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Question

A charge q is placed at the corner of a cube of side a. The electric flux passing through the cube is :
  1. qaε0
  2. qε0a2
  3. q4πε0a2
  4. q8ε0

A
q4πε0a2
B
q8ε0
C
qaε0
D
qε0a2
Solution
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Step 1: Choosing Gaussian Surface [Refer Fig. 1 & 2]
Consider the cube and the charge placed at its corner as shown in figure 1.

For application of Gauss Law, we have to find a gaussian surface inside which this charge(q) 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 2.
Whereas the given smaller cube is one of the part of the larger cube, i.e. cube no. 7.

Combining these 8 cubes, we get a Larger cube of side 2a with charge(q) placed at its center. 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 smallar cube(no. 7) of side a =Total Flux8=q8ϵ0

Hence, Option A is correct.

Additional Info: 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.

2115307_459378_ans_782023be18154ae48f6194b2f7babacb.jpg

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