76 Week 2: Continuous Charge and Gauss’s Law
Ouch! That was a lot of work! And technically, we’re not even done – we should really pick a
point whereR < r(inside the sphere) to prove that the electric fieldvanishesinside. At an interior
point, one has to break the cos(θ) integral up into two pieces withdifferent signsbecause the charge
from the part of the sphere aboveRcreates a field that pointsdown, where the charge from the part
of the sphere belowRpointsup. The integral limits change to:
Ez= 2πr^2 σke{∫R/r− 1...−
∫ 1
R/r}
(88)
(but otherwise all geometry remains the same). Ahhhh, too much work. We’ll rely instead on a
slightly more intuitive argument, one closely tied toGauss’s Law, to show that the field inside a
spherical shell cancels, although (as we will see) it follows trivially from Gauss’s Law itself.
What is this Gauss’s Law of which I speak, you ask? Coming up next...2.2: Gauss’s Law for the Electrostatic Field
Gauss’s Law for the electrostatic field is, as we shall see, one ofMaxwell’s Equations.^35 Maxwell’s
equations are, in turn, the equations of motion for the unifieddynamicelectromagnetic field, laws
of nature, and one of the most beautiful things (mathematically and conceptually speaking) in
all of physics. It is therefore of critical importance that you workhard developing aconceptual
understandingof this law that permits you tovisualizethe relationship between the mathematics of
its expression and the geometry of the field in addition to “just” learning to solve problems with it.
For that reason we will begin this chapter with a derivation of this law from the field equation
of the point charge (which in turn is basically Coulomb’s Law in disguise) and the superposition
principle. Derivations, of course, work both ways and physicists today generally consider Gauss’s
Law the fundamental law of nature and the field of a point charge and Coulomb’s law are rather
consequences to be derived from it instead of the other way around. You will not be responsible for
being able to “do” the derivation yourself in a problem or on an exam, but it is strongly advised
that you work through it a couple of times anyway and get to where you intuitively understand the
relationship between flux integrals and conservation, as we’ll use this idea in a critical way later
when we add the Maxwell Displacement Current to Ampere’s Law in order to be able to show that
light is an electromagnetic wave!
We begin our derivation of Gauss’s Law by considering thefluxof the electrostatic vector field
through a small rectangular patch of surface ∆S. To compute this, we first must understand what
the flux of an arbitrary vector fieldF~through a surfaceSis. Mathematically, the flux of a vector
field through some surface is defined to be:
φf=∫
∆SF~·n dSˆ (89)Note that the word flux meansflow, and this integral measures theflowof the fieldthroughthe
surface. It’s mathematical purpose is to detect theconservation of flowin the vector field. Basically
it takes the magnitude of the fieldF~at all points on the surface, computes the component ofF~
that goesthroughthe surface at right angles (instead of tangent to the surface, which doesn’t really
go “through”), multiplies it times a tiny differential chunk of the area, and then adds up all the
differential chunks thus computed.
Let’s look at this in more detail, specializing to the case of the electric field. Consider figure??,
where we show electric field lines flowing through a small ∆S=abat right angles to the field lines
(so that a unit vectornˆnormal to the surface isparallelto the electric field). ∆Sis small enough
(^35) Wikipedia: http://www.wikipedia.org/wiki/Maxwell’s Equations.