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:
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.

Answer 2

Concept:

• Spot the proportionality first: since the battery keeps $V$ fixed, the energy is just a constant times $x^{-1}$. Differentiating $x^{-1}$ always drops the power by one, so the answer comes from tracking that power shift, without writing out $\epsilon_0$, $A$ or $V$ at all.

Step 1: Write $U$ as a plain power of $x$.
$C = \dfrac{\epsilon_0 A}{x}$ and $U = \dfrac12 CV^2$, so with $V$ held constant by the battery:
$U = \left(\dfrac{\epsilon_0 A V^2}{2}\right)x^{-1} = k\,x^{-1}$, where $k$ is just a constant.

Step 2: Apply the power rule for differentiation directly.
For any $U = k\,x^{n}$, $\dfrac{dU}{dt} = k\,n\,x^{n-1}\dfrac{dx}{dt}$. Here $n=-1$, so this gives an $x^{-2}$ term.

Step 3: Bring in the given speed.
The plates move at uniform speed $v$, so $\dfrac{dx}{dt}=v$, which is just a constant multiplier and does not change the power of $x$.

Step 4: Read off the power of $x$.
$\dfrac{dU}{dt} = -k\,v\,x^{-2}$, so $\dfrac{dU}{dt} \propto x^{-2}$.

Final Answer: $\alpha = -2$