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

Hint:

Since the capacitor stays connected to the battery, it is the voltage V that stays constant here, not the charge Q, and that single detail decides which formula you should differentiate. Start by writing the stored energy U in terms of x alone, using $C = \dfrac{\epsilon_0 A}{x}$ and $U = \dfrac{1}{2}CV^2$, then differentiate that single power of x with respect to time. Watch how each derivative step changes the power of x, since that is what fixes the value of $\alpha$.

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 \(\alpha = -2\).

Answer 2

Concept:

• Use logarithmic differentiation: taking the log of $U$ turns the power of $x$ into a simple multiplying factor, which is often faster than differentiating a fraction directly.

Step 1: Write $U$ in terms of $x$.
Since $V$ stays constant (battery connected) and $C = \dfrac{\epsilon_0 A}{x}$:
$U = \dfrac12 CV^2 = \dfrac{\epsilon_0 A V^2}{2}\cdot\dfrac1x = \dfrac{k}{x}$, where $k=\dfrac{\epsilon_0 A V^2}{2}$ is constant.

Step 2: Take the natural log of both sides.
$\ln U = \ln k - \ln x$

Step 3: Differentiate both sides with respect to time.
$\dfrac{1}{U}\dfrac{dU}{dt} = -\dfrac{1}{x}\dfrac{dx}{dt} = -\dfrac{v}{x}$

Step 4: Solve for $\dfrac{dU}{dt}$ and substitute $U=k/x$ back in.
$\dfrac{dU}{dt} = -\dfrac{Uv}{x} = -\dfrac{k}{x}\cdot\dfrac{v}{x} = -\dfrac{kv}{x^2}$
So $\dfrac{dU}{dt} \propto x^{-2}$.

Final Answer: $\alpha = -2$