Question

Obtain Coulomb’s law from Gauss’s law by considering a spherical surface$\mathit{\delta }$ with centre atq.

Short answer

Coulomb's law is$E=\frac{q}{4\pi {\epsilon }_0{r}^2}\hat{R}$

Step-by-step solution

Given Information.

A Spherical surface$\delta$ with centre atq.

Definition of Coulomb's law.

Coulomb's law states that the force of attraction between the two charged bodies which is directly proportional to the product of their charges or inversely proportional to the square of the distance between the two objects.

Use the Gauss's law.

Consider the Gauss's law$\oint E\cdot nd\sigma =\frac{q_{in}}{{\epsilon }_0}.$

Take the fiction closed surface to be a sphere surrounding the point charge of radius .

$E\cdot 4\pi r^2=\frac{q_{in}}{{\epsilon }_0}$

Now, point radically outward the equation role="math" localid="1659183652852" $E\cdot =\frac{q_{in}}{4\pi r^2{\epsilon }_0}$.

Hence, Coulomb's law is$E\cdot =\frac{q_{in}}{4\pi r^2{\epsilon }_0{r}^2}\hat{r}$