Question

​A parallel plate capacitor with area 200 cm2 and separation between the plates 1.5 cm, is connected across a battery of emf V. If the force of attraction between the plates is 25 x 10-6 N, the value of V is approximately :
1) 250 V 2) 100 V
3) 300 V 4) 150 V

Answer 1

Use the formula parallel plate capacitor

Answer 2

Answer:

(1). The value of V is approximately 250 V.

Explanation:

Given that,

Area $A = 200 cm^2$

Distance d = 1.5 cm

Force $f = 25\times10^{-6}\ N$

We know that,

The force of attraction is

$F=\dfrac{\sigma^2 A}{2\epsilon_{0}}$

$F=\dfrac{1}{2}\epsilon_{0}E^2A$

$F = \dfrac{1}{2}\epsilon_{0}(\dfrac{V}{d})^2A$

$V=d\sqrt{\dfrac{2F}{\epsilon A}}$

$V =1.5\times10^{-2}\sqrt{\dfrac{2\times25\times10^{-6}}{8.85\times10^{-12}\times200\times10^{-4}}}$

$V = 252.11 V$

Therefore,

$V = 250 V$ approximately

Hence, The value of V is approximately 250 V.