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. 12 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

The electric flux \( \Phi_E \) through a surface due to a point charge \( q \) is given by Gauss's Law: \[ \Phi_E = \frac{q}{\varepsilon_0} \] Since the square loop subtends a fraction of the total flux, we need to find the fraction of the flux passing through the loop. The area of the square loop is \( A = 1 \, {m}^2 \). The total flux through a spherical surface surrounding the charge is: \[ \Phi_{{total}} = \frac{q}{\varepsilon_0} = \frac{1}{\varepsilon_0} \] The solid angle subtended by the square loop is proportional to \( \frac{A}{r^2} \). For a square loop placed in front of a point charge, the fraction of the total flux passing through the loop is \( \frac{1}{6} \). Thus, the flux through the square loop is: \[ \Phi_E = \frac{5}{p} \times \frac{1}{\varepsilon_0} \] Comparing this with Gauss's law: \[ \frac{5}{p} = \frac{1}{6} \] Solving for \( p \), we get: \[ p = 30 \] Thus, the value of \( p \) is \( \boxed{30} \).