Question
a. Derive an expression for the energy stored in a parallel plate capacitor of capacitance C when charged up to voltage V. How is this energy stored in the capacitor?
b. A capacitor of capacitance $1 \mu F$ is charged by connecting a battery of negligible internal resistance and emf 10 V across it. Calculate the amount of charge supplied by the battery in charging the capacitor fully.

Answer

a. Work done in adding a charge $dq = dW$
$\begin{array}{l}= Vdq \\ =\frac{q}{c} d q\end{array}$
$\therefore$ Total Amount of work(W) in charging a capacitor
$\begin{array}{l} W =\int d W=\frac{1}{C} \int_0^Q q d q \\ W=\frac{Q^2}{2 C} \\ =\frac{(C V)^2}{2 C}=\frac{1}{2} C V^2\end{array}$
The electrostatic Energy/ potential energy is stored in the electric field between the plates.
b. $C =1 \mu F=1 \times 10^{-6} F ; V =10$ volt
$\begin{array}{l}Q=C V \\ =1 \times 10^{-6} \times 10 \\ =10^{-5} \text { coulomb }\end{array}$
hence, the amount of charge supplied by the battery in charging the capacitor fully is $10^{-5}$ coulomb.

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