Question

A point charge of $2.0\mu C$ is at the centre of a cubic Gaussian surface $9.0cm$ on edge. What is the net electric flux through the surface?

Answer 1

Hint: Since we are given a Gaussian surface here, so we must use Gauss’ law to calculate the net electric flux through the surface.

Formula used
$\begin{gathered}
  \oint\limits_S {\overrightarrow E } \cdot d\overrightarrow S = \dfrac{q}{{{\varepsilon _0}}} \\
   \Rightarrow \phi = \dfrac{q}{{{\varepsilon _0}}} \\
\end{gathered} $
Where $\overrightarrow E $ is the electric field due to a point charge, $q$is the charge enclosed by the surface $S,$$\phi $ is the electric flux flowing through the surface and ${\varepsilon _0}$ is the permittivity of free space.

Complete step by step answer
Gauss’ Law states that the total electric flux over any closed surface is equal to $\dfrac{1}{{{\varepsilon _0}}}$times the total charge enclosed by the surface, where ${\varepsilon _0}$ is the permittivity of free space.
In this question, we are told that a charge of $2.0\mu C$ is enclosed by a cubic Gaussian surface. According to the law, no matter what the shape of the surface is, but if a charge is enclosed by any closed surface, then the flux flowing through the surface must be $\dfrac{1}{{{\varepsilon _0}}}$ times the charged enclosed.
Therefore, according to gauss’ law,
$\begin{gathered}
  \oint\limits_S {\overrightarrow E } \cdot d\overrightarrow S = \dfrac{q}{{{\varepsilon _0}}} \\
   \Rightarrow \phi = \dfrac{q}{{{\varepsilon _0}}} \\
\end{gathered} $
Where $\phi $ is the total electric flux flowing through the Gaussian surface.
We know ${\varepsilon _0} = 8.85 \times {10^{ - 12}}{N^{ - 1}}{m^{ - 2}}{C^2}$
Substituting this value, we get,
$\phi = \dfrac{{2 \times {{10}^{ - 6}}}}{{8.85 \times {{10}^{ - 12}}}}N{m^2}{C^{ - 1}}$
$ \Rightarrow \phi = 2.26 \times {10^5}N{m^2}{C^{ - 1}}$

Therefore, the net electric flux flowing through the surface is $2.26 \times {10^5}N{m^2}{C^{ - 1}}$

Note: The equation used above is the integral form of the Gauss’ law.
If we use divergence theorem on both the sides of the equation, we will get its differential form given by the expression, $\overrightarrow \nabla \cdot \overrightarrow E = \dfrac{\rho }{{{\varepsilon _0}}}$where $\rho $ is defined as the charge per unit volume. This is one of the four Maxwell’s laws of electromagnetic induction.