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.