Question

The plates of a parallel plate capacitor have an area of 90 cm2 eachand are separated by 2.5 mm. The capacitor is charged by connectingit 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 betweenthe plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electricfield E between the plates.

Answer 1

(a)

It is given that the area of a parallel plate capacitor is A=90  cm 2 and separation

between plates is d=2⋅5 mm and the capacitor is charged by 400 V.

The formula of electrostatic energy is,

E 1 = 1 2 εA d V 2

Substitute the values.

E 1 = 8⋅85× 10 −12 ×90× 10 −4 × ( 400 ) 2 2×2⋅5× 10 −3 =2⋅55× 10 −6  J

Thus, the electrostatic energy stored by the capacitor is 2⋅55× 10 −6  J

(b)

It is given that the energy stored in the capacitor per unit volume is u and the electric field between the plates is E.

The volume of the capacitor is,

V'=A×d

Substitute the values of A and d in the above equation.

V'=90× 10 −4 ×2.5× 10 −3 =2⋅25× 10 −4 m 3

The energy stored in the capacitor per unit volume is,

u= E 1 V'

Substitute the values of E 1 and V' in the above equation,

u= 2⋅55× 10 −6  J 2⋅25× 10 −5   m 3 =0⋅113  Jm −3

Thus, the energy stored per unit volume is 0.113  Jm −3 .

The energy stored in the capacitor per unit volume is given as,

u= E 1 V' = εA 2d V 2 Ad = 1 2 ε ( V d ) 2 (1)

The electrical field between plates is given as,

E= V d

Substitute the value of E in equation (1).

u= 1 2 ε E 2

Thus, the relation between energy per unit volume and magnitude of electric field is 1 2 ε E 2 .