The electric field spreads radially outward from the charged rod. The symmetry of the rod of charge allows no special directions, so the electric field strength only depends on the distance, r, from the axis of the rod.

2. C

The total charge in a cylinder of radius r and length L is given by the density rho times the volume of the cylinder. In this case Q = rho * pi r*r L.

3. C.

Choose a Gaussian surface which matches the symmetry of the charge distribution and is at right angles to the elecgtric field lines.

4. B

With this Gaussian surface, Gauss's Law gives E*(area of cylinder) = Q_{enc}/epsilon. But
Q_{enc} = rho pi r*r L and the area of the cylinder is 2 pi r L (don't include the end
caps -- either consider the rod to be infinitely long or notice that the electric field
is *parallel* to the end caps, so there is no flux through them).