W9_parallel_resonance.eps

Week 3: Potential Energy and Potential 115

Let’s step through this.

dl=a dθ (164)

defines a differential chunk of the ring. Its charge is:

dq=λ dl (165)

The differential potential of this chunk at a point on thez-axis is:

dV(z) =

kedq

r =

keλa dθ

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

(166)

We integrate over all of the chunks of charge that make up the ringby integratingθfrom 0 to 2π:

V(z) =

∫

dV =

∫ 2 π

0

keλa dθ

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

=

ke(2πa)λ

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

=

keQ

r

(167)

where we used the fact that 2πaλ=Q, thetotal charge of the ring!

This final answer we can easilyunderstandand might have even guessed without doing an
integral. Allof the charge of the ring is thesame distance rfrom the point of observation, and
potential dependsonlyon this distance (not on direction) so the potential is justketimes the total
charge divided by that distance.

If we do indeed try to find the electric field by differentiating this last result:

Ez = −

d

dz

ke(2πa)λ

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

=

ke(2πa)λz

(z^2 +a^2 )^3 /^2

=

keQz

(z^2 +a^2 )^3 /^2

(168)

Compare this to equation (47) above. Hmmm, looks like they are the same! However, evaluating
the potential integral and then taking its derivative seems (to me,at any rate) to bemuch easier
than doing the integral to find the field directly, with all of its components, and that’sbeforewe
evaluated theExandEyfields explicitly.

Note that we can exploit the insight we gained from this problem in a variety of ways to answer
certain questions concerning the potential “by inspection”. For example:

• A ring of chargeQa distanceR= (a^2 +z^2 )^1 /^2 from the point of observation;


• An arc of chargeQthat has angular widthθand radiusR, at the center of curvature;


• A hemispherical shell of chargeQwith a radiusR, at the center of the (hemi)sphere;


• Six charges each with chargeQ/6 arranged in a hexagon that has a distance 2Rbetween
  opposing corners, at the center;


• A single chargeQa distanceRfrom the point of observation;

all produce a potentialkeQ/Rat the point of observation indicated!In all these cases a total charge
ofQis arranged in various ways a distanceRfrom the point of observation. In potential direction
doesn’t matter, so all of the potentials of all of the charges that make up these systems add to the
one simple result.