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JEE Mains · Physics · STD 12 - Electric charges and fields
  • JEE Mains 2025
  • Easy
  • Numerical

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 _____.

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Answer & Solution

Correct Answer

(A) 48

Step-by-step Solution

Detailed explanation


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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