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The plates of a parallel plate capacitor have an area of $90 \,cm ^{2}$ 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.
Solution
Area of the plates of a parallel plate capacitor, $A=90 \,cm ^{2}=90 \times 10^{-4} \,m ^{2}$
Distance between the plates, $d =2.5\, mm =2.5 \times 10^{-3} \,m$
Potential difference across the plates, $V =400 \,V$
$(a)$ Capacitance of the capacitor is given by the relation, $c=\frac{\epsilon_{0} A}{a}$
Electrostatic energy stored in the capacitor is given by the relation,
$E_{1}=\frac{1}{2} C V^{2}=\frac{1}{2} \frac{\epsilon_{0} A}{d} V^{2}$
Where,
$\epsilon_{0}=$ Permittivity of free space $=8.85 \times 10^{-12} \,C ^{2} \,N ^{-1} \,m ^{-2}$
$\therefore E_{1}=\frac{1 \times 8.85 \times 10^{-12} \times 90 \times 10^{-4} \times(400)^{2}}{2 \times 2.5 \times 10^{-3}}$$=2.55 \times 10^{-6} \,J$
$(b)$ Volume of the given capacitor, $V^{\prime}=A \times d=90 \times 10^{-4} \times 25 \times 10^{-3}$$=2.25 \times 10^{-4} \,m ^{3}$
Energy stored in the capacitor per unit volume is given by, $u=\frac{E_{1}}{V^{\prime}}$
$=\frac{2.55 \times 10^{-6}}{2.25 \times 10^{-4}}=0.113 \,J\,m ^{-3}$
Again, $u=\frac{E_{1}}{V^{\prime}}$
$=\frac{\frac{1}{2} C V^{2}}{A d}=\frac{\frac{\epsilon_{0} A}{2 d} V^{2}}{A d}=\frac{1}{2} \epsilon_{0}\left(\frac{V}{d}\right)^{2}$
Where, $\frac{v}{a}=$ Electric intensity $= E$ Therefore, $U=\frac{1}{2} \epsilon_{0} E^{2}$
