Dipole/quadrupole moments of wires in x-y plane

[quasar_4]

Homework Statement

Two identical, infinitely thin, homogeneously charged cylinders (length L) are placed in the xy plane and form an angle phi (so that they make a V shape). Calculate the angle phi, for which the dipole contribution to the electric field potential at (x0,0,0) is exactly equal to the quadrupole contribution.

Homework Equations

Dipole moment:

$$\vec{p} = \int x' \rho(x') d^3 x'$$

Quadrupole moment tensor

$$Q_{ij} = \int (3 x'_i x'_j - r'^2 \delta_{ij}) \rho(x') d^3 x'$$

where the delta is the Kronecker delta.

The Attempt at a Solution

Each cylinder can be thought of as having a charge density lambda. So I thought, ok, we need to express the charge density in all space. To do this, I expressed the charge density everywhere using delta functions:

$$\rho(x) = \lambda \delta(z) \left[\delta(\phi) + \delta(\phi-\phi_0) \right]$$

and then I tried to find px, py, pz, and the components of Qij by integrating, using the vector x' = (x,y,z). I parameterized x, y, z using x = r cos(phi), y = r sin(phi), z=z, so that I could integrate in cylindrical coordinates.

But it got really complicated when I got to the quadrupole tensor components, and I've been told by my instructor that the answer should be simple, so I think this method might be wrong. Also, I don't know what to do once I *have* the components of p and Qij - I can't just equate them, as they aren't the same quantity physically. How would I use these components to find the angle phi?

Another approach might be to find the dipole and quadrupole terms for each wire separately, then add them together, but I don't quite understand how to do this, either.

Can anyone help? I've been stuck for hours and hours and this is due kind of soon...

 

[gabbagabbahey]

Homework Statement

Two identical, infinitely thin, homogeneously charged cylinders (length L) are placed in the xy plane and form an angle phi (so that they make a V shape). Calculate the angle phi, for which the dipole contribution to the electric field potential at (x0,0,0) is exactly equal to the quadrupole contribution.

Homework Equations

Dipole moment:

$$\vec{p} = \int x' \rho(x') d^3 x'$$

Quadrupole moment tensor

$$Q_{ij} = \int (3 x'_i x'_j - r'^2 \delta_{ij}) \rho(x') d^3 x'$$

Do you really need to calculate the quadrupole moment tensor to find the quadrupole contribution to the potential?

Each cylinder can be thought of as having a charge density lambda. So I thought, ok, we need to express the charge density in all space. To do this, I expressed the charge density everywhere using delta functions:

$$\rho(x) = \lambda \delta(z) \left[\delta(\phi) + \delta(\phi-\phi_0) \right]$$

This doesn't have the correct units...do you get $2\lambda L$ when you integrate this over all space?

 

[quasar_4]

Ah. Ok, here's my new thought. We can express the potential for each rod separately, then use superposition to add them together. This is given in a series of n, so we can find the dipole contribution using n = 1 and the quadrupole contribution using n = 2. Then equating them would give the angle phi. Is that right?

 

[gabbagabbahey]

Ah. Ok, here's my new thought. We can express the potential for each rod separately, then use superposition to add them together. This is given in a series of n, so we can find the dipole contribution using n = 1 and the quadrupole contribution using n = 2. Then equating them would give the angle phi. Is that right?

The dipole term is proportional to 1/r² and the quadrupole term is proportional to 1/r³...other than that, it sounds like a good plan to me...

 

[paranormal]

First of all, it is important to note that the dipole moment and quadrupole moment are two different physical quantities and cannot be directly compared or equated. The dipole moment represents the net charge distribution of a system, while the quadrupole moment represents the distribution of charge asymmetry. Therefore, it is not possible for the dipole contribution to be exactly equal to the quadrupole contribution.

However, it is possible to find the angle phi at which the contributions from both the dipole and quadrupole moments are equal. To do this, you can use the equations given in the problem to calculate the dipole and quadrupole moments for each wire individually. Then, add these moments together to get the total dipole and quadrupole moments for the system.

Next, you can use the electric field potential formula for a dipole and a quadrupole, which can be found in any standard electromagnetism textbook, to find the total electric field potential at the point (x0,0,0). Set these two equations equal to each other and solve for the angle phi. This angle will give you the direction at which the contributions from the dipole and quadrupole moments are equal.

It is important to note that this solution assumes that the two wires are placed symmetrically with respect to the origin. If this is not the case, the solution will be more complicated and involve vector calculus. Additionally, the solution may also be different if the wires are not infinitely thin or if they have a non-uniform charge distribution.

In summary, while it is not possible for the dipole and quadrupole moments to be exactly equal, it is possible to find the angle phi at which their contributions to the electric field potential are equal. This can be done by calculating the individual moments for each wire and then using the electric field potential formula to solve for the angle.