An electric charge 10 -6 μ C is placed at origin (0,0) m of X - Y co-ordinate system. Two points P and Q are situated at (√(3), √(3)) m and (√(6), 0) m respectively. The potential difference between the points P and Q will be :

Physics · Electrostatics

JEE Main 2024 — 27 January, Shift 1 — Question 40

An electric charge $10^{-6} \mu \mathrm{C}$ is placed at origin $(0,0)$ m of $\mathrm{X}-\mathrm{Y}$ co-ordinate system. Two points P and Q are situated at $(\sqrt{3}, \sqrt{3}) \mathrm{m}$ and $(\sqrt{6}, 0) \mathrm{m}$ respectively. The potential difference between the points $P$ and $Q$ will be :

Option A:
$\sqrt{3} \mathrm{~V}$
Option B:
$\sqrt{6} \mathrm{~V}$
Option C:
0 V
Correct
Option D:
3 V

Answer: C

Step-by-step solution

Potential difference

$=\frac{K Q}{r_{1}}-\frac{K Q}{r_{2}}$

$r_{1}=\sqrt{(\sqrt{3})^{2}+(\sqrt{3})^{2}}$

$r_{2}=\sqrt{(\sqrt{6})^{2}+0}$

As $r_{1}=r_{2}=\sqrt{6} \mathrm{~m}$

So potential difference $=0$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 27 January, Shift 1
Subject: Physics
Chapter: Electrostatics
Topic: Electrostatic Potential Energy and Electric Potential

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