Physics · Electrostatics

JEE Main 2026 — 24 January, Morning Shift — Question 38

The electrostatic potential in a charged spherical region of radius $r$ varies as $V=a r^{3}+b$ , where $a$ and $b$ are constants. The total charge in the sphere of unit radius is $\alpha \times \pi \mathrm{a} \in_{0}$ . The value of $\alpha$ is $\_\_\_\_$ . (permittivity of vacuum is $\in_{0}$ )

Option A:
$-12$
Correct
Option B:
$-6$
Option C:
$-9$
Option D:
$-8$

Answer: A

Step-by-step solution

$\mathrm{v}=\mathrm{ar}^{3}+\mathrm{b}$

\begin{aligned} & \mathrm{E}=-\frac{\mathrm{dv}}{\mathrm{dr}}=-3 \mathrm{ar}^{2} & \phi_{\text {closed }}=\frac{\mathrm{q}_{\mathrm{enc}}}{\varepsilon_{0}} & \mathrm{q}_{\mathrm{enc}}=\varepsilon_{0} \cdot \mathrm{E} \cdot \mathrm{~A} & =\varepsilon_{0}\left(-3 \mathrm{a} \cdot(1)^{2}\right) 4 \pi(1)^{2} & =-12 \pi \mathrm{a} \varepsilon_{0} & \therefore \mathrm{x}=-12 \end{aligned}

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Exam: JEE Main 2026
Paper: 24 January, Morning Shift
Subject: Physics
Chapter: Electrostatics
Topic: Electrostatic Potential Energy and Electric Potential

Answer: A

Step-by-step solution

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