Physics · Electrostatics

JEE Main 2025 — 2 April, Evening Shift — Question 50

Consider a circular loop that is uniformly charged and has a radius $a \sqrt{2}$ . Find the position along

the positive $z$ -axis of the cartesian coordinate system where the electric field is maximum if the ring was assumed to be placed in $x y$ plane at the origin

Option A:
$a$
Correct
Option B:
$a / 2$
Option C:
$\frac{a}{\sqrt{2}}$
Option D:
0

Answer: A

Step-by-step solution

$E=\frac{K Q x}{\left(R^{2}+x^{2}\right)^{3 / 2}}$

For $E_{\max }, \frac{d E}{d x}=0$

$x=\frac{R}{\sqrt{2}}$

$R=a \sqrt{2}$

$\therefore \quad x=a$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Evening Shift
Subject: Physics
Chapter: Electrostatics
Topic: Electrostatic Field & motion of charge