Question

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 a function of r> R and r< R.
(r being the distance from the centre of the shell)

Answer 1

Electric field due to a uniformly charged thin spherical shell :
(i) When point P lies outside the spherical shell : Consider a spherical shell of radius R and centre O. Let q be the charge on 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 $\overrightarrow{E}$ be the electric field at point P. Then, the electric flux through area element $\overrightarrow{ds}$ is given by
$d\varphi =\overrightarrow{E}.\overrightarrow{ds}$
Since $\overrightarrow{ds}$ is also along normal to the surface,
$\therefore d\varphi =Eds$
$\therefore$ Total electric flux through the Gaussian surface is given by
$\varphi ={\oint }_{s}E.ds=E\oint dsNow,\oint ds=4\pi r^2\therefore \varphi =E\times 4\pi r^2..........\left(i\right)$
Since the charge enclosed by the Gaussian surface is q, according to Gauss theorem,
$\varphi =\frac{q}{{\epsilon }_0}$ ....... (ii)
From equations (i) and (ii), we get
$E\times 4\pi r^2=\frac{q}{{\epsilon }_0}$
$E=\frac{1}{4\pi {\epsilon }_0}.\frac{q}{r^2}$ (for r > R)

(ii) When point P lies inside the spherical shell : In such a case the Gaussian surface encloses no charge,


According to Gauss law,
$E\times 4\pi r^2=0$
i.e., $E=0(r<R)$
Graph showing the variation of electric field as a function of r.