A charge $$q$$ is placed at one corner of a cube as shown in figure. The flux of electrostatic field $$\vec{E}$$ through the shaded area is:

Corrected Solution & Explanation

1. Total Flux through the Closed Symmetric Structure

By applying Gauss's Law, a charge $$q$$ enclosed fully within a symmetric boundary emits a total flux of $$\Phi = \frac{q}{\varepsilon_0}$$. To completely enclose a charge residing precisely on a corner vertex, we must place $$8$$ identical cubes together. Thus, the total flux passing through our single designated cube is:

$$\Phi_{\text{cube}} = \frac{q}{8\varepsilon_0}$$

2. Flux Distribution Across the Faces

A single cube consists of $$6$$ individual faces:

• Three Adjoining Faces: The 3 faces that intersect to form the corner vertex holding the charge lie completely parallel to the outward electric field lines. Since the field vectors run along the surfaces, the angle with the face area vectors is $$\theta = 90^\circ$$, meaning the net flux through each of these faces is exactly zero ($$\Phi = 0$$).
• Three Opposite Faces: The entire flux entering the cube must exit symmetrically through the 3 remaining outer faces opposite to the charge. Therefore, the total flux passing through one entire opposite square face is:

  $$\Phi_{\text{face}} = \frac{1}{3} \times \Phi_{\text{cube}} = \frac{1}{3} \times \frac{q}{8\varepsilon_0} = \frac{q}{24\varepsilon_0}$$

3. Analyzing the Shaded Geometric Region

Let's look closely at how the shaded area is defined in this classic configuration:

• The shaded area is a triangle formed by connecting the diagonals of the three opposite faces.
• It does not lie flat on a single outer square face; rather, it forms an inclined planar surface that bounds the inner corner.
• By projecting the flux lines passing through the three open outer faces, this specific triangular cross-section intercepts the entire regular flux flow directed through those three outer surfaces combined.

Because any field line passing through the inner shaded triangle must inevitably pass through the three outer symmetric faces to exit the cube, the total flux through the shaded area is exactly equal to the shared contribution of one full face path:

$$\Phi_{\text{shaded}} = \Phi_{\text{face}} = \frac{q}{24\varepsilon_0}$$

Correct Option: B ($$\frac{q}{24\varepsilon_0}$$)