Question

A parallel plate capacitor with area $200c{m}^2$ 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 $25\times 10^{-6}$ N, the value of $V$ is approximately :
$\left({ϵ}_0=8.85\times 10^{-12}\frac{C^2}{N.m^2}\right)$

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

Answer 1

Answer: (A)

Electric field between parallel plate capacitor is given by
$E=\frac{\sigma }{2{\epsilon }_0}=\frac{1}{2{\epsilon }_0}\frac{Q}{A}$
Force between plates is $F=EQ$. Therefore, $F=\frac{1}{2{\epsilon }_0}\frac{Q^2}{A}$. Since $Q=CV=\frac{{\epsilon }_{0}A}{d}V$, therefore,
$F=\frac{1}{2{\epsilon }_0}{\left(\frac{{\epsilon }_{0}A}{d}V\right)}^2\cdot \frac{1}{A}=\frac{{\epsilon }_{0}A{V}^2}{2d^2}$
$\Rightarrow V^2=\frac{2d^{2}F}{{\epsilon }_{0}A}$
$\Rightarrow V=\sqrt{\frac{2d^{2}F}{{\epsilon }_{0}A}}$
Given: $d=1.5cm=1.5\times 10^{-2}m;F=25\times 10^{-6}N;$
${\epsilon }_0=8.85\times 10^{-12}{C}^2/N{m}^2;$
$A=200c{m}^2=200\times 10^{-4}c{m}^2$
Therefore,$V=\sqrt{\frac{2\times {\left(1.5\times 10^{-2}\right)}^2\times 25\times 10^{-6}}{8.85\times 10^{-12}\times 200\times 10^{-4}}}\sim 250$ Volts