Question

*A point charge q is placed at geometrical centre of one of the face of a cube. The total flux through the cubical surface due to the charge is:*

1️⃣ q/ϵ0
2️⃣ q/2ϵ0
3️⃣ 2q/ϵ0
4️⃣ 0​

Answer 1

Answer:

q/2e0

Explanation:

• since guass law is stated only for symmetry shapes with charge at centre

• hence the cube is doubled to make the point in the centre

• since 2 cubes involved charge is halved

hence

• q/2 * e0

hence q/2e0

Answer 2

Given:

Point charge q has been placed at the geometric centre of one of the face of the cube.

To find:

Total flux through the cubical surface ?

Solution:

Now, Let's assume another cube on the side of this cube such that charge appears to be at geometric centre of the two cubes.

• Please see diagram.

Let flux through one cube be $\phi$.

$2 \phi = \dfrac{q}{ \epsilon_{0} }$

$\implies \phi = \dfrac{q}{ 2 \epsilon_{0} }$

So, final answer is:

$\boxed{ \bf \phi = \dfrac{q}{ 2 \epsilon_{0} } }$