Electrostatics

JEE Main level, numerical answer, medium

272 of 311

Question 272 of 311

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}. 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 _____.

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Answer: 48



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=\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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