Problem 128 - A Complete Resource Book in Physics for JEE Main

Question

A parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed. If \(x\) is the separation between the plates, then the rate of change of electrostatic energy of the capacitor is proportional to (A) \(x\) (B) \(x^{2}\) (C) \(\frac{1}{x}\) (D) \(\frac{1}{x^{2}}\)

Step-by-Step Solution

Answer

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

Step 1: Find the Formula for Electrostatic Energy of a Capacitor

The electrostatic energy (U) of a capacitor is given by the formula: \[U = \frac{1}{2}CV^{2}\] where C is the capacitance of the capacitor and V is the voltage across it.

Step 2: Determine the relationship between Capacitance and Separation

The capacitance for a parallel plate capacitor is given by: \[C = \frac{\varepsilon A}{x}\] where \(\varepsilon\) is the permittivity of the dielectric between the plates, A is the area of the plates, and x is the separation between the plates.

Step 3: Substitute the value of Capacitance in the Energy Formula

Now we will substitute the value of C from Step 2 into the formula for the electrostatic energy of the capacitor: \[U = \frac{1}{2} (\frac{\varepsilon A}{x}) V^{2} \] \[U = \frac{\varepsilon AV^{2}}{2x} \]

Step 4: Find the Rate of Change of Electrostatic Energy

To find the rate of change of the electrostatic energy with respect to x, we will differentiate U with respect to x: \[\frac{dU}{dx} = \frac{-\varepsilon AV^{2}}{2x^{2}} \] Since we are interested in the proportionality between the rate of change of electrostatic energy and x, we can remove the constant terms: \[\frac{dU}{dx} \propto \frac{1}{x^{2}} \]

Step 5: Compare with the Given Options

The rate of change of electrostatic energy of the capacitor is proportional to \(\frac{1}{x^{2}}\). Comparing with the given options, we find that the correct option is: (D) \(\frac{1}{x^{2}}\)