Obtain the expression of electric field by charged spherical shell on a point outside it.
Obtain the expression of electric field by charged spherical shell on a point outside it.
Consider a uniformly charged sphere of radius $\mathrm{R}$ having the charge density $\rho$.
Gauss's law is $\int \overrightarrow{\mathrm{E}} \cdot \overrightarrow{d a}=\frac{q}{\varepsilon_{0}}$
Here we imagine one spherical Gaussian surface of radius $r$ which is greater than $R .(r>R)$ Total flux passing through the surface $=4 \pi \mathrm{r}^{2} \mathrm{E}(\mathrm{r})$
Total charge enclosed $q=\frac{4 \pi}{3} \mathrm{R}^{3} \frac{\rho}{\varepsilon_{0}}$
$\therefore 4 \pi r^{2} \mathrm{E}(\mathrm{r})=\frac{4}{3} \pi \mathrm{R}^{3} \frac{\rho}{\varepsilon_{0}}$
$\therefore \mathrm{E}(\mathrm{r})=\frac{\mathrm{R}^{3} \rho}{3 \rho^{2} \varepsilon_{0}}(r>\mathrm{R})$
$\text { This equation gives the valu }$
This equation gives the value of the electric field for the points lying in the region outside the sphere at a distance $r$ from the centre of the sphere $(r>\mathrm{R})$.
Total charge of the sphere $Q=\frac{4}{3} \pi r^{3} \rho$
$\therefore \rho=\frac{3 Q}{4 \pi R^{3}}$
Using this in above equation,
$\therefore \mathrm{E}(\mathrm{r})=\frac{\mathrm{R}^{3}}{3 \mathrm{r}^{2} \epsilon_{0}} \cdot \frac{3 \mathrm{Q}}{4 \pi \mathrm{R}^{3}}$
$\therefore \mathrm{E}(\mathrm{r})=\frac{\mathrm{Q}}{4 \pi \epsilon_{0}} \cdot \frac{1}{\mathrm{r}^{2}}$
Thus, for a point outside the sphere the entire charge of the sphere can be treated as concentrated at its centre.
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