Question

A parallel plate capacitor with area $200 \, cm^2$ 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 :
$\left( \epsilon_0 = 8.85 \times 10^{-12} \frac{C^2}{N.m^2} \right)$

• 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
$E=\frac{\sigma}{2 \varepsilon_{0}}=\frac{1}{2 \varepsilon_{0}} \frac{Q}{A}$
Force between plates is $F=E Q$. Therefore, $F=\frac{1}{2 \varepsilon_{0}} \frac{Q^{2}}{A}$. Since $Q=C V=\frac{\varepsilon_{0} A}{d} V$, therefore,
$F=\frac{1}{2 \varepsilon_{0}}\left(\frac{\varepsilon_{0} A}{d} V\right)^{2} \cdot \frac{1}{A}=\frac{\varepsilon_{0} A V^{2}}{2 d^{2}}$
$ \Rightarrow V^{2}=\frac{2 d^{2} F}{\varepsilon_{0} A} $
$\Rightarrow V=\sqrt{\frac{2 d^{2} F}{\varepsilon_{0} A}}$
Given: $d=1.5 \,cm =1.5 \times 10^{-2} \,m ; F=25 \times 10^{-6}\, N ;$
$ \varepsilon_{0}=8.85 \times 10^{-12} C ^{2} / N m ^{2} ; $
$A=200 \,cm ^{2}=200 \times 10^{-4} \,cm ^{2}$
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