A square loop of sides a=1 ~m is held normally in front of a point charge q=1 C . The flux of the electric field through the shaded region is 5 p × (1)/( _0) Nm^2 C , where the value of p is _____.

Correct answer: 0 Total flux through square = q _0 ((1)/(6)) Lets divide square is 8 equal parts. Flux is same for each part. Flux through shaded portion is (5)/(8) (Total flux) =(5)/(8) × q _0 (1)/(6)=(5)/(48) (1)/( _0) required Ans. is 48

JEE Main 202524 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Morning Shift
Subject: Physics
Chapter: Electrostatics

JEE Main 2025 · 24 January · Morning Shift · Physics · Electrostatics

A square loop of sides a=1 ~m is held normally in front of a point...

A square loop of sides

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 _____. A square loop of sides a=1 ~m is held normally in front of a point charge q=1 C . The flux of the electric field

Correct answer

Step-by-step solution

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{\mathrm{q}}{\epsilon_0} \frac{1}{6}=\frac{5}{48} \frac{1}{\epsilon_0} \therefore required Ans. is 48

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