A square loop of sides $a = 1\ \text{m}$ is held normally in front of a point charge $q = 1\ \text{C}$, the charge lying on the axis of the loop at a distance $a/2$ from its centre. The flux of the electric field through the shaded region is $\dfrac{5}{p}\times\dfrac{1}{\varepsilon_0}\ \dfrac{\text{N m}^2}{\text{C}}$, where the value of $p$ is ______.
Numerical value type. Enter your answer.
Answer: 48
A square of side $a$ with the charge $a/2$ from its centre on the axis is exactly one face of a cube of side $a$ centred on the charge. By symmetry, the flux through the whole square is
$$\phi_{square} = \frac{1}{6}\cdot\frac{q}{\varepsilon_0}$$
The square's four quadrants are identical as seen from the charge, so each carries $\dfrac{1}{24}\dfrac{q}{\varepsilon_0}$. The shaded triangle is half of a quadrant (cut along the diagonal to the centre), which is symmetric too: $\dfrac{1}{48}\dfrac{q}{\varepsilon_0}$.
Shaded region = two quadrants + one half-quadrant:
$$\phi = \left(\frac{2}{24} + \frac{1}{48}\right)\frac{q}{\varepsilon_0} = \frac{5}{48}\cdot\frac{1}{\varepsilon_0}$$
$p = 48$
Solution by Sreeraj P, M.Sc Physics