Page 55 - Engineering Electromagnetics, 8th Edition

CHAPTER 2 Coulomb’s Law and Electric Field Intensity 37

We therefore have found that we have only an E ρ component and it varies only
with ρ.Now to find this component.
We choose a point P(0, y, 0) on the y axis at which to determine the field.
This is a perfectly general point in view of the lack of variation of the field with φ
and z. Applying (10) to find the incremental field at P due to the incremental charge
dQ = ρ L dz ,wehave

ρ L dz (r − r )

dE = 3
4π 0 |r − r |
where
r = ya y = ρa ρ

r = z a z
and

r − r = ρa ρ − z a z
Therefore,
ρ L dz (ρa ρ − z a z )

dE =
4π 0 (ρ + z )
2
2 3/2
Because only the E ρ component is present, we may simplify:
ρ L ρdz
dE ρ =
2
2 3/2
4π 0 (ρ + z )
and
∞ ρ L ρdz

E ρ = 2 2 3/2
−∞ 4π 0 (ρ + z )
Integrating by integral tables or change of variable, z = ρ cot θ,wehave

∞
ρ L 1 z
E ρ = ρ
2
4π 0 ρ 2 ρ + z 2
−∞
and
ρ L
E ρ =
2π 0 ρ
or finally,
ρ L
E = a ρ (16)
2π 0 ρ
We note that the field falls off inversely with the distance to the charged line, as
compared with the point charge, where the field decreased with the square of the
distance. Moving ten times as far from a point charge leads to a field only 1 percent
the previous strength, but moving ten times as far from a line charge only reduces
the field to 10 percent of its former value. An analogy can be drawn with a source of