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Question

The total flux through the faces of the cube with side of length a if a charge q is placed at corner A of the cube is
941198_7e978e303f4d49728d1597b5bf7127d9.png

A
q8ε0
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B
q4ε0
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C
q2ε0
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D
qε0
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Solution

The correct option is A q8ε0

Step 1: Choosing Gaussian Surface [Refer Figure]
For application of Gauss Law, we have to find a gaussian surface inside which this charge(q) at point A lies symmetrically.

Therefore, Imagine 3D picture with the charge q at the origin, and 8 cubes of side a placed such that each having one vertex at the origin, touching the charge q as shown in the figure.

Combining these 8 cubes, we get a Larger cube of side 2a with charge(q) placed at its center A. Assuming it to be Gaussian Surface.

Note: The given cube of side a can not be taken as gaussian surface, as the charge(q) lies on its boundary, therefore Gauss law will not be applicable.

Step 2: Applying Gauss Law
Now, By Gauss Law for the chosen gaussian surface:
Total flux through this larger cube =Charge Enclosedϵ0=qϵ0

Step 3: Finding Flux through given Cube using the symmetry
As the position of the charge is symmetric with respect to all 8 cubes.
Hence, the total flux through the larger cube will be divided equally among all the 8 cubes.
Flux through Given cube of side a =Total Flux8=q8ϵ0

Hence, Option A is correct.

Note: Further, this q/(8ϵ0) flux will be shared equally among the three opposite faces of the cube due to symmetry, Hence flux through each of these 3 faces =13×q8ϵ0=q24ϵ0
And the flux through the 3 adjacent faces of the cube will be zero, as electric field will be parallel to these three faces.

2112095_941198_ans_47ef8ed891704162b728aca53d7372f8.png

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