W9_parallel_resonance.eps

110 Week 3: Potential Energy and Potential

3.4: Examples of Computing the Potential

Example 3.4.1: Potential of a Dipole on thex-axis

−a

+a +q

x

−q

x

z

r

Figure 28: A simple dipole aligned with thez-axis.

This is the same dipole studied in the the chapter on field. Find thepotentialat an

arbitrary point on thex-axis.

This problem is deceptively simple. We know from the superposition principle that the potential
is:

V(x) =

∑^2

i=1

keqi

ri

= keq

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

− keq

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

= 0 (155)

This is absolutely correct – the potential of a dipole vanishes on theentire planethat symmetrically
bisects the line connecting the charges.

The “deception” occurs when we try to compute thefieldby usingE~=−∇~V. We are ever so
tempted to go e.g.:

Ez=−

dV

dz

=−

d 0

dz

= 0 (156)

which is simple, easy, andwrong!The problem is that even though the functionV(x, y, z) is zero at
a point that doesnotmean that itsslopeis zero at the point! We have to use L’Hopital’s Rule to
evaluate a derivative at a point where its lower order derivatives or value are zero.

What this means is that we have to evaluate the function forV(x, y, z)nearbut notonthe point
where the function is zero, take the desired derivative, and then let the parameter that describes
that nearness go to zero. In this case, we need to findV(x, z) for somesmallz(near zero), take the
derivative, and let the value ofzin the derivative go to zero. See if you can draw pictures to verify
the following algebra, for a pointz≪a≪xabove the point on thex-axis.

V(x, z) = keq

(x^2 + (a−z)^2 )^1 /^2

− keq

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

(157)