Question

A point charge $q$ is placed at a distance $\frac{a}{2}$ perpendicular to the above the center of a square of side $a$. The electric flux through the square is:-

• A. $\frac{q}{{ϵ}_0}$
• B. $\frac{q}{\pi {ϵ}_0}$
• C. $\frac{q}{4{ϵ}_0}$
• D. $\frac{q}{6{ϵ}_0}$

Answer 1

The correct option is D $\frac{q}{6{ϵ}_0}$

A point charge is placed by a distance $=a/2$

side $=a$

The electric flux through the square $=?$

The answer is also by the use of Gauss's law. If the charge is at a distance $a/2$ from the centre of a square of side length $a$ then the charge is the centre of a cube of side length $a$.

That means the cube has $6$ equal area faces.

In $V_0$ $K_e$ gauss law which stat is that total flux out of a surface is charge / ${ϵ}_0$

Since, we can taking about $1/6$th of the total surface area hence the net flux should also go down six times.

Hence, $g$ the answer will be $=\frac{q}{6{ϵ}_0}$