Question

Can coulomb’s law be used to derive Gauss’s law? How?

Answer 1

Hint: The Coulomb's law gives us a relation between force between charged particles and the magnitude of charges and distance between them. The gauss law gives us a relation between flux through a closed surface and the charge enclosed in it. The flux is also the dot product of electric field and surface area vectors. Substituting electric field from coulomb’s law, we can derive the gauss law.

Complete step by step answer:
According to Coulomb's law, the force acting between two particles is directly proportional to the product of magnitude of their charge and inversely proportional to the square of the distance between them. Therefore,
$F=\dfrac{{{q}_{1}}{{q}_{2}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$
Here, $F$ is the force acting between the particles
${{q}_{1}}$ is the magnitude of one charge
${{q}_{2}}$ is the magnitude of the other charge
${{\varepsilon }_{0}}$ is the absolute permittivity
$r$ is the distance between the charges
The gauss’s law states that the total flux passing through a closed surface is the ratio of charge enclosed in that surface to the absolute permittivity.

Let us take a charged particle enclosed inside a spherical surface $S$ of radius $r$. The charge is placed at the centre of the sphere.
Consider an element $dS$ on the surface. The lines of force depicting the electric field passing through $dS$ makes an angle $\theta $ with the normal drawn on $dS$. The flux through the small area is given by $\overrightarrow{E}\cdot d\overrightarrow{S}$.
Therefore,
$d\phi =\overrightarrow{E}\cdot d\overrightarrow{S}=EdS\cos \theta $ - (1)
Here, $\phi $ is the flux
According to Coulomb’s law,
$E=\dfrac{q}{4\pi{\varepsilon }_{0} {{r}^{2}}}$
Here, $E$ s the electric field
$q$ is the magnitude of charge on the charge particle
$r$ is the distance of the charged particle from the element on which the electric field acts.
 Substituting the value of electric field in eq (1), we get,
$d\phi =\dfrac{q}{4\pi{\varepsilon }_{0} {{r}^{2}}}dS\cos \theta $
For a sphere, the solid angle is given by-
$d\omega =\dfrac{dS\cos \theta }{{{r}^{2}}}$
Substituting in the above equation, we get,
$d\phi =\dfrac{q}{4\pi{\varepsilon }_{0} }d\omega $
Integrating on both sides of the equation, we get,
$\begin{align}
  & \int{d\phi =\int{\dfrac{q}{4\pi {{\varepsilon }_{0}}}}}d\omega \\
 & \Rightarrow \phi =\dfrac{q}{4\pi {{\varepsilon }_{0}}}\int{d\omega } \\
 & \Rightarrow \phi =\dfrac{q}{4\pi {{\varepsilon }_{0}}}\times 4\pi \\
 & \therefore \phi =\dfrac{q}{{{\varepsilon }_{0}}} \\
\end{align}$
Therefore, it is proved that the flux passing through the spherical surface is equal to $\phi =\dfrac{q}{{{\varepsilon }_{0}}}$.
Thus we can say that, $\int{\overrightarrow{E}\cdot dS=\dfrac{q}{{{\varepsilon }_{0}}}}$.
Therefore, the gauss law can be proved using the coulomb’s law.

Note: The electric lines of forces depict the nature and direction of the electric field. The lines of forces are always perpendicular to the Gaussian surface. A solid angle is the measure of how large an object appears to be from a point.