Page 432 - Advanced engineering mathematics

412 CHAPTER 12 Vector Integral Calculus

Gauss’s laws D · ndσ = q and B · ndσ = 0.

These say that the measure of the normal component of the electric flux density across equals
the total charge in the bounded region, and that the measure of the normal component of the
magnetic flux density across is zero.
We will now carry out arguments similar to that used to derive the heat equation using the
divergence theorem. Begin by applying Stokes’s theorem to Faraday’s law to obtain

∂ϕ
E · dR = ∇× E · ndσ =−
∂t
C
∂ ∂B
=− B · ndσ = − · ndσ.
∂t ∂t
Then
∂B

∇× E + · ndσ = 0.
∂t
Since this holds for any piecewise smooth closed surface within the medium, then the integrand
must be zero, leading to
∂B
∇× E + = 0.
∂t
A similar analysis, using Ampère’s law, yields

∇× H = J.
Maxwell had observed that
∂E
J = σE + .
∂t
Then
∂E
∇× H = σE + .
∂t

Now start on a new tack. Apply Gauss’s theorem to Gauss’s law q = D · ndσ to obtain

D · ndσ = (∇· D)dV = q = QdV.

V V
Again falling back on the arbitrary nature of , we conclude from this that
∇· D = Q.
Now go back to ∇× E =−∂B/∂t and take the curl of both sides:
∂B ∂

∇× (∇× E) =∇ × − =− (∇× B).
∂t ∂t
We were able to interchange ∇ and ∂/∂t here because the curl involves only the space variables.
Since B = μH, then
∂ ∂
∇× (∇× E) =− (∇× μH) =−μ (∇× H).
∂t ∂t
It is a routine calculation to verify that this is the same as
∂
∇(∇· E) − (∇· ∇)E =−μ (∇× H).
∂t

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October 14, 2010 14:53 THM/NEIL Page-412 27410_12_ch12_p367-424