W9_parallel_resonance.eps

Week 2: Continuous Charge and Gauss’s Law 73

s

y

x

α

θ

dq

r

φ

R−r

R

dE

dE dEz

Figure 13: Geometry for finding the field of a uniform sphericalshellof constant charge densityσ
by direct integration, both inside and outside. Note thatθis the angle sweptdownfrom the positive
zaxis (the equivalent of “latitude’, although measured down from thenorth pole and not up from
the equator) andφis the angle to thex-yprojectionsof the point, measured counterclockwise from
the positivex-axis, the equivalent of ‘longitude’). We callφtheazimuthal angle.

sphere:
dA=r^2 sin(θ)dθ dφ=−r^2 d(cosθ)dφ (71)

but if you are not, it is a great time to review them.

For example, from this point on I’m simplifyingall spherical integrals overθby using the
clever identity:
sin(θ)dθ=−d(cos(θ)) (72)

tochange variablesfromθto cos(θ) so that:
∫π

0

f(cos(θ)) sin(θ)dθ=

∫ 1

− 1

f(cos(θ))dcos(θ) =

∫ 1

− 1

f(x)dx (73)

This trick doesn’t always work, but in physics a lot of time it does and when it does it is really
useful!

Consider, then, the small differential chunk of areadAof charge in figure 13. We know from our
usual rule that the charge in the chunk is the charge per unit volumetimes the volume of the chunk,
or:
dq=σdA=σr^2 d(cosθ)dφ (74)

We know that the field ofjust this chunkat the pointPis has a magnitude:

dE=

kedq

s^2

=keσ

(

r^2 d(cosθ)dφ

s^2

)

(75)

Finally, we only care (for the moment, anyway) aboutdEzso we might as well write it down
too:

dEz=dE cos(α) =keσ

(

r^2 d(cosθ)dφ

s^2

)

cos(α) (76)

which we can rewrite using the geometry in figure 14 as:

dEz=keσ

(

r^2 d(cosθ)dφ

s^2

)(

R−rcos(θ)

s

)

(77)

convention, that is, the second of the two pictures on the right.