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Physics Question 40 – JEE-MAIN 2026

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 xa, where a is _______.

The capacitor is a parallel plate air capacitor, and it is connected to a battery, implying the voltage V across it is constant.

Step 1: Express energy in terms of separation✦ Active

The capacitance of a parallel plate capacitor is given by C=ϵ0Ax, where ϵ0 is the permittivity of free space, A is the area of the plates, and x is the separation between them. Since the capacitor is connected to a battery, the voltage V across it remains constant. The electrostatic energy stored in the capacitor is U=12CV2. Substituting the expression for C into the energy formula:

U=12(ϵ0Ax)V2=ϵ0AV22x
Step 2: Calculate the time rate of change of energy○ Expand

To find the time rate of change of electrostatic energy, we differentiate U with respect to time t. Let K=ϵ0AV22 be a constant. Then U=Kx1. Using the chain rule dUdt=dUdxdxdt:

dUdt=ddx(Kx1)dxdt=K(1)x2dxdt
Step 3: Substitute the rate of change of separation and determine proportionality○ Expand

The problem states that the plates are pulled apart at a uniform speed v, which means dxdt=v. Substituting this into the expression for dUdt:

dUdt=Kx2v=(ϵ0AV22)vx2

From this, we can see that the time rate of change of electrostatic energy is proportional to x2. Therefore, comparing with xa, we find a=2.

💡 Teacher's Secret Hint

Ensure to correctly apply the chain rule and identify all constant terms before differentiation.

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