QJEEM24Ph4ea3: JEE Main 2024 Physics - Class 12 JEEM24Ph4ea3 Solution: and are two hollow concentric cubes enclosing charges and…

The electric flux ( $\Phi$ ) through a closed surface, according to Gauss's law, is proportional to the enclosed charge $Q$ , and is given by $\Phi = \frac{Q}{\epsilon_0}$ , where $\epsilon_0$ 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 $2Q$ : $\Phi_{C_1} = \frac{2Q}{\epsilon_0}$

For $C_2$ , enclosing charge $5Q$ : $\Phi_{C_2} = \frac{5Q}{\epsilon_0}$

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

$\frac{\Phi_{C_1}}{\Phi_{C_2}} = \frac{\frac{2Q}{\epsilon_0}}{\frac{5Q}{\epsilon_0}} = \frac{2Q}{5Q} = \frac{2}{5}$

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.