Consider an infinite line of charge with uniform charge density per unit length λ. What is the magnitude of the electric field a distance r from the line?

When we had a finite line of charge we integrated to find the field. We could do that again, integrating from minus infinity to plus infinity, but it's a lot easier to apply Gauss' Law.

First, let's agree that if the charge on the line is positive, the field is directed radially out from the line. If it is negative, the field is directed in.

What is the appropriate gaussian surface to use here? A cylinder of length L and radius r is just what we need, with the axis of the cylinder along the line of charge.

To apply Gauss' Law, we need to answer two questions:
What is the total charge enclosed by the surface?
What is the net electric flux passing through the surface?

The total charge enclosed is q_enc = λL, the charge per unit length multiplied by the length of the line inside the cylinder.

To find the net flux, consider the two ends of the cylinder as well as the side. There is no flux through either end, because the electric field is parallel to those surfaces. On the other hand, the electric field through the side is simply E multiplied by the area of the side, because E has the same magnitude and is perpendicular to the side at all points.

Net flux = E A = E (2 π r) L

By Gauss' Law the net flux = q_enc/ε_o

Therefore E (2 π r) L = λL/ε_o

The factors of L cancel, which is encouraging - the field should not depend on the length we chose for the cylinder. Solving for the magnitude of the field gives:

E = λ/[ 2 π r ε_o ]

Because k = 1/(4π ε_o) this can also be written:

E = 2kλ/r

The electric field is proportional to the linear charge density, which makes sense, as well as being inversely proportional to the distance from the line.