JEE Main 2026 — Physics Electromagnetism

2026mcqmedium

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 _____.

Official previous-year question

Held on 4 Apr 2026 · Verified 6 Jul 2026.

Options

A
$-2$
B
$1$
C
$-1$
D
$2$

Did you get this right?

Solution

The electrostatic energy of a parallel plate capacitor is given by $U = \dfrac{1}{2} C V^2$ .

For a parallel plate capacitor, the capacitance is $C = \dfrac{\varepsilon_0 A}{x}$ , where $x$ is the separation between the plates.

Since the capacitor remains connected to the battery, the potential difference $V$ across the plates is constant.

Substituting $C$ into the energy equation gives:

$U = \dfrac{\varepsilon_0 A V^2}{2x}$

Differentiating $U$ with respect to time $t$ to find the time rate of change of electrostatic energy:

$\dfrac{dU}{dt} = \dfrac{d}{dt} \left( \dfrac{\varepsilon_0 A V^2}{2x} \right) = -\dfrac{\varepsilon_0 A V^2}{2x^2} \dfrac{dx}{dt}$

Given that the plates are pulled apart at a uniform speed $v$ , we have $\dfrac{dx}{dt} = v$ .

$\dfrac{dU}{dt} = -\dfrac{\varepsilon_0 A V^2 v}{2x^2}$

From the above expression, it is clear that $\dfrac{dU}{dt} \propto x^{-2}$ .

Comparing this with $x^\alpha$ , we get $\alpha = -2$ .

Answer: $-2$

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JEE Main 2026 — Physics Electromagnetism

2026mcqmedium

Official previous-year question

Held on 4 Apr 2026 · Verified 6 Jul 2026.

Options

A
$-2$
B
$1$
C
$-1$
D
$2$

Did you get this right?

Solution

The electrostatic energy of a parallel plate capacitor is given by $U = \dfrac{1}{2} C V^2$ .

For a parallel plate capacitor, the capacitance is $C = \dfrac{\varepsilon_0 A}{x}$ , where $x$ is the separation between the plates.

Since the capacitor remains connected to the battery, the potential difference $V$ across the plates is constant.

Substituting $C$ into the energy equation gives:

$U = \dfrac{\varepsilon_0 A V^2}{2x}$

Differentiating $U$ with respect to time $t$ to find the time rate of change of electrostatic energy:

$\dfrac{dU}{dt} = \dfrac{d}{dt} \left( \dfrac{\varepsilon_0 A V^2}{2x} \right) = -\dfrac{\varepsilon_0 A V^2}{2x^2} \dfrac{dx}{dt}$

Given that the plates are pulled apart at a uniform speed $v$ , we have $\dfrac{dx}{dt} = v$ .

$\dfrac{dU}{dt} = -\dfrac{\varepsilon_0 A V^2 v}{2x^2}$

From the above expression, it is clear that $\dfrac{dU}{dt} \propto x^{-2}$ .

Comparing this with $x^\alpha$ , we get $\alpha = -2$ .

Answer: $-2$

Your note

JEE Main Electromagnetism in other years

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Work through every JEE Main Electromagnetism PYQ, year by year.

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JEE Main 2026 — Physics Electromagnetism

Options

Did you get this right?

Solution

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JEE Main Electromagnetism in other years

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JEE Main 2026 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism