Physics · Electrostatics

JEE Main 2025 — 3 April, Morning Shift — Question 56

The electrostatic potential on the surface of uniformly charged spherical shell of radius $R=10 \mathrm{~cm}$ is 120 V . The potential at the centre of shell, at a distance $r=5$ cm from centre, and at a distance $r=15 \mathrm{~cm}$ from the centre of the shell respectively, are:

Option A:
$120 \mathrm{~V}, 120 \mathrm{~V}, 80 \mathrm{~V}$
Correct
Option B:
$40 \mathrm{~V}, 40 \mathrm{~V}, 80 \mathrm{~V}$
Option C:
$0 \mathrm{~V}, 120 \mathrm{~V}, 40 \mathrm{~V}$
Option D:
$0 V, 0 V, 80 \mathrm{~V}$

Answer: A

Step-by-step solution

For $r\le R\text{ }\!\!~\!\!\text{ }V=\frac{kQ}{R}$ For $r>R\text{ }\!\!~\!\!\text{ }V=\frac{kQ}{r}$ (i) For $r=0$ and $r=5\text{ }\!\!~\!\!\text{ cm}$

$(<R=10\text{ }\!\!~\!\!\text{ cm})$

$V=120\text{ }\!\!~\!\!\text{ V}$

(ii) For $r=15\text{ }\!\!~\!\!\text{ cm}V=\frac{kQ}{R}\frac{R}{r}$

$\begin{array}{*{35}{r}}{} & ~=120\times \frac{10}{15} \\{} & ~=80\text{ }\!\!~\!\!\text{ V} \\\end{array}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 3 April, Morning Shift
Subject: Physics
Chapter: Electrostatics
Topic: Electrostatic Potential Energy and Electric Potential