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

- How much electrostatic energy is stored by the capacitor?
- 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.

### Asked by Topperlearning User | 22nd Apr, 2015, 09:59: AM

Area of the plates of a parallel plate capacitor, A = 90 cm^{2} = 90 x 10^{-4}m^{2} Distance between the plates, d = 2.5 mm = 2.5 x 10^{-3} m potential difference across the plates, V = 400 V

- 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 x 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,

V' = A x d

= 90 x 10^{-4} x 2.5 x 10^{-3}

= 2.25 x 10^{-5 }m^{3}

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

Again,

Where,

= Electric intensity = E

(c) When zero of potential energy is taken, d_{1}=1.06

Therefore, Potential energy of the system = Potential energy at d_{1} - Potential energy at d

= 21.73 x 10^{-19}J - 27.2eV

= 13.58eV - 27.2eV

= -13.6eV

### Answered by | 22nd Apr, 2015, 11:59: AM

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