electric-charges-and-fields Question 52
Question: Q. 12. Using Gauss’s law deduce the expression for the electric field due to a uniformly charged spherical conducting shell of radius $R$ at a point (i) outside, and (ii) inside the shell.
Plot a graph showing variation of electric field as function of $r>R$ and $r<R$ ( $r$ being the distance from the centre of the shell).
A [O.D. & O.D. Comptt. I, II, III, 2013;
Delhi I, II, III 2009]
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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.
