Q.40. A parallel plate air capacitor is connected to a battery. The plates are pulled apart at unifo

Q: What is the correct answer for JEE Main Physics Q40?

The correct answer is A. Sol. The capacitance of a parallel plate capacitor is $C = \frac{\epsilon_0 A}{x}$. When connected to a battery, the voltage V across the capacitor is constant. The electrostatic energy stored in the capacitor is $U = \frac{1}{2} C V^2 = \frac{1}{2} \frac{\epsilon_0 A}{x} V^2$. The time rate of change of energy is $\frac{dU}{dt} = \frac{d}{dt} \left( \frac{\epsilon_0 A V^2}{2x} \right) = \frac{\ep

Q: Q.40. A parallel plate air capacitor is connected to a battery. The plates are pulled apart at uniform speed v. If x is the separation between the plates at any instant, then the time rate of change…

The correct answer is A. Sol. The capacitance of a parallel plate capacitor is $C = \frac{\epsilon_0 A}{x}$. When connected to a battery, the voltage V across the capacitor is constant. The electrostatic energy stored in the capacitor is $U = \frac{1}{2} C V^2 = \frac{1}{2} \frac{\epsilon_0 A}{x} V^2$. The time rate of change of energy is $\frac{dU}{dt} = \frac{d}{dt} \left( \frac{\epsilon_0 A V^2}{2x} \right) = \frac{\ep

Question

Q.40. A parallel plate air capacitor is connected to a battery. The plates are pulled apart at uniform speed v. If x is the separation between the plates at any instant, then the time rate of change of electrostatic energy of the capacitor is proportional to , where a is ________

FirstInTest · Accepted Answer

Sol. The capacitance of a parallel plate capacitor is $C = \frac{\epsilon_0 A}{x}$. When connected to a battery, the voltage V across the capacitor is constant. The electrostatic energy stored in the capacitor is $U = \frac{1}{2} C V^2 = \frac{1}{2} \frac{\epsilon_0 A}{x} V^2$. The time rate of change of energy is $\frac{dU}{dt} = \frac{d}{dt} \left( \frac{\epsilon_0 A V^2}{2x} \right) = \frac{\epsilon_0 A V^2}{2} \frac{d}{dt} (x^{-1})$. Since $x$ is changing with time, $\frac{dU}{dt} = \frac{\epsilon_0 A V^2}{2} (-1) x^{-2} \frac{dx}{dt}$. Given that the plates are pulled apart at uniform speed v, so $\frac{dx}{dt} = v$. Therefore, $\frac{dU}{dt} = -\frac{\epsilon_0 A V^2 v}{2} x^{-2}$. Thus, $\frac{dU}{dt}$ is proportional to $x^{-2}$. So, $a = -2$.

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