How to introduce "quandrupole" potential to the students to quickly appreciate the concept
Feb. 15, 2026.
Over the years I always enjoyed teaching the multi-pole expansion of the scalar potential in electrodynamics (PH1020). While I enjoyed it (while teaching a class of 100) and so did some of the first, middle benches; the back benches had real problems with the jargon. Here is the step wise "gyaan" with three STEPS A, B and C for all.
As the Step 0: We know the potential at "r" due to a charge q as kq/r. (1/k is = 4\pi\epsilon_0 ).
STEP A: 3 sub-steps under A:
2. Now, as indicated in the figure, replace that negative charge with a positive q and additionally place a -2q at the origin (0,0,0). Notice you actually have 4 charges. Notice two points (i) the charges are kept along the z-axis, and (ii) this becomes an axially symmetric charge configuration. (The system can also be expressed as two dipoles, but do not bother).
3. By using the superposition, calculate the potential due to this charge configuration at r (this time, for convenience, strictly at a point P along the z-axis). Notice you need distances as r+d and r-d, as well as r. Assume d is small as compared to r. [call this potential at r as VP(quad)] as k(2qd^2)/r^3 ]. You notice two things here: (i). instead of q in the numerator (Step 0) we have 2q (and d^2) and (ii). instead of r in the denominator we have r^3. Compare the dimensions, all well. You have already stumbled upon the "quadrupole" term of the "multi-pole expansion of the potential" here.
Special Comments:
(i). Overall, in general, for a system of charges (whether discrete or continuous) the potential to be proportional to 1/r (single charge, or monopole), 1/r^2 (dipole) and 1/r^3 (a qadrupole) .
(ii). It is extremely helpful to talk in terms of monopole moment, dipole moment and quadrupole moment.
(iii). For example, for a system of two equal charges of same sign, kept equal
and opposite distance from the origin (See figure without -2q), the
dipole moment will be zero. But monopole term is not zero. Thats where the need of "multi-pole expansion comes in!
STEP B (sub step 4, 5 and 6 ):
4. Write a scalar potential V(M) or due to monopole (same as step 0). Monopole moment is a scalar quantity -as it is just the charge q, or sum of charges for several charges.
5. Write a potential (due to dipole) with entry of dipole moment vector p\vec .
p\vec = px x^ + px y^ + pz z^.
I can also write it as a column matrix with three components as px , py and pz.
6. Now comes the Quadrupole moment tensor - can you guess how many components it can have? Well as you saw above (sub STEP 3) for an axially symmetric case of 4 charges, it had only one component. Yes for an axially symmetric case. But for a general case it is a 3x3 matrix. It is known as a Tensor of rank 2. Its written as Qjk with double arrow on top with j and k as x,y,z.
To understand symmetry, you can imagine, if I have a spherical charge distribution (symmetrically kept about origin) with charge q, I will have neither the dipole moment (see special comment (iii) above) nor the components of quadrupole moment (due to symmetry- see below).
7. Now just believe me that when we write an expression for the 9 components of the tensor Q<->, if for for every charge (+q) at x, there is an equal (+q) charge at (-x), then, it happens that Qxy=Qyx and so on (consider it for y and z). Therefor, we have 6 components to find out.
8. We also find that the diagonal elements follow Qxx+Qyy+Qzz=0. This way, we are okay with 5 components for any charge configuration.
9. Now comes a special case for cylindrical symmetric (with cylinder axis along z) case. In this case, we have symmetry about x and y such that the Qxx and Qyy also become equal.
STEP C (sub step 10):
10. Relate the system in substep 9 with the subtep 3. Now is the time to introduce the topic : derive the expression for multipole expansion of the potential as a regular class.
best wishes,
Comments/ critique welcome.
Prem Feb. 16, 2026
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