Week 2: Continuous Charge and Gauss’s Law 85
The uniform ball of charge is the basis for a model of theneutral atom– a positive nucleus
surrounded by a uniform ball of negative charge – that helps us understandpolarizationin a few
weeks. This model is still used (dressed up with damping and a time dependent driving field) in
physics graduate schoolwhere the model is called theLorentz Oscillator Modelfor the atom and
where the result of analyzing the model is understanding ofdispersion– basically time dependent
dielectric response and the absorption of electromagnetic energyby matter! It sounds complicated,
but it isn’t, not really. It isalmostwithin your reach at the end of taking this introductory course
(where we will cover the static part of the result perfectly adequately) – all that separates you is a
bit more work with the damped driven harmonic oscillator to help you manage even the dynamics.
The reward for the effort is that afterwards, you understandmicroscopicallywhy, e.g. rainbows
happen, why the sky is blue, how light from the sun warms the earth,and much more. So keep it
in mind for later.
Example 2.3.3: Cylindrical: A cylindrical shell of charge
LLr 2aSS(^20)
1
σ
n
E
E n
r 1
Figure 20: A cylindrical shell of radiusa, carrying a uniform charge per unit areaσ 0. Two cylindrical
concentricGaussian surfacesS 1 (with radiusr < aandS 2 (with radiusr > a) are shown.
Suppose you are given an infinite cylindrical shell of charge with a uniform charge per
unit areaσ 0 and radiusa. Find the field everywhere in space.
We solve this problemexactlylike we did the sphere. In fact, I block-copied the solution from
above to write this and changed only a few minimal things.
There are two distinct regions, inside the cylinder and outside the cylinder. Draw acylindrical
Gaussian surfaceS 1 of lengthLinside the cylinder (forr < a). We don’t know that the field is on
this surface yet, but we do know that on thecylinderpart it must lie along~rand be constant in
magnitude and perpendicular to the surface at all points on our Gaussian surface from the symmetry
of the distribution. On the end caps the field may well vary withr, but it isparallelto those surfaces
and therefore there is no net flux through the caps. Hence:
φe =
∮
S 1E~·rˆdA= φcaps+Er∫
CyldA= 0 +Er(2πr)L (112)