Page 404 - Engineering Electromagnetics, 8th Edition

386 ENGINEERING ELECTROMAGNETICS

power flow at a point, and many of us think of the Poynting vector as a “pointing”
vector. This homonym, while accidental, is correct. 4
Because S is given by the cross product of E and H, the direction of power flow
at any point is normal to both the E and H vectors. This certainly agrees with our
experience with the uniform plane wave, for propagation in the +z direction was
associated with an E x and H y component,

E x a x × H y a y = S z a z
In a perfect dielectric, the E and H field amplitudes are given by

E x = E x0 cos(ωt − βz)
E x0
H y = cos(ωt − βz)
η
where η is real. The power density amplitude is therefore

2
E x0 2
S z = cos (ωt − βz) (73)
η
In the case of a lossy dielectric, E x and H y are not in time phase. We have
E x = E x0 e −αz cos(ωt − βz)
If we let

η =|η| θ η
then we may write the magnetic field intensity as

E x0 −αz
H y = e cos(ωt − βz − θ η )
|η|
Thus,
E 2
e
S z = E x H y = x0 −2αz cos(ωt − βz) cos(ωt − βz − θ η ) (74)
|η|
Because we are dealing with a sinusoidal signal, the time-average power density,
S z ,is the quantity that will ultimately be measured. To find this, we integrate (74)
over one cycle and divide by the period T = 1/f . Additionally, the identity cos
A cos B = 1/2 cos(A + B) + 1/2 cos(A − B)is applied to the integrand, and we
obtain:
1 T 1 E 2
e
S z = x0 −2αz [cos(2ωt − 2βz − 2θ η ) + cos θ η ] dt (75)
T 0 2 |η|

4 Note that the vector symbol S is used for the Poynting vector, and is not to be confused with the
differential area vector, dS. The latter, as we know, is the product of the outward normal to the surface
and the differential area.