Question

How do you calculate the coulombs law from the gauss theorem?

Answer 1

1. An electric field is produced by a stationary charge. Electric flux is defined as the number of electric field lines traveling through a specific location. The Gauss theorem connects the total amount of charge contained in a closed region to the electric flux that exits that region due to a specific amount of charge.

Gauss Theorem:

1. The electric flux through a closed surface is equal to $\frac{1}{{\epsilon }_0}$ times the total amount of charge present in the area, according to the Gauss theorem. The Gauss theorem's mathematical formulation is as follows: $\oint \overrightarrow{E}.\overrightarrow{dS}=\frac{Q_{net}}{{\epsilon }_0}$

Where $E=$an electric field passes through a certain area $dS$. $Q_{net}=$The total amount of charge. ${\epsilon }_0=$the permittivity of the vacuum.

Derivation of coulomb inverse square law:

1. Consider the case of a charge $+q$at the origin in a vacuum. At a distance r from the charge, we want to determine the electric field caused by that charge. As seen in the illustration below, picture an imaginary sphere with a radius of r surrounding the charge. The Gaussian sphere is this object.

1. Think about the Gaussian sphere's$dS$ element, which has a small area. The following formula can be used to determine the flux through this area element due to charge:$\oint \overrightarrow{E}.\overrightarrow{dS}=\int EdS\cos \left(0^0\right)$

Therefore, the angle between the electric field and area element is $\left(0^0\right)$.

Integrate the area element

The charge-induced electric field is constant

$$\begin{gathered} \Rightarrow \oint \overrightarrow{\mathrm{E}}.\overrightarrow{\mathrm{dS}}=\mathrm{E}\int \mathrm{dS} \\ \Rightarrow \oint \overrightarrow{\mathrm{E}}.\overrightarrow{\mathrm{dS}}=\mathrm{E}\left(4{\mathrm{\pi r}}^2\right) \end{gathered}$$

We obtain the Gauss theorem using this,

$$\begin{gathered} \Rightarrow \mathrm{E}\left(4{\mathrm{\pi r}}^2\right)=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0} \\ \Rightarrow \mathrm{E}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2} \end{gathered}$$

Hence, this is necessary for Coulomb's law discovered by applying the Gauss theorem.