Week 2: Continuous Charge and Gauss’s Law 71
As before with a ring, we can only easily evaluate the field on thez-axis where we know from
symmetry thatEx=Ey= 0. As before, we find the field of a tiny chunk of charge in suitable
coordinates and sum it up using integration.
The coordinate system we choose locates the differential chunk ofcharge at (r, θ) inside the disk.
There we mark out a small chunk of arc lengthr dθas before for the ring, and of widthdr, so its
differential area isdA=r dθ dr. As an exercise:
A=
∫
dA=∫R
0∫ 2 π0rdr dθ=(∫R
0rdr)(∫ 2 π0dθ)
=
R^2
2
(2π) =πR^2 (63)and we’ve evaluated the area of a disk using calculus!
This is animportantexercise, as it shows that the integral can be grouped so that itseparates.
That is, therintegration andθintegration areindependent. We will only do integrals over more
than one coordinate in this course when they separate, so that a student can easily master physics
if they have mastered (a rather small subset of)one-dimensional integration methods. They are
trivially multivariate, so to speak.
At any rate, we can easily finddqfrom our mantra: “The charge of the chunk is the charge per
unit area times the area of the chunk”, or:
dq=σdA=σ rdr dθ=Q
πR^2rdr dθ (64)As before, we find
|dE~|=kedq
(r^2 +z^2 )=keσ rdr dθ
(r^2 +z^2 ) (65)and
dEz=|dE~|cosφ=keσz rdr dθ
(r^2 +z^2 )^3 /^2(66)
Finally:
Ez=∫
dEz=keσz∫R
0∫ 2 π0rdr dθ
(r^2 +z^2 )^3 /^2(67)
Theθintegral is trivial and yields 2π. What’s left is:
Ez = 2πkeσz∫R
0rdr
(r^2 +z^2 )^3 /^2= πkeσz∫R
0(r^2 +z^2 )−^3 /^2 (2rdr)= − 2 πkeσz(r^2 +z^2 )−^1 /^2∣∣
∣
R
0
= 2πkeσ(
1 −
z
(R^2 +z^2 )^1 /^2)
= 2πkeσ(1−cos Φ) (68)where (as was pointed out to me by one of my many clever students)cos Φ =z/
√
R^2 +z^2 where
the angle Φ points fromPto the edge of the disk.
There are two useful limits for us to explore for this problem. One is the limit thatR→ ∞
(which we can also interpret as Φ→π/2). In this limit, the disk of charge isinfinitein extent – it
is an infinite plane of uniform charge. The field is obviously:
Ez(∞) = 2πkeσ (69)and doesn’t depend on the distance from the plane. Again,everypoint is in the middle of an infinite
plane of charge, so the field of an infinite plane (or any large sheet ofcharge wherePis close enough