a. Derive an expression for the energy stored in a parallel plate… - Vidyadip

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 $dq = dW$
$= Vdq$
$=\frac{q}{c} d q$
$\therefore$ Total Amount of work $(W)$ in charging a capacitor
$ W =\int d W=\frac{1}{C} \int_0^Q q d q$
$W=\frac{Q^2}{2 C}$
$=\frac{(C V)^2}{2 C}=\frac{1}{2} C V^2$
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
$Q=C V$
$=1 \times 10^{-6} \times 10$
$=10^{-5} \text { coulomb }$
hence, the amount of charge supplied by the battery in charging the capacitor fully is $10^{-5}$ coulomb.

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