Simple Nature - Light and Matter

d/Example 15: geometry.

e/Example 15: the field on

both sides (forσ>0).

f/A capacitor consisting of

two disks with opposite charges.

Everything inside the integral is a constant, so we have

Ez=

k

b^2 +z^2

cosθ

∫

dq

=

k Q

b^2 +z^2

cosθ

=

k Q

b^2 +z^2

z

r

=

k Qz

(

b^2 +z^2

)3/2

In all the examples presented so far, the charge has been con-
fined to a one-dimensional line or curve. Although it is possible, for
example, to put charge on a piece of wire, it is more common to
encounter practical devices in which the charge is distributed over a
two-dimensional surface, as in the flat metal plates used in Thom-
son’s experiments. Mathematically, we can approach this type of
calculation with the divide-and-conquer technique: slice the surface
into lines or curves whose fields we know how to calculate, and then
add up the contributions to the field from all these slices. In the
limit where the slices are imagined to be infinitesimally thin, we
have an integral.

Field of a uniformly charged disk example 15

.A circular disk is uniformly charged. (The disk must be an in-

sulator; if it was a conductor, then the repulsion of all the charge

would cause it to collect more densely near the edge.) Find the

field at a point on the axis, at a distancezfrom the plane of the

disk.

.We’re given that every part of the disk has the same charge per

unit area, so rather than working withQ, the total charge, it will

be easier to use the charge per unit area, conventionally notated

σ(Greek sigma),σ=Q/πb^2.

Since we already know the field due to a ring of charge, we can

solve the problem by slicing the disk into rings, with each ring

extending fromrtor+ dr. The area of such a ring equals its

circumference multiplied by its width, i.e., 2πrdr, so its charge is

dq= 2πσrdr, and from the result of example 14, its contribution

to the field is

dEz=

k zdq

(

r^2 +z^2

)3/2

=

2 πσk zrdr

(

r^2 +z^2

)3/2

Section 10.3 Fields by superposition 599