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The lower plate of a parallel plate capacitor is supported on a rigid rod. The upper plate is suspended from one end of a balance. The two plates are joined together by a thin wire and subsequently disconnected. The balance is then counterpoised. Now a voltage $V = 5000\, volt$ is applied between the plates. The distance between the plates is $d =5\, mm$ and the area of each plate is $A = 100 cm^2.$ Then find out the additional mass placed to maintain balance.......$g$ [All the elements other than plates are massless and nonconducting] :-
$44$
$4.4$
$0.44$
$440$
Solution
The electric force between the plates will be balanced by the additional weight
hence $\mathrm{mg}=\frac{\mathrm{Q}^{2}}{2 \mathrm{A} \in_{0}}=\frac{\mathrm{C}^{2} \mathrm{V}^{2}}{2 \mathrm{A} \in_{0}}$
${\rm{mg}} = \frac{{{ \in _0}{\rm{A}}{{\rm{V}}^2}}}{{2{{\rm{d}}^2}}}$
${\rm{m}} = \frac{{{ \in _0}{\rm{A}}{{\rm{V}}^2}}}{{2{{\rm{d}}^2}{\rm{g}}}} = \frac{{{ \in _0} \times 100 \times {{10}^{ – 4}}{{(5000)}^2}}}{{2{{\left( {5 \times {{10}^{ – 3}}} \right)}^2} \times 10}}$
$\mathrm{m}=4.425 \mathrm{\,g}$
