W9_parallel_resonance.eps

70 Week 2: Continuous Charge and Gauss’s Law

so

dx=

y dθ

cos^2 θ

(55)

and
r=

y

cosθ

(56)

If we substitute these into the expressions above we get:

dEy=

keλ dx

r^2

cosθ=keλ

(

y dθ

cos^2 θ

)(

cos^2 θ

y^2

)

cosθ=

keλ

y

cosθdθ (57)

which lookseasyto integrate! The limits of integration are the angles to the dotted lines that point
at the ends of the line, which we will callθ 1 on the left,theta 2 on the right. Thus:

Ey=

keλ

y

∫θ 2

θ 1

cosθdθ=

keλ

y

(sinθ 2 −sinθ 1 ) (58)

(where we should carefully note thatθ 1 in the figure above isnegativeas drawn).

If we evaluateExeverything is the sameexcept that there is an overall minus sign and we
integrate over sinθ dθinstead, to get:

Ex=−keλ

y

∫θ 2

θ 1

sinθdθ=keλ

y

(cosθ 2 −cosθ 1 ) (59)

An interesting consequence of this result is that we can easily evaluate the field a distancey
away from aninfiniteline of charge (that still has a uniform charge per unit lengthλ. In that case,
θ 1 =−π/2 andθ 2 =π/2. We get:

Ex(∞) = 0 (60)

Ey(∞) =

2 keλ

y

(61)

where we should recall thateverypointPhas anx-coordinate in the middle of an infinite line of
charge! Remember this result for later, where we will obtain it again using Gauss’s Law.

Example 2.1.3: Circular Disk of Charge

dE dE

z (z + r )^2 2 1/2

θ

φ

θ r

x

y

z

z

dr

r d dA = r dr d θ

σ

R

P

Figure 12: A charged disk with charge per unit areaσ.

In figure 12 above we see a disk of charge with a uniform charge density:

σ=

Q

πR^2

(62)