2.26 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. 

(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. 

Answers (1)

The volume of the capacitor is:

V=A*d=90*10^{-4}25*10^{-3}=2.25*10^{-4}m^3

Now,

Energy stored in the capacitor per unit volume :

 

u =\frac{E}{V}=\frac{2.55*10^{-6}}{2.55*10^{-4}}=0.113per \:m^3

NOW, Relation between u and E.

u =\frac{E}{V}=\frac{\frac{1}{2}CV^2}{Ad}=\frac{\frac{1}{2}(\frac{\epsilon _0A}{d})V^2}{Ad}=\frac{1}{2}\epsilon _0E^2

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