The plates of a parallel plate capacitor have an area of
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 and the magnitude of electric field between the plates.

Solution:

Area of the plates of a parallel plate capacitor, A = 90 cm= 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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