72 Week 2: Continuous Charge and Gauss’s Law
to the sheet so that the angles from it to the edges of the sheet are close toπ/2) is uniform and has
this magnitude, away from the (presumed positive) sheet of charge.
The other is whenz≫R. This limit is a bit tricky. We have to use thebinomial expansionto
evaluate the field to leading order. We get:
Ez = 2πkeσ(
1 −
z
(R^2 +z^2 )^1 /^2)
= 2πkeσ(
1 −
z
z(1 +Rz 22 )^1 /^2)
= 2πkeσ(
1 −(1 +
R^2
z^2)−^1 /^2
)
≈ 2 πkeσ(
1 −(1−^1
2
R^2
z^2+...)
)
≈ πkeσ(
R^2
z^2)
≈
ke(πR^2 σ)
z^2
≈ keQ
z^2(70)
orthe field far away from the disk is the field of a point charge of the same magnitude as the disk.
As we saw in the previous chapter, when we are far away from a charge distribution thedetails
of that distribution are averaged away and we are left with a field whose leading order behavior
is determined by its multipolar moment – if the distribution has a net charge it is monopolar; if
it has no net charge but has a +/−asymmetry it is dipolar; and so on. This means that we can
oftenguessor very simply calculate what field of a charge distribution will look like far away from
the distribution; all we need to know (or calculate) are the total charge and/or the total separated
charge and distance and direction of separation.
Example 2.1.4: Advanced: Spherical Shell of Charge
We will now proceed to set up and find the electric field inside and outside a uniform spherical shell
of charge by direct integration. This is just difficult enough that thissection is marked “Advanced”.
However, even normal humans – that is, humans who don’t plan to major in physics or mathematics
- who probably won’t spend a lot of their lifetime integrating nontrivialfunctions and solving partial
differential equations in spherical coordinate systems might want to look the solution over just to
see how it works and so that they can use it as acheck for Gauss’s Law, which we will cover
next.
We begin by choosing aspherical polar coordinate system, where a point is represented by the
triplet (r, θ, φ). Physicists usually useθandφas represented on the figure above, although in recent
years some mathematics texts (and even a few physics texts) swap them so thatθis the usual polar
angle in thex-yplane. Sadly, I am an ‘old guy’ and learned it so thoroughly the other way that I
just don’t want to change, so we’ll stick with the variable representation as given above.
Because the charge distribution (and hence the field) hasspherical symmetrywe lose nothing by
choosing the pointPwhere we want to evaluate the field on thez-axis and giving it az-coordinate
R(which is also the distance of the point from the origin). Furthermore, although it is not strictly
necessary, we can ignoredE⊥in the figure above because the problem hasazimuthal symmetryand
hence cannot have a total field componentinthex-yplane. I’m assuming that you have some
familiarity with spherical polar coordinates^33 and things like the area element on the surface of a
(^33) Wikipedia: http://www.wikipedia.org/wiki/Spherical Coordinate Systems. Note well that I’m using thephysics