Obtain the expression for the electric field due to a uniformly charged spherical shell at a point outside it.

(N/A) Consider a uniformly charged spherical shell of radius $R$ with a total charge $Q$.
According to Gauss's law,the electric flux through a closed surface is given by $\oint \vec{E} \cdot d\vec{a} = \frac{q_{enclosed}}{\varepsilon_0}$.
For a point outside the shell at a distance $r$ from the center $(r > R)$,we choose a spherical Gaussian surface of radius $r$.
Due to spherical symmetry,the electric field $\vec{E}$ is radial and has the same magnitude at all points on the Gaussian surface.
The flux through the Gaussian surface is $\oint \vec{E} \cdot d\vec{a} = E \oint da = E(4\pi r^2)$.
The total charge enclosed by the Gaussian surface is $Q$.
Applying Gauss's law: $E(4\pi r^2) = \frac{Q}{\varepsilon_0}$.
Therefore,the electric field at a point outside the shell is $E = \frac{Q}{4\pi \varepsilon_0 r^2}$.
This shows that for points outside the shell,the electric field is the same as if all the charge $Q$ were concentrated at the center of the shell.