Spherical Multipoles for q charges at different locations

[MathematicalPhysicist]

Homework Statement

2 q charges are placed at (0,0,a) and (0,0,-a) and a third charge -2q is placed at the origin.
1. find the charge density distribution in spherical coordinates.
2. Find all the spherical multipoles of the above distribution.

Homework Equations

The spherical multiple formula:
$$q_{l,m}=\int r^2 r^l sin(\theta) dr d\theta d\phi Y*_{l,m}(\theta,\phi) \rho$$ where rho is the charge density, Y_l,m are the spherical harmonics.

The Attempt at a Solution

1. Here's what I got by dirac delta function:
$$\rho =-2q\delta (r)\delta (\theta) \delta (\phi)+ q\delta(r-a) \delta(\theta-\pi/2) delta(\phi) +q\delta (r-a) \delta(\theta +\pi/2) \delta (\phi)$$.

2. q_{0,0}=1/sqrt(4pi)[qa^2-qa^2]=0
q_{1,0}=sqrt(3/4pi)[0+0]=0
q_{1,+-1}=-+sqrt(3/8pi)[qa^3+qa^3]=-+2sqrt(3/8pi) qa^3
q_{2,0}=0
and the rest spherical multipoles are vanishing cause Y_{l,m} is a multiplication of P_l(cos(theta)) (Legendre polyonimal) and exp(m\phi) and cosine(+-pi/2)=0

So we are left with only one multiple here, am I right or wrong?

 

[gabbagabbahey]

1. Here's what I got by dirac delta function:
$$\rho =-2q\delta (r)\delta (\theta) \delta (\phi)+ q\delta(r-a) \delta(\theta-\pi/2) delta(\phi) +q\delta (r-a) \delta(\theta +\pi/2) \delta (\phi)$$.

That doesn't look right to me...$\theta=\pm\pi/2$ corresponds to a point in the xy-plane, not on the z-axis...

 

[MathematicalPhysicist]

So theta should be pi and -pi?

If it's pi instead of pi/2 then I get that all the multipoles are zero because sin(+-pi)=0, and this doesn't look right to me either.

 

[gabbagabbahey]

So theta should be pi and -pi?

Yes.

If it's pi instead of pi/2 then I get that all the multipoles are zero because sin(+-pi)=0, and this doesn't look right to me either.

Slow down there...you still have another problem with your charge density...if you integrate each term over all space, you should get the charge of each corresponding point charge...do you?

 

[MathematicalPhysicist]

Yes you are quite right I need to divide by $$4\pi a^2$$ which is the surface area of a sphere with radius a.

 

[MathematicalPhysicist]

No, obviously this isn't right either, I guess I need to divide this by r^2 sin(theta), I mean each term in the density I need to divide by this factor.

 

[gabbagabbahey]

No, obviously this isn't right either, I guess I need to divide this by r^2 sin(theta), I mean each term in the density I need to divide by this factor.

Okay, that should work...now what do you get for the multipoles?