Question

A charge of 2 micro Coulomb is kept inside a cube of side length 10cm.Find the flux through its surface and also flux through one of its face *​

Answer 1

Answer:

flux through the surface is q/e0 and flux through one of its face is q/6e0.

Explanation:

Answer 2

Answer:

The total flux the surface is $22.58\times 10^{4}Nm^{2}C^{-1}$ and flux through each face is $3.76\times 10^{4}Nm^{2}C^{-1}$.

Explanation:  

Given a cube of side of length, $l=10cm$

The charge inside the cube, $q=2\mu C=2\times 10^{-6} C$

According to Gauss's law, the total electric flux out of a closed surface is equal to the charge enclosed divided by the electric permittivity.

Written as,

$\phi_{total}=\dfrac{q}{\varepsilon_0}$

where $\varepsilon_0=8.854 \times 10^{-12} ~C^2N^{-1} m^{-2}$ is the electric permittivity of free space.

Therefore, the total flux through the surface is

$\phi_{total}=\dfrac{2\times 10^{-6}}{8.854 \times 10^{-12}}$

         $=\dfrac{1\times 10^{6}}{4.427 }=0.2258\times 10^{6}$

         $=22.58\times 10^{4}Nm^{2}C^{-1}$

The cube have 6 faces, so the flux through each face is given by

$\phi=\dfrac{\phi_{total}}{6} =\dfrac{q}{6\varepsilon_0}$

Substituting the values,

$\phi=\dfrac{\phi_{total}}{6}=\dfrac{22.58\times 10^{4}}{6}$

                $=3.76\times 10^{4}Nm^{2}C^{-1}$

Therefore, the total flux the surface is $22.58\times 10^{4}Nm^{2}C^{-1}$ and flux through each face is $3.76\times 10^{4}Nm^{2}C^{-1}$.

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