivalentcapacitonce=\frac{900\times 200}{900+200}=\frac{1800}{11}pF \\ sonowenergystoredinthiscombination=\frac{q^2}{2C_{eq}}=\frac{{\left(90nC\right)}^2}{2\times \frac{1800}{11}pF} \\ =\frac{8100\times {10}^{-18}\times 11}{1800}=49.5\mu J \end{gathered}$$
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