The area of each plate of a parallel plate capacitor is $20\,cm^2$ and separation between the plates is $2\,mm$. If dielectric strength of air is $3 \times 10^6\,V/m,$ the maximum possible value of emf of the battery, which can be connected across the plates of this capacitor and the corresponding charge on the plates is
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Dielectric strength of air $=\mathrm{E}_{0}=3 \times 10^{6} \mathrm{\,V} / \mathrm{m}$
Therefore, maximum potential difference that can be produced in between the plates of the capacitor of plate separation $\mathrm{d}$ is
$\mathrm{V}_{\mathrm{m}} =\mathrm{E}_{\mathrm{m}} \cdot \mathrm{d} $
$=3 \times 10^{6} \times 2 \times 10^{-3} $
$=6 \times 10^{3} \mathrm{\,V}$
Also the capacitance of the capacitor is $\mathrm{C},$ then
the maximum possible charge on the plates of the capacitor
$\mathrm{q}_{\mathrm{m}}=\mathrm{CV}_{\mathrm{m}}=\frac{\varepsilon_{0} \mathrm{A}}{\mathrm{d}} \mathrm{V}_{\mathrm{m}}$
$=\frac{8.86 \times 10^{-12} \times 20 \times 10^{-4}}{2 \times 10^{-3}} \times 6 \times 10^{3}=53.16 \times 10^{-9}$
$\approx 53 \mathrm{\,nC}$
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