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:
Since the capacitor is connected to a battery, the potential difference ($V$) remains constant. As the distance $x$ changes, the capacitance $C$ changes, which in turn changes the stored electrostatic energy $U$.

Step 2: Key Formula or Approach:
1. Capacitance $C = \frac{\epsilon_0 A}{x}$. 2. Energy $U = \frac{1}{2} C V^2$. 3. Rate of change of energy is $\frac{dU}{dt}$.

Step 3: Detailed Explanation:
1. Express energy in terms of $x$: $U = \frac{1}{2} \left( \frac{\epsilon_0 A}{x} \right) V^2$. 2. Differentiate with respect to time $t$: $\frac{dU}{dt} = \frac{1}{2} \epsilon_0 A V^2 \frac{d}{dt} (\frac{1}{x})$. 3. Using the chain rule: $\frac{d}{dt}(\frac{1}{x}) = -\frac{1}{x^2} \frac{dx}{dt}$. 4. Given the plates are pulled at a uniform speed $v$, $\frac{dx}{dt} = v$ (constant). 5. So, $\frac{dU}{dt} = -\frac{\epsilon_0 A V^2 v}{2 x^2}$. 6. Therefore, $\frac{dU}{dt} \propto \frac{1}{x^2}$ or $x^{-2}$.

Step 4: Final Answer:
The value of $\alpha$ is -2.