Question 140 of 180

A parallel plate capacitor with area $200 \mathrm{~cm}^2$ and separation between the plates $1.5 \mathrm{~cm}$, is connected across a battery of emf V. If the force of attraction between the plates is $25 \times 10^{-6} \mathrm{~N}$, the value of $V$ is approximately:
$$\left.\left(\varepsilon_0=8.85 \times 10^{-12} \frac{\mathrm{C}^2}{\text { N.m }}\right)^2\right)$$

A$150 \mathrm{~V}$B$100 \mathrm{~V}$C$250 \mathrm{~V}$D$300 \mathrm{~V}$

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Answer: (C) $250 \mathrm{~V}$

Given area of Parallel plate capacitor, $A=$ $200 \mathrm{~cm}^2$
Separation between the plates, $d=1.5 \mathrm{~cm}$ Force of attraction between the plates, $F=$ $25 \times 10^{-6} \mathrm{~N}$ $F=Q E$
$F=\frac{Q^2}{2 A \epsilon_0} \quad$ (E due to parallel plate $\left.=\frac{\sigma}{2 \epsilon_0}=\frac{Q}{A 2 \epsilon_0}\right)$
But $Q=C V=\frac{\in_0 A(V)}{d}$


$$\begin{gathered} \left.\therefore F=\frac{\left(\epsilon_0 A V^2\right)}{\mathrm{d}^2 \times 2 \mathrm{~A} \epsilon_0}\right) \\ =\frac{\left(\epsilon_0 A\right)^2 \times V^2}{d^2 \times 2 \times\left(A \epsilon_0\right)}=\frac{\left(\epsilon_0 A\right) \times V^2}{d^2 \times 2} \\ \text { or, } 25 \times 10^{-6}=\frac{\left(8.85 \times 10^{-12}\right) \times\left(200 \times 10^{-4}\right) \times V^2}{2.25 \times 10^{-4} \times 2} \\ \Rightarrow V=\frac{25 \times 10^{-6} \times 2.25 \times 10^{-4} \times 2}{8.85 \times 10^{-12} \times 200 \times 10^{-4}} \approx 250 \mathrm{~V} \end{gathered}$$