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,
$\begin{array}{l}
\quad \overrightarrow{ E } \cdot \overrightarrow{ S }=\frac{q}{\varepsilon_0} \\
\therefore \quad E \cdot S \cos 0=\frac{q}{\varepsilon_0} \\
\therefore \quad ES =\frac{q}{\varepsilon_0} \\
\text { But } S = 4 \pi r^2 \text { (area of Gaussian surface) } \\
\text { and } q= \sigma A (\text { Where } A \text { - area of spherical shell) } \\
= \sigma\left(4 \pi R ^2\right)
\end{array}$ 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),
$\begin{array}{l}
\therefore E \left(4 \pi r^2\right)=\frac{\sigma\left(4 \pi R ^2\right)}{\varepsilon_0} \\
\therefore E =\frac{\sigma R ^2}{\varepsilon_0 r^2}
\end{array}$
►But $\sigma=\frac{q}{A}=\frac{q}{4 \pi R ^2}$,
$\therefore \quad E =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}$ (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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