Question

An electric charge q is placed at the centre of a cube of side $\alpha$. The electric flux on one of its faces will be

• A. $\frac{q}{6{ϵ}_0}$
• B. $\frac{q}{{ϵ}_0{a}^2}$
• C. $\frac{q}{4\pi {ϵ}_0{a}^2}$
• D. $\frac{q}{{ϵ}_0}$

Answer 1

Answer: (A)

Step 1: Given that:

The charge placed at the centre of the cube= $q$

side of the cube= $a$

Step 2: Formula used:

1. According to Gauss theorem, " the electric flux linked with a closed surface due a charged enclosed by the surface is equal to $\frac{1}{{\epsilon }_0}$ times the charge enclosed by the surface.
2. That is $\text{Φ}=\frac{q_{enclosed}}{{\epsilon }_0}$ , where $\text{Φ}$ is the electric flux.​

Step 3: Calculation of the electric flux linked with one face of the cube:

The electric flux linked with the whole surface of the cube which encloses the charge $q$ is given by

${\text{Φ}}_t=\frac{q}{{\epsilon }_0}$

As there are $6$ faces in a cube and each havce same surface area.

The electric flux linked with each face of the cube will be

${\text{Φ}}_f=\frac{1}{6}\times {\text{Φ}}_t$

${\text{Φ}}_f=\frac{1}{6}\times \frac{q}{{\epsilon }_0}$

${\Phi }_f=\frac{q}{6{\epsilon }_0}$

Thus,

The electric flux linked with one of the surfaces of the cube will be $\frac{q}{6{\epsilon }_0}$ .

Hence,

Option C) $\frac{q}{6{\epsilon }_0}$ is the correct option.