Question

the plates of a parallel capacitor have a area of 90cm squareseparated by 2.5 Umma the capacitor is charged by connecting to a 400 supply calculate the energy stored in the capacitor energy density and the magnitude of the electric field E​

Answer 1

Answer:

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Explanation:

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=

d

ϵ

0

A

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}C=

d

ϵ

0

A

also, V = E × d [a you Know, electric potential = electric field × separation between plates ]

so, U = $$\begin{lgathered}\frac{1}{2}\times\frac{\epsilon_0A}{d}\times(E.d)^2\\U=\frac{1}{2}\epsilon_0AE^2d\:or,\frac{U}{Ad}=\frac{1}{2}\epsilon_0E^2\end{lgathered}$$

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