Physics · Electrostatics

JEE Main 2025 — 2 April, Morning Shift — Question 60

A point charge $+q$ is placed at the origin. A second point charge $+9 q$ is placed at $(d, 0,0)$ in Cartesian coordinate system. The point in between them where the electric field vanishes is:

Option A:
$(d / 4,0,0)$
Correct
Option B:
$(d / 3,0,0)$
Option C:
$(3 d / 4,0,0)$
Option D:
$(4 d / 3,0,0)$

Answer: A

Step-by-step solution

\begin{aligned} & \frac{9}{x^{2}}=\frac{2 q}{(d-x)^{2}} &\\ d-x=3 x &\\ x=\frac{d}{4} \end{aligned}

$\therefore E=0$ at $(d / 4,0,0)$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Morning Shift
Subject: Physics
Chapter: Electrostatics
Topic: Properties of Electric Charge and Coulomb's Law

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