Calculating Quadrupole Moment of Sphere w/ Surface Charge

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The discussion centers on calculating the quadrupole moment of a spherical shell with a surface charge density defined as sigma = sigma_o * cos(theta). The user is questioning the integration process and the role of the (r')^2 factor in the quadrupole moment equation. They note that the vector product yields zero, indicating that higher multipole moments, including the quadrupole, may also be zero. The user concludes that only the dipole moment is expected to be non-zero due to the orthogonality of spherical harmonics, which suggests that the quadrupole moment does not contribute in this case. The focus remains on the implications of charge density and spherical harmonics in determining multipole moments.
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1.

Looking for quadrupole moment of spherical shell with surface charge (sigma)=(sigma_o)*cos(theta) and (sigma_o) => constant
Sphere is at the origin with radius=a.

2.

Well, I think I am using the right equation, by integration of the quadrapole moment taken from the quadrupole term, but I am questioning my [(r')^2] factor. 3.

I use vector product I get a zero. I guess a vector crossed with itself is always zero because of no span. I have already found the monopole and dipole (monopole =0), and I think I take the vector product of r', and not the dot product, because the dot product is a scalar... and I found by another theorem that whether it is shell or sphere, the quadrapole and higher terms are zero. So is this where the quadrapole zeroes out?? from the [(r')^2] factor?
 
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i'm pretty sure only the dipole moment should be non-zero. you can write all the moments in terms of the charge density integrated against spherical harmonics. E.g., Y_{0,0} for monopole term, Y_{1,m} for dipole term, Y_{2,m} for quadrupole term, etc. But since the charge density itself is proportional to Y_{1,0} only the l=1 multipole moments should be non-zero due to orthogonality of spherical harmonics.
 
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