Find electric potential and field from Electric Quadrupole

[dreamspace]

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.

1. Show that potential from a position R can be written as

2. The quadrupole moment can be written as

where

Show that

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

We now change the Linear quadrupole out with a quadratic quadrupole

Show that the quadrupole moment can be written as

where

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.

 

[BvU]

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)

 

[Taaben]

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?

 

[BvU]

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 ?

 

[HalfLostAndSearching]

Hello,

Thank you for reaching out for help with your homework problem. I understand that you are having difficulty with the concept of tensors and how they apply to the problem of finding the electric potential and field from an electric quadrupole.

First, let's define what a tensor is. A tensor is a mathematical object that describes the relationships between different coordinate systems. In the case of electromagnetism, tensors are used to describe the relationship between electric and magnetic fields in different reference frames.

In this problem, we are dealing with a linear electric quadrupole, which consists of two positive charges on the z-axis and two negative charges at the origin. To find the potential from a position R, we can use the principle of superposition and add up the potentials from each individual charge. This results in the first equation you provided:

V(R) = (1/4πε0) * [(q1/R1) + (q2/R2) + (q3/R3) + (q4/R4)]

Where q1, q2, q3, q4 are the charges and R1, R2, R3, R4 are the distances from each charge to the position R.

Now, let's move on to the quadrupole moment. The quadrupole moment is a measure of the charge distribution of the quadrupole. It is defined as the second moment of the charge distribution, which is given by the equation you provided:

Qij = ∑qk * (Rik * Rjk)

Where qk is the charge at position Rk.

To show that Qij = 0 for a linear quadrupole, we need to use the fact that the charges are arranged symmetrically along the z-axis. This means that R1 = -R2 and R3 = -R4, and Rik = 0 for all other combinations. Substituting these values into the equation for the quadrupole moment, we get:

Qij = q1 * (-Ri * -Rj) + q2 * (Ri * -Rj) + q3 * (-Ri * Rj) + q4 * (Ri * Rj)

= q1 * (Ri * Rj) + q2 * (-Ri * Rj) + q3 * (-Ri * Rj) + q4