Question

A parallel plate capacitor with area $200{cm}^2$ and separation between the plates $1.5cm$, is connected across a battery of emf $V$. If the force of attraction between the plates is $25\times {10}^{-6}N$, the value of $V$ is approximately:
$({ϵ}_0=8.85\times {10}^{--12}\frac{C^2}{N.m^2})$

• A. $150V$
• B. $100V$
• C. $250V$
• D. $300V$

Answer 1

Answer: (C)

$250V$

$A=200c{m}^2$

$d=1.5cm$

$F=25\times {10}^{-6}N$

$\because E=\frac{\sigma }{2{\in }_o}=\frac{Q}{2A{\in }_o}$

$F=QE$

$F=\frac{Q^2}{2A{\in }_o}$

But $Q=CV=\frac{{\in }_{o}A\left(V\right)}{d}$

$\therefore F=\frac{({\in }_{o}AV{)}^2}{d^2\times 2A{\in }_o}$

$=\frac{({\in }_{o}A{)}^2\times V^2}{d^2\times 2\times \left(A{\in }_o\right)}$

$=\frac{\left({\in }_{o}A\right)V^2}{d^2\times 2}$

$25\times {10}^{-6}=\frac{(8.85\times {10}^{-12})\times (200\times {10}^{-4})\times V^2}{2.25\times {10}^{-4}\times 2}$

$V=\sqrt{\frac{25\times {10}^{-6}\times 2.25\times {10}^{-4}\times 2}{8.85\times {10}^{-12}\times 200\times {10}^{-4}}}$

Here, on solving, $v\approx 250V$