Question

A parallel plate capacitor has circular plates of $10cm$ radius separated by an air gap of $1mm$. It is charged by connecting the plates to a $100volt$ battery. Find the change in energy stored in the capacitor when the plates are moved to a distance of $1cm$ and the plates are maintained in connection with the battery.

Answer 1

$\Delta E=\frac{1}{2}{V}^2(C_1-C_2)$ $d=1mm$

$A=\pi \times 1\times 1=1\times {10}^{-6}{m}^2$

$\Delta E=\frac{1}{2}{V}^2\frac{{ϵ}_{0}A}{1}(\frac{1}{d}-\frac{1}{d+10})$

$\Delta E=\frac{1}{2}{V}^2{ϵ}_{0}A\left(\frac{10}{d(d+10)}\right)$

$\Delta E=\frac{1}{2}\times {100}^2\times 8.824\times {10}^{-12}\times {10}^{-12}\times \pi \times \frac{10}{11\times {10}^{-3}}$

$\Delta E=1.2623\times {10}^{-6}J$

$\Delta E=1.2623\mu J$