Question

The plates of a parallel plate capacitor have an area of $90c{m}^2$ each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.

(a) How much electrostatic energy is stored by the capacitor?

(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.

Answer 1

$\text{Step 1: Capacitance of the capacitor(C)}$

Capacitance of parallel plate capacitor is given by:

$C=\frac{{\in }_{0}A}{d}$ $....\left(1\right)$

$\text{Step 2: Electrostatic energy stored by the capacitor(U)}$

$U=\frac{1}{2}C{V}^2$$=\frac{1}{2}\frac{A{\in }_0}{d}{V}^2$ $....\left(2\right)$

$=\frac{1}{2}\times \frac{90\times {10}^{-4}{m}^2\times 8.85\times {10}^{-12}\times (400V{)}^2}{(2.5\times {10}^{-3}m)}$

$=2.55\times {10}^{-6}J$

$\text{Step 3: Energy per unit volume(u)}$

Energy per unit volume : $u=\frac{U}{V^{\prime }}=\frac{U}{Ad}$ $....\left(3\right)$

where $V^{\prime }=$ Volume of the capacitor

$\therefore u=\frac{2.55\times {10}^{-6}J}{90\times {10}^{-4}{m}^2\times 2.5\times {10}^{-3}m}=0.113J{m}^{-3}$

$\text{Step 4: Relation between u and electric field (E)}$

From eq $\left(2\right)$ and $\left(3\right)$

$u=\frac{U}{V^{\prime }}$$=\frac{\frac{1}{2}\frac{A{\in }_0}{d}{V}^2}{Ad}$

$\Rightarrow u=\frac{1}{2}{\in }_0{\left(\frac{V}{d}\right)}^2$

$\frac{V}{d}=$ Electric field ntensity $=E$

$\therefore u=\frac{1}{2}{\in }_0{E}^2$