**Question:**

The plates of a parallel plate capacitor have an area of 90 cm2 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 cm2 = 90 × 10−4 m2

Distance between the plates, *d* = 2.5 mm = 2.5 × 10−3 m

Potential difference across the plates, *V* = 400 V

(a) Capacitance of the capacitor is given by the relation,

$C=\frac{\in_{0} A}{d}$

Electrostatic energy stored in the capacitor is given by the relation, $E_{1}=\frac{1}{2} C V^{2}$

$=\frac{1}{2} \frac{\in_{0} A}{d} V^{2}$

Where,

$\epsilon_{0}=$ Permittivity of free space $=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$

$\therefore E_{1}=\frac{1 \times 8.85 \times 10^{-12} \times 90 \times 10^{-4} \times(400)^{2}}{2 \times 2.5 \times 10^{-3}}=2.55 \times 10^{-6} \mathrm{~J}$

Hence, the electrostatic energy stored by the capacitor is $2.55 \times 10^{-6} \mathrm{~J}$.

(b) Volume of the given capacitor,

$V^{\prime}=A \times d$

$=90 \times 10^{-4} \times 25 \times 10^{-3}$

$=2.25 \times 10^{-4} \mathrm{~m}^{3}$

Energy stored in the capacitor per unit volume is given by,

$u=\frac{E_{1}}{V^{\prime}}$

$=\frac{2.55 \times 10^{-6}}{2.25 \times 10^{-4}}=0.113 \mathrm{~J} \mathrm{~m}^{-3}$

Again, $u=\frac{E_{1}}{V^{\prime}}$

$=\frac{\frac{1}{2} C V^{2}}{A d}=\frac{\frac{\epsilon_{0} A}{2 d} V^{2}}{A d}=\frac{1}{2} \in_{0}\left(\frac{V}{d}\right)^{2}$

Where,

$\frac{V}{d}=$ Electric intensity $=E$

$\therefore u=\frac{1}{2} \in_{0} E^{2}$

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