JEE Main 2024 — Electrostatics Question with Solution

When considering the electric flux linked with a cube due to a charge placed at one of its surfaces, it's important to apply Gauss's law. Gauss's law states that the total electric flux through a closed surface is equal to $$\frac{q_{\text{enc}}}{{ϵ}_0}$$, where $$q_{\text{enc}}$$ is the charge enclosed by the surface and $${ϵ}_0$$ is the permittivity of free space.

In the given scenario, the charge $$q$$ is placed at the center of one of the surfaces of the cube. Conceptually, we can think of this arrangement as part of a larger situation where if we had a larger, imaginary cube that encompasses the entire setup such that the point charge is at its geometric center, the charge would then be uniformly distributing its flux through all six faces of this larger cube. However, since the actual setup involves only one cube with one face adjacent to the charge, we essentially have only one-half of the total possible geometry through which the flux from this charge can emerge - implying the flux through our actual cube is a fraction of the total flux that would emanate from the charge if it were centrally placed within a larger, encompassing cube.

Therefore, only half of the flux emanating from the charge will pass through the actual cube because the charge is placed directly on one of its surfaces, effectively distributing its influence through 180 degrees (half of the space around it) instead of the full 360 degrees. Hence, the flux linked with the cube will be half of the total flux $$\frac{q}{{ϵ}_0}$$, which is calculated using Gauss's law for a point charge.

Therefore, the correct answer is $$\frac{q}{2{ϵ}_0}$$.

Option A $$\frac{q}{2{ϵ}_0}$$ is the correct choice.