Question

A charged particle q is placed at the centre O of cube of length L (A B C D E F G H). Another same charge q is placed at a distance L from O. Then the electric flux through ABCD is

• A. $q/4\pi \in_0 L$
• B. zero
• C. $q/2\pi \in_0 L$
• D. $q/3\pi \in_0 L$

Answer 1

The Correct Option is B

The flux for both the charges exactly cancels the effect of each other.

Answer 2

Given:
There's a cube with sides of length L.
A charge q is at the very center O of the cube.
Another identical charge q is placed at a distance L away from O.

1. Electric Field Due to Charge at O:
The charge q at the center O creates an electric field that spreads out equally in all directions.
This electric field is the same at every point on the face ABCD because of the cube's symmetry.

2. Electric Field Due to Charge at Distance L:
The charge q placed L units away from O also creates an electric field.
However, the effect of this charge on the face ABCD might be negligible due to how the cube is set up.

3. Total Electric Field at Face ABCD:
Because of the cube's symmetry and the charge q at O being at the center, the electric field through face ABCD is likely uniform and points directly outward or inward, depending on the sign of the charge.

4. Electric Flux Through Face ABCD:
The electric flux \(\Phi_E\) through the face, ABCD is calculated by taking the electric field strength E and multiplying it by the area A of the face:
\(\Phi_E = E \cdot A\)
However, due to the symmetric placement of charges and the nature of the problem described, the flux through face ABCD is zero. This is because any flux outward from O is balanced by flux inward from the charge at distance L, resulting in no net flux through the face.

5. Conclusion:
Therefore, the electric flux through the face ABCD of the cube is 0.

So, the correct option is (B): Zero