W9_parallel_resonance.eps

Week 2: Continuous Charge and Gauss’s Law 79

S′, and which subtends thesame solid angleand hence the same “stream of flow” of the field from
the charge, that the flux through it is the same:

∆φ′e=E~·nˆ′∆S′=|E~|cosθ

r^2 ∆Ω

cosθ

=|E~|r^2 ∆Ω = ∆φe (96)

In the differential limit, then, we can compute the flux through a small chunk of the arbitrary
surfaceS′as:

dφe = E~·nˆdS′

= |E~|r^2 dΩ

=

keq

r^2

r^2 dΩ

= keq dΩ (97)

which isindependentof the shape ofS′and involves only the differential solid angle swept out from
the charge as one does the integral. If we integrate both sides, noting that the complete solid angle
(in, say, spherical polar coordinates) is:
∫
dΩ =

∫π

0

∫ 2 π

0

sin(θ)dθ dφ= 4π steradians (98)

we get:

φe=

∮

S′

E~·nˆdS= 4πkeq (99)

independent of the shape of the closed surface that we integrateover that encloses the chargeq!

This isalmostGauss’s Law. To complete our statement, we have to note first, that if the charge
qisoutsidethe closed surfaceS′, the net flux throughS′iszero. There are a variety of ways to see
this, but the easiest one is to considerS′itself to be part of a larger surface that inclosesq. This
creates two surfaces: one that includes the “outside” ofS′and one that includes the “inside” ofS′.
The net flux through the two must be the same, and by changing onlythe sign ofnˆon the inner
surface we can immediately see that the net flux throughS′must vanish.

Second, we have to use the superposition principle. If we enclose more than one charge byS′,
we just add up the fluxes so that thetotalflux is produced by thetotalcharge inS′, no matter how
it is distributed! Putting all this together, and getting rid of the prime onS(because it is no longer
needed – the flux is the same for all closed surfaces that inclose a certain amount of charge) we get:

Gauss’s Law for the Electric Field

∮

S/V

E~·ˆndA= 4πke

∫

V

ρ dV =Qin S

ǫ 0

(100)

or in words, the flux of the electric field through a closed surfaceSequals the total charge insideS
divided byǫ 0 , the permittivity of the electric field. This is the first one ofMaxwell Equation’sthat
we’ve covered so far. Only three more to go!

I used integration to compute the total charge of a continuous distribution, but of course I could
equally well have summed over a bunch of discrete charges instead.The integral form will be very
useful later on if you continue in physics, as it helps to transform this integral expression of Gauss’s
Law into a differential expression that is more useful still.

So, what’s it good for? Lots! But for the moment, we’llstartbut using Gauss’s law to easily
evaluate the electric field for charge density distributions that have the symmetry of a coordinate
system that we’d otherwise have to evaluate using painful direct integration. We will also use it to
help us reason about things like the distribution of charge on a conductor in electrostatic equilibrium.
And don’t forget, we consideritto be the actual Law of Nature for the electrostatic field, so things