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
B
x
C
x1
D
x2

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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