Question

A parallel-plate capacitor with the plate area $100c{m}^2$ and the separation between the plats $1.0cm$ is connected across a battery of emf $24$ volts. Find the force of attraction between the plates.

Answer 1

The surface area of the capacitor plates $A=100c{m}^2={10}^{}{m}^2$

The separation between the plates is $d=1cm={10}^{-2}m$

The emf of the battery is $V=24V_0$

$\therefore$ The capacitance od the capacitor is given as:

$C=\frac{{ϵ}_{0}A}{d}$

$=\frac{8.85\times {10}^{-12}\times {10}^{-2}}{{10}^{-2}}$

$=8.85\times {10}^{-12}$

$\therefore$ The energy stored in the capacitor is given as:

$E=\left(1/2\right)C{V}^2$

$=\left(1/2\right)\times {10}^{-12}\times (24{)}^2=2548.8\times {10}^{-12}$

$\therefore$ The forced attraction between the plates $=\frac{E}{d}=\frac{2548.8\times {10}^{-12}}{{10}^{-2}}=2.54\times {10}^{-7}N$