Question

A charge $q$ is placed at the centre of the cubical box of side a with top open. The flux of the electric field through the surface of the cubical box is:
A) $\text{Zero }\\!\\!~\\!\\!\text{ }\\!\\!~\\!\\!\text{ }$
B) $\dfrac{q}{{{\varepsilon }_{0}}}$
C) $\dfrac{q}{6{{\varepsilon }_{0}}}$
D) $\dfrac{5q}{6{{\varepsilon }_{0}}}$

Answer 1

Net electric flux through a cube is the sum of electric flux through its six faces. But here one side is open; hence the net electric flux will sum electric flux through five faces. We can use the formula which relates the total electric flux and charge enclosed to find the flux through the surfaces of the cubical box.
Formula used:
${{\phi }_{c}}=\dfrac{q}{{{\varepsilon }_{0}}}$

Complete answer:
We have,
$~\phi =\dfrac{\text{charge inside}}{{{\varepsilon }_{0}}}$
Where,
$\phi -Electric\text{ }flux$
${{\text{ }\\!\\!\varepsilon\\!\\!\text{ }}_{\text{0}}}\text{- Permittivity of free space}$
Charge $q$ is placed at the centre of the cube. Then,
$\text{Electricflux through the cube, }{{\phi }_{\text{c}}}\text{=}\dfrac{\text{q}}{{{\text{ }\\!\\!\varepsilon\\!\\!\text{ }}_{\text{0}}}}$
$Flux\text{ }through\text{ }1\text{ }surface,{{\phi }_{1}}=\dfrac{q}{6{{\varepsilon }_{0}}}$
Since one side is open, flux through the open surface is zero.
Then,
$Total\text{ }electric\text{ }flux\text{ }through\text{ }5\text{ }surface,{{\phi }_{5}}=\dfrac{5q}{6{{\varepsilon }_{0}}}$

So, the correct answer is “Option D”.

Additional Information:
The net flux of an electric field in a closed surface is directly proportional to the charge enclosed. This is known as Gauss’s law. Electric flux is defined as the electric field passing through a given area multiplied by the area of a surface plane which is perpendicular to the field. Or in other words, Gauss’s law can be given as; the net flux of an electric field through a given surface, divided by the enclosed charge is always a constant.

Note:
Even though electric flux is not affected by charges that are not within the closed surface, the net electric field, E, can be affected by charges that lie outside the closed surface. But the electric flux through any closed surface depends only on the amount of charge enclosed. It does not depend on the shape and size of the surface.