Question

Question 2.26: 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.

Class 12 - Physics - Electrostatic Potential And Capacitance Electrostatic Potential And Capacitance Page-90

Answer 1

cross section area of parallel plate capacitor , A = 90 cm² = 90 × 10^-4 m²
separation between plates , d = 2.5mm = 2.5 × 10^-3 m
$C=\frac{\epsilon_0A}{d}$
C = (8.85 × 10^-12 × 90 × 10^-4)/(2.5 × 10^-3)
= 32 × 10^-12 F = 32pF

(a) electrostatic energy is stored by the capacitor , U = 1/2 CV²
U = 1/2 × 32 × 10^-12 × (400)²
= 1/2 × 32 × 10^-12 × 160000
= 16 × 10^-12 × 16 × 10^4
= 256 × 10^-8 = 2.56 × 10^-6 J

(b) volume , V = A × d = 90 × 10^-4 × 2.5 × 10^-3
= 2.25 × 10^-5 m³
so , energy per unit volume = Energy/volume
= (2.56 × 10^-6)/(2.25 × 10^-5) J/m³
= 1.13 × 10^-1 = 0.113 J/m³

well, energy = 1/2 CV²
here, $C=\frac{\epsilon_0A}{d}$
also, V = E × d [a you Know, electric potential = electric field × separation between plates ]
so, U = $\frac{1}{2}\times\frac{\epsilon_0A}{d}\times(E.d)^2\\U=\frac{1}{2}\epsilon_0AE^2d\:or,\frac{U}{Ad}=\frac{1}{2}\epsilon_0E^2$

hence, energy per unit volume = $\bf{\frac{1}{2}\epsilon_0E^2}$

Answer 2

Answer:

this is your 1 st answer