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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.

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#### Solution

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,

C = `(in_0"A")/"d"`

Electrostatic energy stored in the capacitor is given by the relation, `"E"_1 = 1/2 "CV"^2`

= `1/2 (in_0"A")/"d""V"^2`

Where,

`in_0` = Permittivity of free space = 8.85 × 10^{−12 }C^{2 }N^{−1} m^{−2}

∴ `"E"_1 = (1 xx 8.85 xx 10^-12 xx 90 xx 10^-4 xx (400)^2)/(2 xx 2.5 xx 10^-3) = 2.55 xx 10^-6 "J"`

Hence, the electrostatic energy stored by the capacitor is `2.55 xx 10^-6 "J"`

**(b) **Volume of the given capacitor,

`"V'" = "A" xx "d"`

= `90 xx 10^-4 xx 25 xx 10^-3`

= `2.25 xx 10^-4 "m"^3`

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

`"u" = "E"_1/("V'")`

= `(2.55 xx 10^-6)/(2.25 xx 10^-4) = 0.113 "J m"^-3`

Again, u = `"E"_1/("V'")`

= `((1/2)"CV"^2)/("Ad") = ((in_0"A")/(2"d")V^2)/("Ad") = 1/2in_0("V"/

"d")^2`

Where,

`"V"/"d"` = Electric intensity = E

∴ `"u"=1/2 in_0"E"^2`

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