Physics - Capacitor Question with Solution | TestHub

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, $F=$ $25\times 10^{-6}\mathrm{N}$ $F=QE$ $F=\frac{Q^2}{2A{ϵ}_0}$ (E due to parallel plate $=\frac{\sigma }{2{ϵ}_0}=\frac{Q}{A2{ϵ}_0})$ But $Q=CV=\frac{{\in }_{0}A(V)}{d}$
$$\begin{array}{c}\therefore F=\frac{\left({ϵ}_{0}A{V}^2\right)}{{\mathrm{d}}^2\times 2\mathrm{A}{ϵ}_0})\\ =\frac{{\left({ϵ}_{0}A\right)}^2\times V^2}{d^2\times 2\times \left(A{ϵ}_0\right)}=\frac{\left({ϵ}_{0}A\right)\times V^2}{d^2\times 2}\\ \text{ or, }25\times 10^{-6}=\frac{\left(8.85\times 10^{-12}\right)\times \left(200\times 10^{-4}\right)\times V^2}{2.25\times 10^{-4}\times 2}\\ \Rightarrow V=\frac{25\times 10^{-6}\times 2.25\times 10^{-4}\times 2}{8.85\times 10^{-12}\times 200\times 10^{-4}}\approx 250\mathrm{V}\end{array}$$