Physics · Electrostatics

JEE Main 2025 — 24 January, Morning Shift — Question 67

A square loop of sides $\mathrm{a}=1 \mathrm{~m}$ is held normally in front of a point charge $\mathrm{q}=1 \mathrm{C}$ . The flux of the electric field through the shaded region is $\frac{5}{\mathrm{p}} \times \frac{1}{\varepsilon_{0}} \frac{\mathrm{Nm}^{2}}{\mathrm{C}}$ , where the value of p is $\qquad$

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Answer: 48

Numerical answer — enter this value.

Step-by-step solution

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Total flux through square $=\frac{\mathrm{q}}{\epsilon_{0}}\left(\frac{1}{6}\right)$

Lets divide square is 8 equal parts.

Flux is same for each part.

$\therefore$ Flux through shaded portion is $\frac{5}{8}$

(Total flux)

$=\frac{5}{8}\times \frac{\text{q}}{{{\epsilon }_{0}}}\frac{1}{6}=\frac{5}{48}\frac{1}{{{\epsilon }_{0}}}$

$\therefore$ required Ans. is 48

Note : Distnace of charge from square loop is not mentioned we have assume it as $\frac{a}{2}$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 24 January, Morning Shift
Subject: Physics
Chapter: Electrostatics
Topic: Electric flux and Gauss's Law