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NCERT Class XII Chapter
Electrostatic Potential and Capacitance
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Question : 26 of 37
Marks: +1, -0
The plates of a parallel plate capacitor have an area of 90 cm2\mathrm{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:  
A = 90 cm2\mathrm{cm}^2 = 90 × 10−4 m210^{-4}\,\mathrm{m}^2 , d = 2.5 mm = .5 × 10−310^{-3} m
C = ε0Ad\frac{\varepsilon_0 A}{d} = 8.85×10−12×9×10−32.5×10−3\frac{8.85 \times 10^{-12} \times 9 \times 10^{-3}}{2.5 \times 10^{-3}} or C = 32 pF
(a) U = 12CV2\frac{1}{2} C V^2 = 12\frac{1}{2} × 32 × 10−1210^{-12} × 4002400^2 or UI = 2.56 µJ
(b) V = A × d = 2.25 × 10−4 m310^{-4}\,\mathrm{m}^3 = EnergyVolume\frac{\text{Energy}}{\text{Volume}} = 2.56×10−62.25×10−4\frac{2.56 \times 10^{-6}}{2.25 \times 10^{-4}} = 0.113 J/m3\mathrm{m}^3
U = 12CV2\frac{1}{2} C V^2 = 12×ε0Ad\frac{1}{2} \times \frac{\varepsilon_0 A}{d} × (E⋅d)2(E \cdot d)^2 = 12ε0AE2d\frac{1}{2} \varepsilon_0 A E^2 d or UAd\frac{U}{A d} = 12ε0E2\frac{1}{2} \varepsilon_0 E^2
or Energy per unit volume u = 12ε0E2\frac{1}{2} \varepsilon_0 E^2
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