2—Infinite Series 53Ifc aorbthen this is really not very different from the preceding case, whereaandbare zero.
Ifa= 0this isFz→Q 1 Q 2 c
2 π^2 0∫π/ 20dθ
[
c^2 +b^2] 3 / 2 =
Q 1 Q 2 c
2 π^2 0π/ 2
[
c^2 +b^2] 3 / 2 =
Q 1 Q 2 c
4 π 0[
c^2 +b^2] 3 / 2 (24)
The electric field on the axis of a ring is something that you can compute easily. The only component of the
electric field at a point on the axis is itself along the axis. You can prove this by assuming that it’s false. Suppose
that there’s a lateral component ofE~ and say that it’s to the right. Rotate everything by 180 ◦about the axis
and this component ofE~ will now be pointing in the opposite direction. The ring of charge has not changed
however, soE~must be pointing in the original direction. This supposed sideways component is equal to minus
itself, and the only thing that’s equal to minus itself is zero.
All the contributions toE~ except those parallel the axis add to zero. Along the axis each piece of charge
dqcontributes the component
bcdq
4 π 0 [c^2 +b^2 ].√ c
c^2 +b^2The first factor is the magnitude of the field of the point charge at a distancer=
√
c^2 +b^2 and the last factor
is the cosine of the angle between the axis andr. Add all thedqtogether and you getQ 1. Multiply that byQ 2
and you have the force onQ 2 and it agrees with the expressions Eq. ( 24 )
Ifc→ 0 thenFz→ 0 in Eq. ( 23 ). The rings are concentric and the outer ring doesn’t push the inner ring
either up or down.
But wait. In this case, wherec→ 0 , what ifa=b? Then the force should approach infinity instead of
zero because the two rings are being pushed into each other. Ifa=bthen
Fz=Q 1 Q 2 c
2 π^2 0∫π/ 20dθ
[
c^2 + 4a^2 sin^2 θ