c
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

\(=\frac{\sigma}{2 \epsilon_{0}}=\frac{Q}{A 2 \epsilon_{0}})\)

But \(Q=C V=\frac{\epsilon_{0} A(V)}{d}\)

\(\therefore \,F = \,\frac{{({\epsilon_0}A{V^2})}}{{{{\text{d}}^2} \times 2{\text{A}}{\epsilon_0}}}\)

\( = \frac{{{{\left( {{\epsilon_0}A} \right)}^2} \times {V^2}}}{{{d^2} \times 2 \times \left( {A{\epsilon_0}} \right)}} = \frac{{\left( {{\epsilon_0}A} \right) \times {V^2}}}{{{d^2} \times 2}}\)

or, \(25 \times {10^{ - 6}} = \frac{{\left( {8.85 \times {{10}^{ - 12}}} \right) \times \left( {200 \times {{10}^{ - 4}}} \right) \times {V^2}}}{{2.25 \times {{10}^{ - 4}} \times 2}}\)

\( \Rightarrow V\, = \frac{{25 \times {{10}^{ - 6}} \times 2.25 \times {{10}^{ - 4}} \times 2}}{{8.85 \times {{10}^{ - 12}} \times 200 \times {{10}^{ - 4}}}}\) \( \approx 250\,V\)