Page 136 - Electromagnetics

Then J 1 cos θ 1 = J 2 cos θ 2 by (3.39), while σ 2 J 1 sin θ 1 = σ 1 J 2 sin θ 2 by (3.40). Hence

σ 2 tan θ 1 = σ 1 tan θ 2 . (3.42)
It is interesting to consider the case of current incident from a conducting material onto
an insulating material. If region 2 is an insulator, then J 2n = J 2t = 0; by (3.39)we have
J 1n = 0. But (3.40)does not require J 1t = 0; with σ 2 = 0 the right-hand side of (3.40)
is indeterminate and thus J 1t may be nonzero. In other words, when current moving
through a conductor approaches an insulating surface, it bends and flows tangential to
the surface. This concept is useful in explaining how wires guide current.
Interestingly, (3.42)shows that when σ 2 σ 1 we have θ 2 → 0; current passing from a
conducting region into a slightly-conducting region does so normally.

3.2.3 Uniqueness of the electrostatic field
In § 2.2.1 we found that the electromagnetic field is unique within a region V when
the tangential component of E is specified over the surrounding surface. Unfortunately,
this condition is not appropriate in the electrostatic case. We should remember that
an additional requirement for uniqueness of solution to Maxwell’s equations is that the
field be specified throughout V at some time t 0 . For a static field this would completely
determine E without need for the surface field!
Let us determine conditions for uniqueness beginning with the static field equations.
Consider a region V surrounded by a surface S. Static charge may be located entirely
or partially within V , or entirely outside V , and produces a field within V . The region
may also contain any arrangement of conductors or other materials. Suppose (D 1 , E 1 )
and (D 2 , E 2 ) represent solutions to the static field equations within V with source ρ(r).
We wish to find conditions that guarantee both E 1 = E 2 and D 1 = D 2 .
Since ∇· D 1 = ρ and ∇· D 2 = ρ, the difference field D 0 = D 2 − D 1 obeys the
homogeneous equation
∇· D 0 = 0. (3.43)

Consider the quantity

∇· (D 0 0 ) = 0 (∇· D 0 ) + D 0 · (∇ 0 )
where E 0 = E 2 − E 1 =−∇ 0 =−∇( 2 − 1 ). We integrate over V and use the
divergence theorem and (3.43)to obtain

0 [D 0 · ˆ n] dS = D 0 · (∇ 0 ) dV =− D 0 · E 0 dV. (3.44)
S V V
Now suppose that 0 = 0 everywhere on S, or that ˆ n · D 0 = 0 everywhere on S, or that
0 = 0 over part of S and ˆ n · D 0 = 0 elsewhere on S. Then

D 0 · E 0 dV = 0. (3.45)
V
Since V is arbitrary, either D 0 = 0 or E 0 = 0. Assuming E and D are linked by the
constitutive relations, we have E 1 = E 2 and D 1 = D 2 .
Hence the fields within V are unique provided that either , the normal component
of D, or some combination of the two, is specified over S. We often use a multiply-
connected surface to exclude conductors. By (3.33)we see that specification of the

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