Page 101 - Electromagnetics

describes the electromagnetic energy entering the resistor and the thermodynamic flux
describes the heat dissipation. For the sum of the two subsystems conservation of energy
requires
∇· (S em + S th ) =−J · E + P th = 0.

To compute the heat dissipation we can use

P th = J · E =−∇ · S em
and thus either use the boundary fields or the fields and current internal to the resistor
to find the dissipated heat.
Boundaryconditions on the Poynting vector. The large-scale form of Poynting’s
theorem may be used to determine the behavior of the Poynting vector on either side
of a boundary surface. We proceed exactly as in § 2.8.2. Consider a surface S across
which the electromagnetic sources and constitutive parameters are discontinuous (Figure
2.6). As before, let ˆ n 12 be the unit normal directed into region 1. We now simplify the
notation and write S instead of S em . If we apply Poynting’s theorem

∂D ∂B
J · E + E · + H · dV + S · n dS = 0
V ∂t ∂t S
to the two separate surfaces shown in Figure 2.6, we obtain
∂D ∂B

J · E + E · + H · dV + S · n dS = ˆ n 12 · (S 1 − S 2 ) dS. (2.303)
V ∂t ∂t S S 10
If on the other hand we apply Poynting’s theorem to the entire volume region including
the surface of discontinuity and include the contribution produced by surface current, we
get

∂D ∂B
J · E + E · + H · dV + S · n dS =− J s · E dS. (2.304)
V ∂t ∂t S S 10
Since we are uncertain whether to use E 1 or E 2 in the surface term on the right-hand side,
if we wish to have the integrals over V and S in (2.303) and (2.304) produce identical
results we must postulate the two conditions
ˆ n 12 × (E 1 − E 2 ) = 0

and
ˆ n 12 · (S 1 − S 2 ) =−J s · E. (2.305)

The first condition is merely the continuity of tangential electric field as originally postu-
lated in § 2.8.2; it allows us to be nonspecific as to which value of E we use in the second
condition, which is the desired boundary condition on S.
It is interesting to note that (2.305) may also be derived directly from the two pos-
tulated boundary conditions on tangential E and H. Here we write with the help of
(B.6)

ˆ n 12 · (S 1 − S 2 ) = ˆ n 12 · (E 1 × H 1 − E 2 × H 2 ) = H 1 · (ˆ n 12 × E 1 ) − H 2 · (ˆ n 12 × E 2 ).
Since ˆ n 12 × E 1 = ˆ n 12 × E 2 = ˆ n 12 × E,wehave

ˆ n 12 · (S 1 − S 2 ) = (H 1 − H 2 ) · (ˆ n 12 × E) = [−ˆ n 12 × (H 1 − H 2 )] · E.

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