W9_parallel_resonance.eps

Week 2: Continuous Charge and Gauss’s Law 67

Example 2.1.1: Circular Loop of Charge

dθ

dE

r

dEz

z

a

dl = a dθ

φ

λ

x

y

z

Figure 10: A charged ring with charge per unit lengthλ.

In figure??above we see a circular ring of charge of radiusaand uniform charge per unit length:

λ=Q

L

= Q

2 πa

(39)

Our job is to find the electric field at an arbitrary point on thez-axis, a point with sufficient
symmetry to make the evaluation fairly straightforward^32.

We begin by finding a small chunk of charge on the ring expressed in some coordinate we can
integrate over. In this case the best possible coordinate system to use is (fairly obviously)cylindrical
coordinates, so that we can locate a small chunk on the ring at an angleθswung around in the
counterclockwise direction from the positivex-axis. The angular width of the chunk is thendθ, and
the length of the arc subtended isdℓ=a dθ.

From the previous section we recall that we need to find the chargeof this little chunk of arc,
repeating the litany: “the charge in the chunk is the charge per unitlength, times the length of the
chunk”. That is:

dq=λ dℓ=λa dθ=Q

dθ

2 π

(40)

where the last form is clearly thefractionof the total charge that lies inside the tiny subtended arc.
The magnitude of the field produced by this little chunk of charge at the pointzon the axis is:

|dE~|=kedq

r^2

=keλadθ

z^2 +a^2

(41)

where we have used the pythagorean theorem to evaluater=

√

z^2 +a^2 as drawn in the figure.
This vector has three components. All we need to worry about is thez-component from the
symmetry of the ring. The field at a point on the axis cannot change as we rotate the coordinate
system around thez-axis because the ring of charge looks the same as we do. Therefore it cannot
havexorycomponents as these wouldchangeas we rotated the coordinate system. However, for
the sake of completeness (and to give you something to figure out on the picture) I’ll put down the
xandycomponents as well:

dEx = −|dE~|sinφcosθ (42)

dEy = −|dE~|sinφsinθ (43)

dEz = |dE~|cosφ (44)

(^32) Wecoulduse the same general approach to find the field at an arbitrary point in space, but thecalculusand
geometryrequired to get an actual would become very difficult – so difficult that in real life one would be very likely
to concede finding an analytic solution as too difficult and resort to the use of a computer instead.