Capacitors

JEE Main level, single correct, medium

4 of 180

Question 4 of 180

A parallel plate air capacitor is connected to a battery. The plates are pulled apart at uniform speed vv. If xx is the separation between the plates at any instant, then the time rate of change of electrostatic energy of the capacitor is proportional to xαx^\alpha, where α\alpha is _____.

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Answer: (A) −2-2

The electrostatic energy of a parallel plate capacitor is given by U=12CV2U = \dfrac{1}{2} C V^2.

For a parallel plate capacitor, the capacitance is
C=ε0AxC = \dfrac{\varepsilon_0 A}{x}, where xx is the separation between the plates.

Since the capacitor remains connected to the battery, the potential difference
VV across the plates is constant.

Substituting
CC into the energy equation gives:

U=ε0AV22xU = \dfrac{\varepsilon_0 A V^2}{2x}

Differentiating
UU with respect to time tt to find the time rate of change of electrostatic energy:

dUdt=ddt(ε0AV22x)=−ε0AV22x2dxdt\dfrac{dU}{dt} = \dfrac{d}{dt} \left( \dfrac{\varepsilon_0 A V^2}{2x} \right) = -\dfrac{\varepsilon_0 A V^2}{2x^2} \dfrac{dx}{dt}

Given that the plates are pulled apart at a uniform speed
vv, we have dxdt=v\dfrac{dx}{dt} = v.

dUdt=−ε0AV2v2x2\dfrac{dU}{dt} = -\dfrac{\varepsilon_0 A V^2 v}{2x^2}

From the above expression, it is clear that
dUdt∝x−2\dfrac{dU}{dt} \propto x^{-2}.

Comparing this with
xαx^\alpha, we get α=−2\alpha = -2.

Answer:
−2-2

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