Question

A point charge $+ 20\mu C$ is at a distance $6\,cm$ directly above the centre of a square of the side $12\,cm$ as shown in the figure. The magnitude of electric flux through the square is

A) $2.5 \times {10^6}\,N{m^2}{C^{ - 1}}$
B) $3.8 \times {10^6}\,N{m^2}{C^{ - 1}}$
C) $4.2 \times {10^6}\,N{m^2}{C^{ - 1}}$
D)$2.9 \times {10^6}\,N{m^2}{C^{ - 1}}$

Answer 1

In this solution, we will use the relation of Gauss’s law to determine the flux through the surface given to us. The surface gives us one of the sides of a cube through which the entire flux will pass.
Formula used: In this solution, we will use the following formula:
Flux due to a charge $q$: $\phi = \dfrac{q}{{{\varepsilon _0}}}$

Complete step by step answer:
In the configuration given to us, a charge $q$ is placed above a surface given to us. The flux given from this charge will be determined from Gauss’s law as:
$\phi = \dfrac{q}{{{\varepsilon _0}}}$
Now, this flux will be emitted in all directions equally. To find the flux through the surface given to us, we can think of the charge being surrounded by a cube of side 12 centimeters. Then the flux through the cube will be
$\phi = \dfrac{q}{{{\varepsilon _0}}}$
Since the flux is equally distributed, the flux through the one surface will be $1/6th$ of the total flux as the cube has six sides.
This implies that the flux through one surface will be
$\phi = \dfrac{q}{{6{\varepsilon _0}}}$
Substituting the value of charge $q = 20\,\mu C = 20 \times {10^{ - 6}}C$ and ${\varepsilon _0} = 8.85 \times {10^{ - 12}}$, we get
$\phi = \dfrac{{20 \times {{10}^{ - 6}}}}{{6\left( {8.85 \times {{10}^{ - 12}}} \right)}}$
This can be simplified to:

$\phi = 3.8 \times {10^5}\,N{m^2}{C^{ - 1}}$ which corresponds to option (B).

Note: To take the flux surface as a cube, the charge must lie at the centre of the cube. In this case, the charge is 6 centimeters away from one of the surfaces of the cube which is where the centre of the cube lies. Gauss’s law enables us to use this symmetry to directly calculate the flux through the surface as $1/6th$ of the total flux.