Multipole Expansion of Dipole on Z-Axis w/ Spherical Harmonics

[shikhapunia]

given a dipole on z-axis(+q at z=a and -q at z= -a) , find out the non vanishing multipoles using spherical harmonics.
can somebody tell me how to do this problem using spherical harmonics..because when we write charge density using dirac delta function in spherical polar coordinates. then we get
phi = tan^-1 (0/0) because x and y coordinates for both the charges are zero.

 

[vela]

Please show your work. I can't really tell what you're doing.

 

[shikhapunia]

In spherical polar coordinates charge density can be written as Ʃqi*δ(r-ri)*δ(θ-θi)*δ(∅-∅i).
where ∅=tan^-1(y/x) . since its a dipole on z-axis therefore ∅'=tan^-1(0/0) . i don't know how to deal with this form of ∅.

 

[vela]

You can pick an arbitrary angle since it doesn't matter anyway when $\theta = 0\text{ or }\pi$.

 

[shikhapunia]

does that mean its ∅ independent? i.e. it has azimuthal symmetry.

 

[vela]

Yes, the charge distribution is azimuthally symmetric.

 

[shikhapunia]

ok..thanks a lot

 

[shikhapunia]

Generally a surface which is symmetric about z-axis is s.t.b azimuthally symmetric like a sphere as we can easily see its symmetric. How can a dipole along z-axis be azimuthally symmetric?

 

[vela]

Does the dipole look different if you rotate the system about the z-axis?

 

[shikhapunia]

ok..now its clear.
thankyou.

 

[sunjin09]

In spherical polar coordinates charge density can be written as Ʃqi*δ(r-ri)*δ(θ-θi)*δ(∅-∅i).
where ∅=tan^-1(y/x) . since its a dipole on z-axis therefore ∅'=tan^-1(0/0) . i don't know how to deal with this form of ∅.

a point charge in spherical coordinates is given by $\rho(r,\theta,\phi)=q\delta(r-r_q)\delta(\theta-\theta_q)\delta(\phi-\phi_q)/(r^2\sin\theta)$

 

[shikhapunia]

oh! yeah..thanks