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(A) The charge $q$ is placed at a distance $a/2$ from the center of the square loop of side $a$. By symmetry,we can enclose the charge $q$ in a cube o

Vedclass · Accepted Answer

$48$

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(A) The charge $q$ is placed at a distance $a/2$ from the center of the square loop of side $a$. By symmetry,we can enclose the charge $q$ in a cube of side $a$ such that the square loop forms one of the faces of the cube. The total electric flux through the entire cube is $\Phi_{total} = \frac{q}{\varepsilon_0}$. Since the cube has $6$ identical faces,the flux through the square loop (which is one face) is $\Phi_{square} = \frac{1}{6} \frac{q}{\varepsilon_0}$. The square loop can be divided into $8$ identical triangular regions by drawing lines from the center to the corners and midpoints of the sides. Due to symmetry,the flux through each of these $8$ identical triangular regions is equal. Therefore,the flux through each triangular region is $\Phi_{triangle} = \frac{1}{8} \Phi_{square} = \frac{1}{8} 	imes \frac{1}{6} \frac{q}{\varepsilon_0} = \frac{1}{48} \frac{q}{\varepsilon_0}$. The shaded region in the figure consists of $5$ such identical triangular regions. Thus,the flux through the shaded region is $\Phi_{shaded} = 5 	imes \Phi_{triangle} = 5 	imes \frac{1}{48} \frac{q}{\varepsilon_0} = \frac{5}{48} \frac{q}{\varepsilon_0}$. Given $q = 1 \ C$,the flux is $\frac{5}{48} \frac{1}{\varepsilon_0}$. Comparing this with $\frac{5}{p} 	imes \frac{1}{\varepsilon_0}$,we get $p = 48$.

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