Page 154 - Engineering Electromagnetics, 8th Edition

136 ENGINEERING ELECTROMAGNETICS

and the division of this equation by (39) gives

tan θ 1 1
= (41)
tan θ 2 2
In Figure 5.11 we have assumed that 1 > 2 , and therefore θ 1 >θ 2 .
The direction of E on each side of the boundary is identical with the direction of
D, because D = E.
The magnitude of D in region 2 may be found from Eq. (39) and (40),

2

2
2 2 (42)
D 2 = D 1 cos θ 1 + sin θ 1
1
and the magnitude of E 2 is

2
2 1 2
E 2 = E 1 sin θ 1 + cos θ 1 (43)
2
An inspection of these equations shows that D is larger in the region of larger permit-
tivity (unless θ 1 = θ 2 = 0 where the magnitude is unchanged) and that E is larger
◦
in the region of smaller permittivity (unless θ 1 = θ 2 = 90 , where its magnitude is
◦
unchanged).

EXAMPLE 5.5
Complete Example 5.4 by finding the fields within the Teflon ( r = 2.1), given the
uniform external field E out = E 0 a x in free space.
Solution. We recall that we had a slab of Teflon extending from x = 0to x = a,
as shown in Figure 5.12, with free space on both sides of it and an external field
E out = E 0 a x .We also have D out = 0 E 0 a x and P out = 0.
Inside, the continuity of D N at the boundary allows us to find that D in = D out =
0 E 0 a x . This gives us E in = D in / = 0 E 0 a x /( r 0 ) = 0.476E 0 a x .To get the polar-
ization field in the dielectric, we use D = 0 E + P and obtain

P in = D in − 0 E in = 0 E 0 a x − 0.476 0 E 0 a x = 0.524 0 E 0 a x
Summarizing then gives

D in = 0 E 0 a x (0 ≤ x ≤ a)
E in = 0.476E 0 a x (0 ≤ x ≤ a)
P in = 0.524 0 E 0 a x (0 ≤ x ≤ a)
A practical problem most often does not provide us with a direct knowledge of
the field on either side of the boundary. The boundary conditions must be used to help
us determine the fields on both sides of the boundary from the other information that
is given.