A) \[{{x}^{2}}\]
B) \[x\]
C) \[\frac{1}{x}\]
D) \[\frac{1}{{{x}^{2}}}\]
Correct Answer: D
Solution :
\[U=\frac{1}{2}C{{V}^{2}}=\frac{1}{2}\frac{{{\varepsilon }_{0}}A}{d}{{V}^{2}}\] At any instant, let the separation between plates be\[x\]. So, \[U=\frac{1}{2}\frac{{{\varepsilon }_{0}}A}{x}{{V}^{2}}\] \[\therefore \] \[\frac{dU}{dt}=\frac{1}{2}{{\varepsilon }_{0}}A{{V}^{2}}(-1)\frac{1}{{{x}^{2}}}=\frac{dx}{dt}\] \[=-\frac{1}{2}\frac{{{\varepsilon }_{0}}A{{V}^{2}}}{{{x}^{2}}}(v)\] i.e., potential energy decreases as\[(1/{{x}^{2}})\].
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