Find electric potential and field from Electric Quadrupole

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Homework Statement



So the problem revolves around a Linear Electric Quadrupole, with two positive charges on z-axis (in positions s and -s from origo) and two negative charges in origo.

Hand-drawn diagram of a linear electric quadrupole on the z-axis: charges labelled q at height s and q at -s, and -2q at the origin. A vector r runs from the origin at angle theta from the z-axis, with vectors s and -s marked along the axis.


1. Show that potential from a position R can be written as
Formula for electric potential V(r) = (1/(4 pi epsilon0)) (q / r^3) (3 (s . r-hat)^2 - s^2).


2. The quadrupole moment can be written as
Formula for electric potential V(r) = (1/(4πε₀)) · (r̂ · Q · r̂)/r³, where Q is the quadrupole moment tensor and r̂ is the unit vector.

where
Quadrupole moment tensor Q = q s^2 times a 3x3 matrix with rows (-1, 0, 0), (0, -1, 0), (0, 0, 2).


Show that
Equation: r̂ · Q · r̂ = q (3 (s · r̂)^2 − s^2), with r̂ · Q · r̂ on the left and q(3(s·r̂)^2 − s^2) on the right.


Show that the electric field of the linear quadrupole can be written as

Equation for the electric field E(r) = (1/(4 pi epsilon_0)) (3q/r^4) [ (3(s . r-hat)^2 - s^2) r-hat - 2 (s . r-hat)(s . theta-hat) theta-hat ]


We now change the Linear quadrupole out with a quadratic quadrupole

Diagram of a linear electric quadrupole on x and z axes: charges labelled -q at z = s and z = -s, and +q at x = s and x = -s, with spherical position vector r at angle theta from the z-axis, and the separations labelled 2s.


Show that the quadrupole moment can be written as

Formula for electric potential V(r) = (1/(4πε0)) (r̂·Q·r̂)/r^3, with Q the quadrupole tensor and r̂ the unit vector of position r.

where
Equation Q = 3q s^2 times a 3x3 matrix with diagonal entries 1, 0, -1 and all off-diagonal entries 0, representing a quadrupole moment tensor.


Homework Equations





The Attempt at a Solution



I've done number 1. It was pretty straight forward, and I did it with Law of Cosines and some series expansion, but I'm completely stuck at number 2.

After some googling, it seems that there's tensors involved in this one, but I can't remember the professor talking about tensors for more than maybe 1 min, with a very brief example on Ohms Law (I think).

I just can't figure out where to start, as I have no intuitive image of how tensors work, or how they're to be applied in this problem.

As for the electric field, it should just be the negative Gradient of the potential.

But again, I can't get my head around tensors, or how to derive them from the picture or equations I've been given.
 
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Must have been quite some work, this number 1. Can you show it ?

In return I show how I work out ##\hat r \cdot Q \cdot \hat r## in cartesian coordinates: begin with Q.r (factor ##q\,s^2## I leave out and add at the end)
(I assume you are familiar with matrix multiplication) :
$$ \left (\begin{array}{cc}-1&0&0\\0&-1&0\\0&0&2\end{array}\right )\, \left (\begin{array}{cc}\cos\phi \sin\theta\\ \sin\phi \sin\theta \\ \cos\theta\end{array}\right )=\left (\begin{array}{cc}-\cos\phi \sin\theta\\ -\sin\phi \sin\theta \\ 2\cos\theta\end{array}\right ) $$ Then
$$ \left (\begin{array}{cc}\cos\phi \sin\theta& \sin\phi \sin\theta & \cos\theta\end{array}\right ) \,\left (\begin{array}{cc}-\cos\phi \sin\theta\\ -\sin\phi \sin\theta \\ 2\cos\theta\end{array}\right ) = -\cos^2\phi \sin^2\theta - \sin^2\phi \sin^2\theta + 2 \cos^2\theta = - \sin^2\theta + 2 \cos^2\theta = 3 \cos^2\theta - 1$$ So that with ##(s\cdot \hat r)^2= s^2 \cos^2\theta ## you get $$\hat r \cdot Q \cdot \hat r = q\,\left (3(s\cdot \hat r)^2 -s^2 \right )$$

And for now, no need for tensors of higher rank (tensors are arrays, matrices. Rank 0 is a scalar, rank 1 a vector, rank 2 a matrix -- I think I remember)
 
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I'm doing the exact same problem, and I've gotten to the quadratic quadrupole... Anyone know how to derive that stuff? Is it just rule of cosines for all of the charges, approximation with Taylor and plugging?
 
Hello Taaben, and welcome to PF.
The derivation of this expansion can be googled, or found in books like Griffiths, Jackson, ...

It expands ##1/|\vec r - \vec r' |## in spherical harmonics. The nice thing is that for symmetries in the charge distributions (monopole, dipole, quadrupole) the series expansion of ##\int \rho /|\vec r - \vec r' |## terminates.

Sho how is our dreamspacer doing ?