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QJEEM25Ph8f56: JEE Main 2025 Physics - Class 12 JEEM25Ph8f56 Solution: A square loop of sides is held normally in front of a point…

A square loop of sides a=1 ma=1 \mathrm{~m} is held normally in front of a point charge q=1C\mathrm{q}=1 \mathrm{C} at a distance a2\frac{\mathrm{a}}{2} . The flux of the electric field through the shaded region is 5p×1ε0Nm2C\frac{5}{\mathrm{p}} \times \frac{1}{\varepsilon_0} \frac{\mathrm{Nm}^2}{\mathrm{C}} , where the value of p is ________ .

JEE Main 2025 (Online) 24th January Morning Shift Physics - Electrostatics Question 5 English

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JEE Main 2025 (Online) 24th January Morning Shift Physics - Electrostatics Question 5 English Explanation

Total flux through square =qϵ0(16)=\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 58\frac{5}{8} (Total flux)

=58×qϵ016=5481ϵ0

∴\therefore Required Ans. is 48

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