Question

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. 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 1

Answer:

Energy per unit volume = $0.113\ J/m^3$

Explanation:

Given,

• Area of parallel plate, A = 90 cm²

                                        $=\ 0.009\ m^2$

• Distance between the plates, d = 2.5 mm

                                                     = 0.0025 m

• Voltage, v = 400 v

Volume between the plates, V = 0.009 x 0.0025

                                                   $=\ 2.25\times 10^{-5}\ m^3$

$Capacitance,c\ =\ \dfrac{\epsilon_0.A}{d}$

                             $=\ \dfrac{8.84\times 10^{-12}\times 0.009}{0.0025}$

                             $=\ 3.18\times 10^{-11}\ F$

The energy stored in the capacitor can be given by,

$E\ =\ \dfrac{1}{2}cv^2$

    $=\ \dfrac{1}{2}\times 3.18\times 10^{-11}\times (400)^2$

   $=\ 2.54\times 10^{-6}\ J$

Energy per unit volume can be given by,

$E_V\ =\ \dfrac{E}{V}$

        $=\ \dfrac{2.54\times 10^{-6}\ J}{2.25\times 10^{-5}\ m^3}$

        $=\ 0.113 J/m^3$

Hence, energy per unit volume is $0.113 J/m^3.$