Solution:

Ans. Electric field due to a uniformly charged thin spherical shell :

(i) When point $P$ lies outside the spherical shell : Suppose that we have to calculate electric field at the point $P$ at a distance $r(r>R)$ from its centre. Draw the Gaussian surface through point $P$ so as to enclose the charged spherical shell. The Gaussian surface is a spherical shell of radius $r$ and centre $O$. Let $\vec{E}$ be the electric field at point $P$, then the electric flux through area element $\overrightarrow{d S}$ is given by,

$$ \Delta \phi=\vec{E} \cdot \overrightarrow{\Delta S} $$

Since $\overrightarrow{\Delta S}$ is also along normal to the surface,

$$ \Delta_{\phi}=E . d S $$

$\therefore$ Total electric flux through the Gaussian surface is given by.

Now,

$\oint d S=4 \pi r^{2}$

$\therefore \quad \phi=E \times 4 \pi r^{2}$

Since the charge enclosed by the Gaussian surface is $q$ according to the Gauss’s law,

$$ \begin{equation*} \phi=\frac{q}{\varepsilon_{0}} \tag{ii} \end{equation*} $$

From equations (i) and (ii), we obtain

$$ \begin{aligned} E \times 4 \pi r^{2} & =\frac{q}{\varepsilon_{0}} \ E & =\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r^{2}} \quad(\text { for } r>R) \mathbf{1} \end{aligned} $$

(ii) When point $P$ lies inside the spherical shell : In such a case, the Gaussian surface encloses no charge. According to the Gauss’s law,

$$ \begin{aligned} E \times 4 \pi r^{2} & =0 \ \text { i.e., } \quad E & =0 \quad(r<\mathrm{R}) \end{aligned} $$

A graph showing the variation of electric field as a function of $r$ is shown in figure.