Page 58 - Engineering Electromagnetics, 8th Edition

40 ENGINEERING ELECTROMAGNETICS

density, or charge per unit length, is ρ L = ρ S dy , and the distance from this line

2

2
charge to our general point P on the x axis is R = x + y . The contribution to
E x at P from this differential-width strip is then
ρ S dy ρ S xdy
dE x = cos θ =
2
2
2π 0 x + y 2 2π 0 x + y 2
Adding the effects of all the strips,
∞
∞
ρ S xdy ρ S −1 y ρ S
E x = 2 2 = tan =
2π 0 −∞ x + y 2π 0 x −∞ 2 0
If the point P were chosen on the negative x axis, then
ρ S
E x =−
2 0
for the field is always directed away from the positive charge. This difficulty in sign
is usually overcome by specifying a unit vector a N , which is normal to the sheet and
directed outward, or away from it. Then
ρ S
E = a N (17)
2 0
This is a startling answer, for the field is constant in magnitude and direction.
It is just as strong a million miles away from the sheet as it is right off the surface.
Returning to our light analogy, we see that a uniform source of light on the ceiling of
avery large room leads to just as much illumination on a square foot on the floor as it
does on a square foot a few inches below the ceiling. If you desire greater illumination
on this subject, it will do you no good to hold the book closer to such a light source.
If a second infinite sheet of charge, having a negative charge density −ρ S ,is
located in the plane x = a,we may find the total field by adding the contribution of
each sheet. In the region x > a,
ρ S ρ S
E + = a x E − =− a x E = E + + E − = 0
2 0 2 0
and for x < 0,
ρ S ρ S
E + =− a x E − = a x E = E + + E − = 0
2 0 2 0
and when 0 < x < a,
ρ S ρ S
E + = a x E − = a x
2 0 2 0
and

ρ S
E = E + + E − = a x (18)
0
This is an important practical answer, for it is the field between the parallel plates
of an air capacitor, provided the linear dimensions of the plates are very much greater
than their separation and provided also that we are considering a point well removed