Question

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 \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is:

• A. 15 N/m²
• B. 10 N/m²
• C. 48 N/m²
• D. 8 N/m²

Hint:

When dealing with flux through a surface in front of a point charge, remember that the flux is proportional to the solid angle subtended by the surface. For a square loop, the flux is a fraction of the total flux, which can be found using the formula for the solid angle.

Answer 1

The Correct Option is C

Given:

- The total flux through the square is given by: \[ \Phi_{\text{total}} = \frac{q}{\epsilon_0} \times \frac{1}{6}, \] where \( q \) is the charge and \( \epsilon_0 \) is the permittivity of free space.

Step 1: Divide the square into 8 equal parts

The square is divided into 8 equal parts, and the flux is the same for each part. Therefore, the flux through the shaded portion is: \[ \Phi_{\text{shaded}} = \frac{5}{8} \times \Phi_{\text{total}} = \frac{5}{8} \times \frac{q}{\epsilon_0} \times \frac{1}{6}. \] Simplifying this expression: \[ \Phi_{\text{shaded}} = \frac{5}{48} \times \frac{q}{\epsilon_0}. \]

Final Answer:

The required answer is 48.