Why the electric field just outside the surface of a conductor is normal to the surface?

A spherical shell of radius $\mathrm{R}$ is uniformly charged with charge $+\mathrm{q}$. By Gauss's theorem, find the electric field intensity at a point $\mathrm{p}$.

Inside the spherical shell.

In the figure, a very large plane sheet of positive charge is shown. $P_1$ and $P_2$ are two points at distance $l$ and $2l$ from the charge distribution. If $\sigma$ is the surface charge density, then the magnitude of electric fields $E_1$ and $E_2$ at $P_1$ and $P_2$ respectively are 

A spherical shell of radius $\mathrm{R}$ is uniformly charged with charge $+\mathrm{q}$. By Gauss's theorem, find the electric field intensity at a point $\mathrm{p}$. Outside the spherical shell.

A spherically symmetric charge distribution is considered with charge density varying as

$\rho \left(r\right)=\left\{\begin{array}{ll}{\rho }_0\left(\frac{3}{4}-\frac{r}{R}\right)& \text{ for }r\le R\\ \mathrm{Zero}& & & \mathrm{for} r>R\end{array}\right.$

Where, $r\left(r<R\right)$ is the distance from the centre $O$ (as shown in figure). The electric field at point $P$ will be :

A long cylindrical volume contains a uniformly distributed charge of density $\rho \mathrm{C} {\mathrm{m}}^{-3}$. The electric field inside the cylindrical volume at a distance $x=\frac{2{\epsilon }_0}{\rho } \mathrm{m}$ from its axis is _____ $\mathrm{V} {\mathrm{m}}^{-1}$.

Electric intensity outside a charged cylinder having the charge per unit length $\lambda$ at a distance $r$ from its axis is ____.