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[/color] 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...