Question

If the right plate of a parallel plate capacitor is pulled with constant velocity as shown in the figure, find how the rate of change of energy stored in the capacitor depends on the separation \(x\).

• A. \(x^{-2}\)
• B. \(x^{-3}\)
• C. \(x^{-1}\)
• D. \(x^{2}\)

Answer 1

The Correct Option is A

Concept:
Capacitance of a parallel plate capacitor is given by \[ C = \frac{\varepsilon_0 A}{x} \] where \(x\) is the separation between the plates. Energy stored in a capacitor connected to a constant voltage source is \[ U = \frac{1}{2}CV^2 \] Since the capacitor remains connected to the battery, the voltage \(V\) remains constant. Step 1: Substitute the capacitance expression into energy formula. \[ U = \frac{1}{2}\left(\frac{\varepsilon_0 A}{x}\right)V^2 \] \[ U = \frac{1}{2}\frac{\varepsilon_0 A V^2}{x} \] Step 2: Differentiate with respect to time. \[ \frac{dU}{dt} = \frac{1}{2}\varepsilon_0 A V^2 \frac{d}{dt}\left(\frac{1}{x}\right) \] \[ \frac{dU}{dt} = \frac{1}{2}\varepsilon_0 A V^2 \left(-\frac{1}{x^2}\right)\frac{dx}{dt} \] Step 3: Use constant velocity condition. Since the plate moves with constant velocity, \[ \frac{dx}{dt} = \text{constant} \] Thus, \[ \frac{dU}{dt} \propto \frac{1}{x^2} \]