Question

Using Gauss’s law obtain the expression for the electric field due to the uniformly charged thin spherical shell of radius $R$ at a point outside the shell. Draw a graph showing the variation of electric field with $r$, for $r>R$ and $r<R$.

Answer 1

Hint A uniformly charged sphere is spherically symmetric i.e. all points around the sphere are identical. Therefore, choose a spherical Gaussian surface whose radius is equal to the distance$r$ from the center of the sphere.
Formula used: $\oint \overrightarrow{E}\cdot d\overrightarrow{A}={\displaystyle \frac{Q_{enc}}{{\epsilon }_0}}$ where $\overrightarrow{E}$ is the electric field vector, $d\overrightarrow{A}$ is the infinitesimal area vector, $Q_{enc}$ is the charge enclosed in the Gaussian surface and ${\epsilon }_0$ is the permittivity of free space, $\oint \overrightarrow{E}\cdot d\overrightarrow{A}$ is the total flux going through the Gaussian surface.

Complete step by step answer

Gauss law allows us to easily find the electric field of a charge distribution by taking advantage of a possible symmetry in its arrangement. From integral form of Gauss law, we have
$\Rightarrow \oint \overrightarrow{E}\cdot d\overrightarrow{A}={\displaystyle \frac{Q_{enc}}{{\epsilon }_0}}$
For a uniform spherical charge, the equation becomes
$\Rightarrow E\times A={\displaystyle \frac{Q_{enc}}{{\epsilon }_0}}$ because the electric fields flowing through every infinitesimal area of our Gaussian surface are equal.
Then,
$\Rightarrow E\times 4\pi r^2={\displaystyle \frac{Q_{enc}}{{\epsilon }_0}}$ (since $A=4\pi r^2$)
Rearranging for $E$, we get
$\Rightarrow E={\displaystyle \frac{Q_{enc}}{4\pi {\epsilon }_0{r}^2}}$
$\therefore E\propto {\displaystyle \frac{1}{r^2}}$
The electric field for a uniformly charged thin spherical shell at radius $r<R$ is zero (since there would be no charge enclosed).
Hence, the graph of electric field with distance is as shown below.

Note
For $r<R$ we said that this is due to the fact that there would be no charged enclosed. For further understanding consider the constant form of Gauss’s law
$\Rightarrow E\times A={\displaystyle \frac{Q_{enc}}{{\epsilon }_0}}$
For $r<R$, we pick a Gaussian surface that is inside the spherical shell. Since the charges are distributed around the shell, the charge enclosed by the Gaussian surface is zero, i.e. $\Rightarrow Q_{enc}=0$. Hence, from
$\Rightarrow E\times A={\displaystyle \frac{Q_{enc}}{{\epsilon }_0}}$
$\Rightarrow E=0$
Also, $E$ is maximum at $r=R$ because $Q_{enc}=\sigma 4\pi R^2$ where $\sigma$ is the surface charge density. Thus, from $E={\displaystyle \frac{Q_{enc}}{4\pi {\epsilon }_0{r}^2}}$ we replace $Q_{enc}$ as $\sigma 4\pi R^2$. Then, we have
$\Rightarrow E={\displaystyle \frac{\sigma 4\pi R^2}{4\pi {\epsilon }_0{r}^2}}$.
For $r=R$,
$\Rightarrow E={\displaystyle \frac{\sigma }{{\epsilon }_0}}$