A parallel plate capacitor with area $200\ \text{cm}^2$ and separation between the plates $1.5\ \text{cm}$, is connected across a battery of emf $V$. If the force of attraction between the plates is $25\times10^{-6}\ \text{N}$, the value of $V$ is approximately: $\left(\varepsilon_0 = 8.85\times10^{-12}\ \dfrac{\text{C}^2}{\text{N m}^2}\right)$
Answer: (C) $250\ \text{V}$
Force between the plates:
$$F = \frac{Q^2}{2\varepsilon_0A} = \frac{\varepsilon_0AV^2}{2d^2}$$
$$V^2 = \frac{2Fd^2}{\varepsilon_0A} = \frac{2\times25\times10^{-6}\times(1.5\times10^{-2})^2}{8.85\times10^{-12}\times200\times10^{-4}} = \frac{1.125\times10^{-8}}{1.77\times10^{-13}} \approx 6.36\times10^4$$
$$V \approx 252\ \text{V} \approx 250\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics