Page 116 - Engineering Electromagnetics, 8th Edition

98 ENGINEERING ELECTROMAGNETICS

The final result is then

Qd cos θ
V = (33)
4π 0 r 2
Again, we note that the plane z = 0(θ = 90 )isat zero potential.
◦
Using the gradient relationship in spherical coordinates,

∂V 1 ∂V 1 ∂V

E =−∇V =− a r + a θ + a φ
∂r r ∂θ r sin θ ∂φ
we obtain

Qd cos θ Qd sin θ
E =− − a r − a θ (34)
2π 0 r 3 4π 0 r 3
or
Qd
E = (2 cos θ a r + sin θ a θ ) (35)
4π 0 r 3
These are the desired distant fields of the dipole, obtained with a very small
amount of work. Any student who has several hours to spend may try to work the
problem in the reverse direction—the authors consider the process too long and de-
tailed to include here, even for effect.
To obtain a plot of the potential field, we choose a dipole such that
2
Qd/(4π 0 ) = 1, and then cos θ = Vr . The colored lines in Figure 4.9 indicate
equipotentials for which V = 0, +0.2, +0.4, +0.6, +0.8, and +1, as indicated.
The dipole axis is vertical, with the positive charge on the top. The streamlines for
the electric field are obtained by applying the methods of Section 2.6 in spherical
coordinates,

E θ rdθ sin θ
= =
E r dr 2 cos θ
or

dr
= 2 cot θ dθ
r
from which we obtain
2
r = C 1 sin θ

The black streamlines shown in Figure 4.9 are for C 1 = 1, 1.5, 2, and 2.5.
The potential field of the dipole, Eq. (33), may be simplified by making use of
the dipole moment. We first identify the vector length directed from −Q to +Q as d