Question

A point charge 𝑞 q is placed at a distance 𝑑 d from the center of a square of side 𝑎 a. Find the flux formula through the square.

Answer 1

$$\Phi = \frac{q}{\pi\varepsilon_0}\arctan\left(\frac{a^2}{2d\sqrt{4d^2+a^2}}\right)$$

Answer 2

The electric flux through any surface from a point charge is given by:

$$\Phi = \frac{q}{\varepsilon_0}\cdot\frac{\Omega}{4\pi}$$

where $\Omega$ is the solid angle subtended by the surface at the location of the charge.

For a square of side $a$ with its center at the origin and the charge at $(0,0,d)$, the solid angle is known to be:

$$\Omega = 4\arctan\left(\frac{a^2}{2d\sqrt{4d^2+a^2}}\right)$$

Thus, the flux through the square is:

$$\Phi = \frac{q}{\varepsilon_0}\cdot\frac{1}{4\pi}\cdot4\arctan\left(\frac{a^2}{2d\sqrt{4d^2+a^2}}\right)$$

which simplifies to:

$$\Phi = \frac{q}{\pi\varepsilon_0}\arctan\left(\frac{a^2}{2d\sqrt{4d^2+a^2}}\right)$$