W9_parallel_resonance.eps

Week 10: Maxwell’s Equations and Light 349

Note that I’ve chosen two simple surfacesS 1 andS 2 bounded byC– in fact, they are

both parts of a sphere, and together they make aclosed surface, one that encloses a

volumeV (inside the sphere). The current densityJ~flows in through surfaceS 1 , but

notallof it flows out throughS 2. Some of it is building up in a charge distribution

ρinside the sphere. So the total currentIflowinginto the sphere is larger than the

total current flowingout. None of this – the choice of a sphere, the particular curveC

or surfacesS 1 orS 2 – is important; we just choose them to make the result easy to see.

If charge is conserved, theratethat charges builds up inside the closed surfaceS=S 1 +S 2

will equal thedifferencethe the flux of the current densities:

∫

S 1 /C

J~·nˆdA−

∫

S 2 /C

J~·ˆndA= d

dt

∮

V/S

ρdV (818)

In this equation, the normalsnˆin the two integrals on the left are directed from the

left to the right, in the direction of the current’s apparent flow. Note that we first

derived/discussed this law as equation 300 way back in the week where we first derived

and discussed current and resistance and defined current density in the first place, so

thisshouldn’tbe a complete mystery even if you’ve forgotten the first pass through it.

The integral on the left looks strangely familiar. In fact, it is part ofGauss’s Law for

Electricity! Using Gauss’s Law we canreexpressit:

∮

V/S

ρdV=ǫ 0

∮

S

E~·nˆ′dA (819)

Now we’ll break this up into two pieces – we can surely integrate this overS 1 andS 2

separately, as long asS 1 +S 2 =S. However, I’m going to make one more change. The

normalnˆ′in the Gauss’s Law expression is (recall) theoutward directed normal. This

goes from left to right onS 2 , but onS 1 it goes from right to left! I want to makenˆ

exactly the same as in the integrals on the left hand side of my expression of charge

conservation, so I have to change the sign of theS 1 integral:

∮

V/S

ρdV = ǫ 0

∮

S

E~·ˆn′dA

= −ǫ 0

∫

S 1 /C

E~·nˆdA+ǫ 0

∫

S 2 /C

E~·nˆdA (820)

Now we substitute this back into our original equation to get:

∫

S 1 /C

J~·nˆdA−

∫

S 2 /C

J~·ˆndA=−ǫ 0 d

dt

∫

S 1 /C

E~·nˆdA+ǫ 0 d

dt

∫

S 2 /C

E~·nˆdA (821)

Last, we move all of theS 1 integrals to the left, and all of theS 2 integrals to the right:

∫

S 1 /C

J~·nˆdAǫ 0 d

dt+

∫

S 1 /C

E~·nˆdA=

∫

S 2 /C

J~·ˆndA+ǫ 0 d

dt

∫

S 2 /C

E~·nˆdA (822)

The left side only depends onS 1 /C. The right depends only onS 2 /C. We used no special

properties of these curves or surfaces beyond the fact that any two non-coincident open

surfaces bounded by the same closed curveCenclose a volume. The two sides are thus

invariantunder any possible change in the curvesCor surfacesS. We thus define the

invariant currentto be:

Iinvariant, through C=

∫

S/C

J~·ˆndA+ǫ 0 d

dt

∫

S/C

E~·ˆndA (823)

where the result now holds foranysurfaceSbounded by any give closed curveC!