Question

State Gauss law in electrostatics. Using the law derive an expression for electric field due to a uniformly charged thin spherical shell at a point outside the shell.

Answer 1

$\text{Gauss Law:}$

Gauss law states that electric flux through any closed surface is equal to the net charge enclosed inside the surface divided by the permittivity of vacuum.

$\text{Electric field due to a uniformly charged thin shell:}$

In the diagram, O is the centre of the spherical shell of radius R, and charge enclosed by the spherical shell is $q$.

Let a point A is outside the spherical shell, where we have to calculate the electric field.The distance between the centre of spherical shell and the point A.

$\varphi =E.Scos\theta$

So, $\varphi =E.4\pi r^2--------------\left(1\right)$

As per gauss law,

$\varphi =\frac{q}{{ϵ}_o}---------\left(2\right)$

From equation (1) and (2)

$E.4\pi r^2=\frac{q}{{ϵ}_o}$

Now,

$E=\frac{q}{4\pi {ϵ}_o{r}^2}$

This is the electric field at the distance r, outside from the shell.