Question

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: 

 

• A. $\frac{q}{4\epsilon_0}$
• B. $\frac{q}{8\epsilon_0}$
• C. $\frac{q}{24\epsilon_0}$
• D. (Option seems to be missing from OCR)

Hint:

For a charge placed at a corner of a cube, remember this hierarchy: Total flux through 8 enclosing cubes is $q/\epsilon_0$. Flux through one cube is $q/(8\epsilon_0)$. Flux through one of the three non-adjacent faces is $q/(24\epsilon_0)$. Flux through the three faces adjacent to the charge is 0.

Answer 1

The Correct Option is C

To find the flux through the cube, we imagine the charge 'q' at the corner to be at the center of a larger cube of side 2L, which is composed of 8 such smaller cubes.
By Gauss's law, the total flux through this large imaginary cube is $\Phi_{total} = \frac{q}{\epsilon_0}$.
Due to symmetry, this total flux is shared equally among the 8 smaller cubes.
The flux through a single cube due to the charge at its corner is $\Phi_{cube} = \frac{\Phi_{total}}{8} = \frac{q}{8\epsilon_0}$.
Now, consider the three faces of the cube that meet at the corner where the charge 'q' is placed. The electric field lines are parallel to the surface of these three faces.
Therefore, the electric flux through these three faces is zero.
The entire flux of $\frac{q}{8\epsilon_0}$ must pass through the other three faces (the shaded faces) which do not touch the charge.
Due to the symmetry of the situation with respect to these three faces, the flux is distributed equally among them.
The shaded area in the figure appears to be a single face opposite to the corner. Assuming the question asks for the flux through one of these faces:
$\Phi_{shaded\_face} = \frac{1}{3} \Phi_{cube} = \frac{1}{3} \left(\frac{q}{8\epsilon_0}\right) = \frac{q}{24\epsilon_0}$.