88 Week 2: Continuous Charge and Gauss’s Law
Finally, we write out Gauss’s law and solve forEz:φe= 2EzA = QS
ǫ 0=σ^0 A
ǫ 0
Ez =σ 0
2 ǫ 0
= 2πkσ 0 (120)where we note that the field isuniform– it doesn’t depend onz, and of course it cannot depend on
xandyeither as every point is in the middle of an infinite plane! This last result isvery important.
Note well: The parameterA(which youmade upwhen you drew your Gaussian surface)cancels
from the problem. Also note that this is exactly the result we got forthe field on the axis of a disk
of charge when we let the radius go to∞. This gives us confidence that Gauss’s Lawworks!
As before, in lecture your instructor will probably do a few more problems, perhaps a slab of
charge of finite thickness or the field produced bytwoinfinite sheets of charge, one with chargeσ 0
and the other with charge−σ 0 (a model for a parallel plate capacitor that we will study in great
detail shortly).
2.4: Gauss’s Law and Conductors
Properties of Conductors
E = 0
Q = 0
E = 0
S
A
Figure 22: An arbitrary chunk of conducting material in electrostatic equilibrium can have no field
inside, or else it wouldn’t be in equilibrium. It can have no field tangent toits surface, or it
wouldn’t be in equilibrium. From these facts we can deduce several useful things about conductors
in electrostatic equilibrium using Gauss’s Law.
A conductor is a material that contains many “free” charges thatarebound to the materialso
that they cannot easily jump from the conductor into a surrounding insulating material (where a
vacuum is considered an insulator for the time being, as is air) butfree to movewithin the material
itself if any e.g. electrical field exerts a force on them.