W9_parallel_resonance.eps

Week 2: Continuous Charge and Gauss’s Law 69

Example 2.1.2: Long Straight Line of Charge

dx

θ

dEx

dEy

dE

y r

x

λ

x

y

θ P

Figure 11: A straight line of charge with uniform charge per unit lengthλ.

In figure 11 we see a long straight line of charge. As before, we haveto choose a coordinate
system in terms of which to do the integral to add up the field components produced by all the little
chunks of charge that make up the line.

At first glance, it seems as though cartesian components are a natural choice for the problem, so
we start by using them. We want to find the field at an arbitrary pointP in space, so we pick one
and draw ay-axis through it such thatPis a (shortest) distanceyfrom the line. We pick a chunk
of charge of lengthdx, a distancexout from the origin. The charge of our chunk isagaingiven
by our magic spell: “The charge of the chunk is the charge per unit length of the chunk times the
length of the chunk”, or:

dq=λ dx (49)

Finally, the magnitude of the field is given by:

|dE~|=

kedq

r^2

=

keλ dx

(x^2 +y^2 )

(50)

We need in this case to evaluatebothdExanddEy, asExandEywill in general both be nonzero
(unlessP happens to be in the middle of the line, in which case we expectEx= 0. From the
triangles in the figure it is pretty obvious that:

dEx = −|dE~|sinθ (51)

dEy = |dE~|cosθ (52)

where we will assume that theθwe have drawn ispositivewhen swung out to the right in the positive
xdirection, and negative when it swings out in the direction of negativex. Noting that cosθ=y/r
we get:

dEy=

keλ dx

r^2

cosθ=

keλ dx

(x^2 +y^2 )

cosθ=keλy

dx

(x^2 +y^2 )^3 /^2

(53)

(for example). This, unfortunately, doesn’t look terribly easy to integrate!

In fact, this is one of the most difficult integrals we have to do in this course, not because it is
particularlydifficult but because it is one of the few times we have to integrate something other than
xndx, a simple trig function, or an exponential function. The problem is that as we varyx, bothr
andθvary as well! It turns out that this problem is easier to do if we convert it into atrigonometric
form using nothing buty(which is fixed) andθas ouronevariable. Thus:

x=ytanθ (54)