Question

A charge $q$ is located at the centre of a cube. The electric flux through any face is

• A. $ \frac{ \pi q}{ 6 ( 4 \pi \varepsilon_0 ) } $
• B. $ \frac{ q}{ 6 ( 4 \pi \varepsilon_0 ) } $
• C. $ \frac{ 2 \pi q}{ 6 ( 4 \pi \varepsilon_0 ) } $
• D. $ \frac{ 4 \pi q}{ 6 ( 4 \pi \varepsilon_0 ) } $

Answer 1

Answer: (D)

Gauss's law states that, "the net electric flux through any closed surface is equal to the net charge inside the closed surface divided by $\varepsilon_{0} "$
From Gauss's law
$ \phi = \frac{ q}{ \varepsilon_0 } $
This is the net flux coming out of the cube.
Since, a cube has $6$ sides so electric flux through any face is
$ \phi ' = \frac{ \phi }{ 6 } = \frac{ q}{ 6 \varepsilon_0 } =\frac{ 4 \pi q}{ 6 ( 4 \pi \varepsilon_0 ) } $