5. Preparing for Maxwell’s Equations 5.1 The incomplete set.
Before Maxwell could unify the fundamental equations governing electromagnetism he had to add a crucial piece to one of them. We have already collected three complete Maxwell equations (available either in differential or integral forms): Gauss’ law (1.21)
∇.D = ρ f ;
Faraday’s law of induction (1.32)
∇×E = −
∂B ∂t
“No magnetic monopoles” (1.28) ∇.B = 0 . But Ampere’s law is incomplete in the forms we have seen so far, e.g. (2.12) ∇×H = J f . This only describes a system with steady currents but is inadequate to cope even with such a simple time-dependent problem as the charging of a capacitor see section 5.3 below. 5.2 What is missing from Ampere’s Law (formally) Take the divergence of both sides of (2.12)
∇.∇ × H = ∇.J f
(5.1)
.But for any vector (see Tools) ∇.∇ × A = 0 , so if (2.12) is true the divergence of the free
current density J f is everywhere zero. That would forbid a small local concentration of negative charge (say) from streaming outward into the space around it. Such an outflow must happen because of the Coulomb repulsion between like charges, and it has to satisfy the conservation of charge equation ∂ρ ∇.J f = − f , (5.2) ∂t sometimes called the continuity equation. Free charge densities must be able to change with time, so current densities can have divergence, so (2.12) must be incomplete. This paradox implies there is something missing from the physics. [ Derivation of (5.2): For any volume τ , the net outflow of current through its surface S ∂Q ∂ is equal to the rate of decrease of charge Q inside it; v∫ J.dS = − = − ∫ ρ f dτ . ∂t ∂t τ S
Gauss’ divergence theorem is
∂
v∫ A.dS = ∫τ ∇.Adτ . So v∫ J.dS = ∫τ ∇.Jdτ = − ∫τ ∂t ρ S
2B29 Prep’n for Maxwell’s Equations. Spring 2004
f
dτ .
S
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This has to be true for any arbitrarily chosen small volume, so the integrands of the two volume integrals must be equal; hence (5.2)] Differentiating Gauss’ law ∇.D = ρ f we get ∂ ∂D ∂ρ f (∇.D) = ∇. . (5.3) = ∂t ∂t ∂t The rate of change of charge density is equal to the divergence of the rate of change of the displacement D. If we add (5.2) to (5.3) we get ∂D ∇.J f + ∇. =0 ∂t ∂D or ∇. J f + (5.4) =0 ∂t Maxwell suggested that we should modify Ampere’s law by using the quantity in brackets in (5.4) instead of just J f , giving in the differential form of the law
∇×H = J f +
∂D ∂t
(5.5)
∂D is called the displacement current (strictly, the displacement current ∂t density). (5.5) is what we use as the fourth Maxwell equation. Note that taking the divergence of both sides now gives zero, even when there are varying charge densities.
The quantity
5.3 What was missing from Ampere’s Law (practical illustration) Imagine a parallel plate capacitor with current I (t ) flowing into it. This will be producing increased charge on one plate and a matching negative charge on the opposite plate.
I I P
The incomplete version of Ampere’s law can be written in integral form, equation (2.13), v∫ H.dl = ∫ J f .dS P
SP
The integration path is a loop P around the wire connected to the capacitor. The surface Sp is joined on to P all around its edges. We naturally think of Sp as a flattish disc cutting through the wire. Then the integrated flux of J f over the area of Sp is I, so we have
v∫ H.dl =I
(5.6)
P
This should be true for any surface Sp connected at its edges to S.
2B29 Prep’n for Maxwell’s Equations. Spring 2004
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But we can draw such a surface which instead of cutting the wire is distorted to pass between the plates of the capacitor. In the sketch, SP1 represents the flattish disc surface through the wire, but SP2 goes through the I I gap of the capacitor. There is SP1 no free current in the gap, so there can be no contribution to ∫ J f .dS . SP2 P SP The incomplete version of Ampere’s law says that I = ∫ J f .dS ≠ ∫ J f .dS = 0 . S P1
SP 2
This is an alternative version of the paradox described at the beginning of section 5.2 above. It is resolved in an exactly parallel way, by introducing the displacement current into the integral form of Ampere’s law, giving ∂D (5.7) v∫P H.dl = S∫ J f + ∂t .dS P The J f term still does what is needed when we do the surface integral over SP1, through the wire, but when we integrate through the capacitor gap on suface SP2 we get a ∂D contribution from ∫ .dS . This will be nonzero inside a charging capacitor. We saw ∂t SP in section 1.4 above that the uniform displacement inside a large parallel plate capacitor is D = Q / A . So ∂D ∂D ∂Q (5.8) ∫S ∂t .dS = A ∂t = ∂t = I ; P2 The integral of the displacement current density over the area of the capacitor is equal to the current flowing onto the plates. When there is dielectric inside the gap we can think of a large part of the displacement current as coming from the change of the polarisation in the material; movements of the polarisation charge as the tiny dipoles stretch under the influence of the E field. But when the gap is empty ∂D ∂E there is a displacement current in the vacuum, . (5.9) = ε0 ∂t vac ∂t
2B29 Prep’n for Maxwell’s Equations. Spring 2004
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