W9_parallel_resonance.eps

Week 2: Continuous Charge and Gauss’s Law 87

where I’ve used the fact thatλ 0 =QS/L= 2πaσ 0 to help show thatthe field of a cylindrically
symmetric charge distribution outside that distributionis the same as that of aline of charge with
the same net charge per unit length on its axis.

Note well: The parameterL(which youmade upwhen you drew your Gaussian surface)cancels
from the problem. Of course it does! And a good thing, too!

In lecture your instructor will probably do a few more difficult problems – perhaps a solid
cylinder of charge, or multiple cylindrical shells, or even a solid cylinderwith a charge distribution
likeρ(r) =ArwhereAis a constant! You should be able to doanyproblem with a cylindrical
distribution of charge that you can integrate or sum inside any givenGaussian cylinder using this
method.

Example 2.3.4: Planar: A sheet of charge

n

E

n

En

E

σ A

z

0

Figure 21: An (infinite) plane sheet of uniform charge per unit areaσ 0. The Gaussian surface in
this case is a simple “pillbox” symmetrically drawn so it intersects the sheet as drawn.

Suppose you are given an infinite sheet of charge with a uniform charge per unit area

σ 0. Find the field everywhere in space.

We solve this problemexactlylike we did the two above. You (by now) should know the drill.
Here we only need to draw a single Gaussian surface as indicated in figure??above. We will
again draw acylindricalGaussian surfaceof lengthz, but this time it must be symmetrically located
so that itsymmetrically intersectsthe plane of charge withz/2 of its length above and below the
plane. This cylinder has an end-cap area ofAwhich (likeLin the previous problem) willcancel
when we go to evaluate the field. We don’t know what the field is on this surface yet, but we do
know that on theend-capsit must lie parallel to~zand be constant in magnitude and perpendicular
to the end caps at all points. On the side of the cylinder the field may well vary withr, but it is
parallelto this surface and therefore there is no net flux through it. Hence:

φe =

∮

S

E~·ˆzdA

= φside+ 2EzA

= 2EzA (118)

where you should note that we havetwoend caps, each of which contributesEzAto the flux.

The total charge inside this Gaussian surface is trivial:

QS=

∫

A

σ 0 dA=σ 0 A (119)

where there really isn’t much of anything to integrate or evaluate.