Question
Get the electric field due to a uniformly charged thin spherical shell.
✓
Answer
$►$ Let $\sigma$ be the uniform surface charge density of a thin spherical shell of radius $R ($Fig$.).$
$(i)$ Field outside the shell :
$►$ Consider a point $P$ outside the shell with radius vector $r$ Fig. $(a)$.
$►$ To calculate $E$ at $P ,$ we take the Gaussian surface to be a sphere of radius $r$ and with centre $O$, passing through $P$ .
All points on this sphere are equivalent relative to the given charged configuration.
$►$ According to Gauss's law,
$\overrightarrow{ E } \cdot \overrightarrow{ S }=\frac{q}{\varepsilon_0}$
$\therefore E \cdot S \cos 0=\frac{q}{\varepsilon_0}$
$\therefore ES =\frac{q}{\varepsilon_0}$
But $ S = 4 \pi r^2 \ ($area of Gaussian surface$)$
$\text { and } q= \sigma A \ ($ Where $ A - $ area of spherical shell$)$
$= \sigma\left(4 \pi R ^2\right)$
But $S =4 \pi r^2 \ ($area of Gaussian surface$)$
and $q=\sigma A \ ($Where $A -$ area of spherical shell$)$
$($Total charge on spherical shell.$)$
$►$ From equation $(1),$
$\therefore E \left(4 \pi r^2\right)=\frac{\sigma\left(4 \pi R ^2\right)}{\varepsilon_0}$
$\therefore E =\frac{\sigma R ^2}{\varepsilon_0 r^2}$
$►$ But $\sigma=\frac{q}{A}=\frac{q}{4 \pi R ^2}$,
$\therefore E =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}\ ($From Eq. $(2))$
$►$ The electric field is directed outward if $q>0$ and inward if $q<0$.
This, however, is exactly the field produced by a charge $q$ placed at the centre $O$ .
Thus for points outside the shell, the field due to a uniformaly charged shell is as if the entire charge of the shell is concentrated at its centre.
$(ii)$ Field inside the shell :
$►$ In Fig. $(b),$ the point $P$ is inside the shell.
The Gaussian surface is again a sphere through $P$ centred at $O$ .
$►$ Gaussian surface encloses no charge. From Gauss's law then gives the field due to a uniformly charged thin shell is zero at all points inside the shell.
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