Question

State Gauss theorem and use it to derive the Coulomb’s inverse square law.

Answer 1

Hint:To solve the given question we must know how to find the electric flux through a surface. Gauss theorem relates the flux theorem through a closed surface and the total charge enclosed in it.

Formula used:
$\phi =\dfrac{{{q}_{enclosed}}}{{{\varepsilon }_{0}}}$
$\phi =\int{\overrightarrow{E}.\overrightarrow{dA}}$

Complete step by step answer:
Gauss theorem states that the net electric flux through a closed surface is equal to the total or net charge enclosed by the closed surface divided by the permittivity of the medium. If the electric field is present in vacuum then the mathematical equation for the Gauss theorem is $\phi =\dfrac{{{q}_{enclosed}}}{{{\varepsilon }_{0}}}$ …. (i).
Here, $\phi $ is the net flux through the closed surface, ${{q}_{enclosed}}$ is the net charge inside the closed surface and ${{\varepsilon }_{0}}$ is the permittivity of vacuum.

Let us now use the above theorem and derive the Coulomb’s inverse square law. The Coulomb’s inverse square law gives the expression for the electric field (E) produced by a charge q at a distance r from it.
i.e. $E=\dfrac{q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$.

To prove it, consider a point charge q. Draw a sphere (closed surface) of radius r with the location of q as its centre. From symmetry we can say that the electric field will be uniform at all the points on the sphere.

The direction of the electric field at a point due to a positive point charge is always pointing away from the charge along the line joining the charge and the point . Therefore, in this case the electric field lines will be coming out of the sphere radially (parallel to the area vector of the sphere). The electric flux through a surface is given as,
$\phi =\int{\overrightarrow{E}.\overrightarrow{dA}}$. (dA is the small area of the surface of the sphere).
Since the electric field and the area vector are parallel, $\overrightarrow{E}.\overrightarrow{dA}=EdA$.
$\Rightarrow \phi =\int{EdA}$.
Here, E is constant.

$\phi =E\int{dA}=EA$ …. (ii).
The surface of a sphere is given as $A=4\pi {{r}^{2}}$.
Substitute the value of A in (ii).
$ \phi =E\int{dA}\\
\Rightarrow \phi =E\left( 4\pi {{r}^{2}} \right)$ …. (iii)
From (iii) and (i) we get that
$\Rightarrow \dfrac{{{q}_{enclosed}}}{{{\varepsilon }_{0}}}=E\left( 4\pi {{r}^{2}} \right)$
In this case, the net charge within the sphere is q.
Therefore, ${{q}_{enclosed}}=q$.
$\Rightarrow \dfrac{q}{{{\varepsilon }_{0}}}=E\left( 4\pi {{r}^{2}} \right)$
$\therefore E=\dfrac{q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$.

Hence, we derived Coulomb's square law using the Gauss law.

Note: Note that the equation $\phi =\dfrac{{{q}_{enclosed}}}{{{\varepsilon }_{0}}}$ is true only when the medium is vacuum because different mediums have different values of permittivity. Also, the charges that are located outside the closed surface are not considered in the equation. Meaning, the flux due to the charges outside the closed surface is zero.