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.
- Should I use the Cartesian coordinates system? Cylindrical? Spherical?
- How do I find the electric field from the electric potential?
- 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?
Best Answer
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.)