Question

A parallel plate capacitor with area 200 cm^2 and seperation between the plates 1.5 cm is connected across a battery of emf if the force of attraction between the plates is 25*10_6 then the value of v

Answer 1

This guide provides clear, practical and step-by-step explanation for the complete study of an electrical installation, according to IEC 60364 and other relevant IEC Standards.

Answer 2

Answer :

The value of V is 250 Volts

Step-by-step explanation :

Given,

area ( A ) = 200 cm²

Seperation between plates, that is, d = 1.5 cm

Force of Attraction ( F ) = 25 * 10⁶ N

Since we know,

F = σ ² A / 2ε₀

where,

σ is Surface charge density

ε₀ = 8.85 * $10^{-12}$ $\frac{C^{2}}{Nm^{2}}$

We know,

Electrical field between plates ( E ) = σ / ε₀

σ =  E . ε₀

So the above formula can be rewritten as :

F = E² ε₀² . A / 2ε₀

F = E² ε₀ . A / 2

We also know,

E = $\frac{V}{D}$

F = $(\frac{V}{D} )^{2}$ ε₀ A / 2

V = √ (2F / ε₀ A) * d

Substituting the values,

V = $\sqrt{\frac{2*25*10^{-6}}{8.85*10^{-12}*200*10^{-4}} } * 1.5 * 10^{-2}$

V = $\sqrt{\frac{25}{8.85}*10^{8} }* 1.5 * 10^{-2}$

V =  $\frac{1.5*10^{-2}*5*10^{4}}{3}$

V = 2.5 * $10^{2}$ V = 250 V

Hence,

The value of V is 250 Volts