A) \[\frac{1}{d}\]
B) \[\frac{1}{{{d}^{2}}}\]
C) \[{{d}^{2}}\]
D) \[d\]
Correct Answer: B
Solution :
\[C=\frac{{{\varepsilon }_{0}}A}{x};\] \ \[\frac{dC}{dt}={{\varepsilon }_{0}}A\frac{d}{dt}\left( \frac{1}{x} \right)\]\[=\frac{-{{\varepsilon }_{0}}A}{{{x}^{2}}}\left( \frac{dx}{dt} \right)=\frac{-{{\varepsilon }_{0}}A}{{{d}^{2}}}\left( \frac{dx}{dt} \right)\] Þ \[\left| \frac{dC}{dt} \right|=\frac{{{\varepsilon }_{0}}A}{{{d}^{2}}}v\] i.e. \[\left| \frac{dC}{dt} \right|\propto \frac{1}{{{d}^{2}}}\]You need to login to perform this action.
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