Potential of uniformly charged disk off axis

[finalnothing]

Homework Statement

As per Griffiths 3.21, I am given the on axis potential a distance r from a uniformly charged disk of radius R as a function of \sigma. Using this and the general solution for laplace's equation in spherical coordinates with azimuthal symmetry, calculate the first three terms in the general solution. Assume r>R.

Homework Equations

V(r,\theta)=\sum^\infty_{l=0}{(A_lr^l +\frac{B_l}{r^{l+1}})P_l(\cos\theta)}

V(r,0)=\frac{\sigma}{2\epsilon_0}(\sqrt{r^2+R^2}-r)

The Attempt at a Solution

As I know that V goes to 0 at infinity, all the A terms must be 0. Applying the boundary condition V(r, 0) and nothing that P_l(1)=1 leaves me with

V(r,0)=\sum^\infty_{l=0}{\frac{B_l}{r^{l+1}}}

Now, my major idea was to rewrite V(r, 0) in terms of u=\frac{R}{r} and then perform a taylor series expansion in terms of u. This will generate successive terms of the form \frac{C_l}{r^{l+1}}, then I simply assign my B variables equal to the C variables. However, I hit a moderate snag that I was not able to reason out. What point should I expand my taylor series about? Should I expand it about u=0 (r >> R) or about u=1 (r=R)? Or should I take the average of the two expansions? Each expansion does give different C values, when the problem expressed suggests this should not be the case. Am I simply going in the wrong direction or does this problem expect an approximate answer? Any help would be much appreciated.

 

[gabbagabbahey]

However, I hit a moderate snag that I was not able to reason out. What point should I expand my taylor series about? Should I expand it about u=0 (r >> R) or about u=1 (r=R)?

If you Taylor expand a function f(u) about the point u=u_0, you get terms in powers of (u-u_0). So, if you want terms in powers of u, you expand about u=0.

 

[finalnothing]

Thanks for your reply. However that is something I already considered. For the case where u=1 and I have terms (u-1)^n I simply expanded that into a polynomial of degree n in u. I then grouped all of the u^n terms together for my final polynomial. Here are the equations and results I have.

V(u)=\frac{u\sigma}{2\epsilon_0R}(\sqrt{1+u^2} - 1) where u=\frac{R}{r}

I expand to the fourth term of the taylor series. Expanding about point u = 0 gives
V(r, 0)=\frac{u^3\sigma}{4\epsilon_0R} as most of the terms in the derivatives of V are a product with u. However, when I expand about u=1 I get (for 3 terms, really messy)
V(r, 0)=\frac{\sigma}{2\epsilon_0R}(\frac{3\sqrt{2}}{4}-(1+\sqrt{2})u+\frac{5\sqrt{2}}{4}u^2)
This is starting to get really messy. However, the difference in approximations is quite drastic since nearly all terms of V(u) are product of u and thus disappear in the u=0 approximation but remain in the u=1 approximation. At this point I'm not sure if expanding V(r, 0) in terms of r^{-l} is the correct response. If anyone has any suggestions on another route to take this problem that would be great.

//note, preview is showing latex images for my previous post, not sure if my tex is correct

 

[gabbagabbahey]

However that is something I already considered. For the case where u=1 and I have terms (u-1)^n I simply expanded that into a polynomial of degree n in u.

That's fine, in theory. However in order to find the coefficient of say u^2 in that manner, you'd have to expand out every term in the infinite Taylor series (terms like (u-1)^{677} will contain multiples of every power of u from 1 to 677, including u^2... You would quite literally be calculating coefficients for eternity.