Q: Electric Field inside a long cylinder with uniform surface charge density Why is the field inside a uniformly charged long hollow cylinder zero?

A: Consider our cylinder to be of radius $b$. Now construct another cylinder coaxial to the original one with radius $r<b$. If we calculate the flux through this cylinder we get
$$\oint\vec E\cdot d\vec a=\frac{Q_{enclosed}}{\epsilon_0}=0\tag{01}$$
From symmetry we can say that the field must be radial in direction thus we can pull out electric field from the integral.
But since the radius of the constructed cylinder is arbitrary we see that within the original cylinder $E=0$.
