Question

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 \(x^\alpha\), where \(\alpha\) is ______.

• A. -2
• B. 1
• C. -1
• D. 2

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Because the capacitor remains connected to the battery, the potential difference (\(V\)) across the plates is constant. As the plates are pulled apart, the capacitance \(C\) changes over time, causing the electrostatic energy to change.
Step 2: Key Formula or Approach:
The capacitance of a parallel plate capacitor is \(C = \frac{\epsilon_0 A}{x}\), where \(x\) is the plate separation.
The electrostatic energy stored in the capacitor at constant voltage is \(U = \frac{1}{2} C V^2\).
The rate of change of energy is \(\frac{dU}{dt}\), which we can evaluate using the chain rule: \(\frac{dU}{dt} = \frac{dU}{dx} \cdot \frac{dx}{dt}\).
The plates are pulled apart at uniform speed \(v\), so \(\frac{dx}{dt} = v\).
Step 3: Detailed Explanation:
First, express the energy \(U\) explicitly in terms of \(x\):
\(U = \frac{1}{2} \left( \frac{\epsilon_0 A}{x} \right) V^2 = \frac{\epsilon_0 A V^2}{2} x^{-1}\).
Now, differentiate \(U\) with respect to time \(t\):
\(\frac{dU}{dt} = \frac{d}{dt} \left( \frac{\epsilon_0 A V^2}{2} x^{-1} \right)\).
\(\frac{dU}{dt} = \frac{\epsilon_0 A V^2}{2} \left( -x^{-2} \frac{dx}{dt} \right)\).
Since \(\frac{dx}{dt} = v\), we get:
\(\frac{dU}{dt} = -\frac{\epsilon_0 A V^2 v}{2} \cdot \frac{1}{x^2}\).
Since \(\epsilon_0, A, V,\) and \(v\) are all constants, we can see that:
\(\frac{dU}{dt} \propto \frac{1}{x^2} \propto x^{-2}\).
Step 4: Final Answer:
By comparing with \(x^\alpha\), we find that \(\alpha = -2\).