Question 294: Electrostats & Capacitance

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 $$25 	imes {10^{ - 6}}N,$$ the value of V is approximately : $$\left( {{ \in _o} = 8.85 	imes {{10}^{ - 12}}{{{C^2}} \over {N.{m^2}}}} ight)$$

Accepted Answer

Given area of Parallel plate capacitor, A = 200 cm
2
Separation between the plates, d = 1.5 cm
Force of attraction between the plates, F = 25 ÃÂ 10
Ã¢ÂÂ6
N
F = QE
$$ \Rightarrow $$ F = $${{{Q^2}} \over {2A{ \in _0}}}$$
[ As E due to parallel plate = $${\sigma  \over {2{ \in _0}}} = {Q \over {2A{ \in _0}}}$$]
Also we know, Q = CV = $${{{ \in _0}AV} \over d}$$
$$ 	herefore $$ F = $${{{{\left( {{ \in _0}AV} ight)}^2}} \over {{d^2} 	imes 2A{ \in _0}}}$$
$$ \Rightarrow $$ V = d$$\sqrt {{{2F} \over {{ \in _0}A}}} $$
$$ \Rightarrow $$ V = $$1.5 	imes {10^{ - 2}}\sqrt {{{2 	imes 25 	imes {{10}^{ - 6}}} \over {8.85 	imes {{10}^{ - 12}} 	imes 2 	imes {{10}^{ - 2}}}}} $$
= $$1.5 	imes {10^{ - 2}}\sqrt {{{25} \over {8.85}}} $$ = 250 V

Answer

250 V

Answer

100 V

Answer

300 V

Answer

150 V

Electrostats & Capacitance

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