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Question

A parallel plate capacitor is connected to a battery. The plate are pulled apart with a uniform speed. If $$'x'$$ is the separation between the plates, then the time rate of change of electrostatic energy of the capacitor is proportional to :


A
x2
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B
x
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C
x1
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D
x2
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Solution

The correct option is D $$x^{-2}$$
Parallel plate capacitance at instant time t , $$C=\dfrac{\epsilon_0 A}{x}$$

Also potential $$(V)$$  will remain same (constant) because battery is attached.
Energy $$U=\dfrac{1}{2}CV^2=\dfrac{1}{2}\times \dfrac{\epsilon_0 A}{x}\times V^2$$

$$U = \dfrac{1}{2} \epsilon_0 A V^2(\dfrac{1}{x})$$

Thus, $$U \propto \dfrac{1}{x}$$

$$\dfrac{dU}{dt} =  \dfrac{1}{2} \epsilon_0 A V^2 (-\dfrac{1}{x^2} \dfrac{dx}{dt})$$

$$\dfrac{dU}{dt} \propto \dfrac{1}{x^2}$$
Thus rate of change of U is proportional to  $$x^{-2}$$.

Physics

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