JEE Main 2024 — Electrostatics Question with Solution

The electric flux ($$\Phi$$) through a closed surface, according to Gauss's law, is proportional to the enclosed charge $$Q$$, and is given by $$\Phi =\frac{Q}{{ϵ}_0}$$, where $${ϵ}_0$$ is the vacuum permittivity. This is regardless of the shape of the surface, as long as it is closed and it encloses the charge fully.

In this case, both cubes $$C_1$$ and $$C_2$$ are closed surfaces, and they are concentric, meaning they share the same center. Each cube encloses a different charge, $$2Q$$ for $$C_1$$ and $$3Q$$ + $$2Q$$ = $$5Q$$ for $$C_2$$. According to Gauss's law, the electric flux through each cube is directly proportional to the enclosed charge.

For $$C_1$$, enclosing charge $$2Q$$: $${\Phi }_{C_1}=\frac{2Q}{{ϵ}_0}$$

For $$C_2$$, enclosing charge $$5Q$$: $${\Phi }_{C_2}=\frac{5Q}{{ϵ}_0}$$

Now, we want the ratio of electric flux through $$C_1$$ to that through $$C_2$$:

$$\frac{{\Phi }_{C_1}}{{\Phi }_{C_2}}=\frac{\frac{2Q}{{ϵ}_0}}{\frac{5Q}{{ϵ}_0}}=\frac{2Q}{5Q}=\frac{2}{5}$$

Therefore, the ratio of electric flux passing through $$C_1$$ to that passing through $$C_2$$ is $$\frac{2}{5}$$, which corresponds to option C.