A parallel plate capacitor is connected to a battery. The plates are pulled apart with uniform speed. If x is the separation between the plates, then the rate of change of electrostatic energy of the capacitor is proportional to

Options

A. `x^2`
B. `x`
C. `1/x`
D. `1/x^2`

Explanation

To solve the problem, we need to analyze the relationship between the electrostatic energy of a parallel plate capacitor and the separation between its plates. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Capacitance of a Parallel Plate Capacitor
The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\varepsilon_0 A}{x} \]
where:
- \( \varepsilon_0 \) is the permittivity of free space,
- \( A \) is the area of one of the plates,
- \( x \) is the separation between the plates.

### Step 2: Energy Stored in the Capacitor
The energy \( U \) stored in a capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]
Since the capacitor is connected to a battery, the voltage \( V \) remains constant. Substituting the expression for capacitance into the energy formula gives:
\[ U = \frac{1}{2} \left( \frac{\varepsilon_0 A}{x} \right) V^2 = \frac{1}{2} \frac{\varepsilon_0 A V^2}{x} \]

### Step 3: Differentiate the Energy with Respect to Time
To find the rate of change of energy \( \frac{dU}{dt} \), we differentiate \( U \) with respect to time \( t \):
\[ \frac{dU}{dt} = \frac{d}{dt} \left( \frac{1}{2} \frac{\varepsilon_0 A V^2}{x} \right) \]
Using the quotient rule for differentiation, we get:
\[ \frac{dU}{dt} = -\frac{1}{2} \frac{\varepsilon_0 A V^2}{x^2} \frac{dx}{dt} \]
where \( \frac{dx}{dt} \) is the rate at which the plates are being pulled apart.

### Step 4: Analyze the Proportionality
From the expression derived, we see that:
\[ \frac{dU}{dt} \propto -\frac{1}{x^2} \frac{dx}{dt} \]
This indicates that the rate of change of electrostatic energy \( \frac{dU}{dt} \) is inversely proportional to the square of the separation \( x \).

### Conclusion
Thus, the rate of change of electrostatic energy of the capacitor is proportional to \( \frac{1}{x^2} \).

### Final Answer
The rate of change of electrostatic energy of the capacitor is proportional to \( \frac{1}{x^2} \).

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