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 \(25\times10^{-6}\,N\), the value of \(V\) is approximately........\(V\) \(\left( {{\varepsilon _0} = 8.85 \times {{10}^{ - 12}}\,\frac{{{C^2}}}{{N{m^2}}}} \right)\)

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

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