Question 26

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.

Answer

Area of the plates of a parallel plate capacitor, A = 90 cm^{2} = 90 × 10^{ - 4 }m^{2}

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,

Electrostatic energy stored in the capacitor is given by the relation,

Where,

= Permittivity of free space = 8.85 × 10 ^{- 12} C^{2} N ^{- 1} m^{ - 2 }

Hence, the electrostatic energy stored by the capacitor is 2.55 x 10^{-6} J

(b) Volume of the given capacitor,

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

Where,

= Electric intensity = *E*

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2019-08-04 13:55:10

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