In the following problem, I have already solved for the value of the potential, and I would like to tackle the extra exercise, which asks for the electric field of a point quadrupole:

At every point in the $yz$ plane, find the electric potential of the quadrupole with three collinear charges $(+q, -2q , +q)$ that lie in the $z$ axis, with the middle charge at the origin. Consider $r \gg a$, $\vec{Q}$ is the quadrupole moment, and $\theta$ is the angle between $\vec{r}$ and $\vec{Q}$.

Hint: the quadrupole moment is $Q=\frac{1}{2} \sum_i q_i (3z_i^2-r_i^2)$.

As an extra exercise, use that result to find the electric field everywhere in the $yz$ plane.

With the help of some notes from Dr. J Tatum, I was able to understand where that value of Q comes from. So, I understand the use of Legendre polynomials and the binomial expansion to find the electric potential of a quadrupole. So the potential is given by: $$ V=\frac{1}{4\pi \varepsilon_0} \frac{qa^2}{r^3} (3\cos^2 \theta -1) = \frac{1}{4\pi \varepsilon_0} \frac{2qa^2}{r^3} (P_2(\cos \theta))$$ Where $P_n(x)$ represents the Legendre polynomials. (As a side note, I think the equations 3.12.3 and 3.12.4 in Tatum's notes are wrong).

To find the electric field I tried to use $r^3=(x^2+y^2+z^2)^{3/2}$ and then calculate $E=(-\frac{\partial V}{\partial x}, -\frac{\partial V}{\partial y}, -\frac{\partial V}{\partial z} )$, but I feel that's wrong, although I don't know why.

  1. Should I use the Cartesian coordinates system? Cylindrical? Spherical?
  2. How do I find the electric field from the electric potential?
  3. In case this problem's suggestion is not the most standard way to find the electric field, which method is? Is there a way to find the electric field directly without having to find the electric potential first?
  • 1
    $\begingroup$ As an aside: it is extremely misleading to treat the quadrupole moment as a vector. The quadrupole moment is a symmetric traceless tensor (i.e. matrix) with five independent components, and the moment used in the question is only one of those components. Just something to keep in mind for the future so that you don't carry a misconception away from this question's usage. $\endgroup$ – Emilio Pisanty Sep 21 '18 at 23:13
  • $\begingroup$ @EmilioPisanty thank you! I think that kind of clarifications is quite important. Where can I find more information about those five components? Any link with content that a college sophomore can understand? $\endgroup$ – evaristegd Sep 21 '18 at 23:20

To find the electric field I tried to use $r^3=(x^2+y^2+z^2)^{3/2}$ and then calculate $E=(-\frac{\partial V}{\partial x}, -\frac{\partial V}{\partial y}, -\frac{\partial V}{\partial z} )$, but I feel that's wrong, although I don't know why.

This is pretty close to the best approach, particularly if you make systematic use of the identity $$ \frac{\partial r}{\partial x_j} = \frac{x_j}{r} $$ for the cartesian partial derivatives of the radial coordinate.

The other simplification comes from noting that the cosine in the potential also appears in the coordinate $$ z=r\cos(\theta), $$ and that therefore it pays to multiply the top and bottom by an extra factor of $r^2$: \begin{align} V & \propto \frac{3\cos^2(\theta)-1}{r^3} \\ & = \frac{3r^2\cos^2(\theta)-r^2}{r^5} \\ & = \frac{3z^2-r^2}{r^5} \\ & = \frac{2z^2-x^2-y^2}{r^5} . \end{align} This gets you the potential as an explicit rational function - a polynomial in the Cartesian components divided by a simple function whose derivatives are also of that form. (Which means: you are therefore guaranteed an expression for each of the electric field components in the form $p_3(x,y,z)/r^7$, where $p_3$ is some homogeneous cubic polynomial.)

  • $\begingroup$ Thank you for the answer. Why is it more convenient to get an $r^5$ in the denominator? When I see $r$ and I want to differentiate, I just feel the immediate need to replace $r$ with $(x^2+y^2+z^2)^{1/2}$, because $r$ is a function of $x$ (and y and z). So, getting $r^5$ just seems to make things more complicated. Also, how do you get $r^7$ from $r^5$ in the denominator? $\endgroup$ – evaristegd Sep 22 '18 at 0:08
  • 1
    $\begingroup$ Basically, $r^5$ and $r^3$ are equally handleable from a differentiation perspective, but you trade an ugly cosine for a nice polynomial. Don't replace $r$ with its explicit value - calculate the partial derivatives once (as in this answer) and then use that identity exclusively when calculating the derivatives of the denominator. $\endgroup$ – Emilio Pisanty Sep 22 '18 at 0:12
  • $\begingroup$ As for the appearance of the $r^7$ denominator - that's an excellent exercise ;-). $\endgroup$ – Emilio Pisanty Sep 22 '18 at 0:12
  • $\begingroup$ I was getting an $r^6$ on the denominator when doing a mental calculation, but I guess I'll get that $r^7$ if I do it more carefully. Thank you for the help and your prompt replies! $\endgroup$ – evaristegd Sep 22 '18 at 0:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.