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........$V$ $\left( {{\varepsilon _0} = 8.85 \times {{10}^{ - 12}}\,\frac{{{C^2}}}{{N{m^2}}}} \right)$
JEE MAIN 2018, Diffcult
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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$
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