Previous Year NEET Question- Applications Of Gauss’S Law

The answer is (B) $\frac{Q}{4\pi\varepsilon_0}$.

The electric flux through a surface is given by:

$\phi = \frac{Q}{4\pi\varepsilon_0}$

In this case, the charge is placed at the centre of the cube, so the electric field is uniform and the flux through each face is equal. Therefore, the total flux through the cube is:

$\phi = \frac{Q}{\varepsilon_0} \times \frac{1}{6} = \frac{Q}{6\varepsilon_0}$

2. The answer is (C) $\frac{q}{8\varepsilon_0}$.

The electric flux through a surface is given by:

$\phi = \frac{Q}{4\pi\varepsilon_0}$

In this case, the charge is placed at the centre of the cube, so the electric field is uniform and the flux through each face is equal. Therefore, the total flux through the cube is:

$\phi = \frac{q}{\varepsilon_0} \times \frac{1}{6} = \frac{q}{6\varepsilon_0}$