Potential of uniformly charged disk off axis

In summary, the problem at hand involves finding the first three terms in the general solution for a uniformly charged disk, using the given on axis potential and the general solution for Laplace's equation in spherical coordinates with azimuthal symmetry. The major idea is to rewrite the potential in terms of u=R/r and perform a Taylor series expansion, but there is a dilemma as to which point to expand the series about. Expanding about u=0 gives vastly different results compared to expanding about u=1, and expanding about u=1 would require expanding every term in the infinite series. Another approach may be necessary to solve this problem.
  • #1
finalnothing
2
0

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 [tex]\sigma[/tex]. 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 [tex]r>R[/tex].

Homework Equations



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

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

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 [tex]P_l(1)=1[/tex] leaves me with

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

Now, my major idea was to rewrite V(r, 0) in terms of [tex]u=\frac{R}{r}[/tex] and then perform a taylor series expansion in terms of u. This will generate successive terms of the form [tex]\frac{C_l}{r^{l+1}}[/tex], 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.
 
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  • #2
finalnothing said:
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 [itex]f(u)[/itex] about the point [itex]u=u_0[/itex], you get terms in powers of [itex](u-u_0)[/itex]. So, if you want terms in powers of [itex]u[/itex], you expand about [itex]u=0[/itex].
 
  • #3
Thanks for your reply. However that is something I already considered. For the case where u=1 and I have terms [tex](u-1)^n[/tex] I simply expanded that into a polynomial of degree n in u. I then grouped all of the [tex]u^n[/tex] terms together for my final polynomial. Here are the equations and results I have.

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

I expand to the fourth term of the taylor series. Expanding about point u = 0 gives
[tex]V(r, 0)=\frac{u^3\sigma}{4\epsilon_0R}[/tex] 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)
[tex]V(r, 0)=\frac{\sigma}{2\epsilon_0R}(\frac{3\sqrt{2}}{4}-(1+\sqrt{2})u+\frac{5\sqrt{2}}{4}u^2)[/tex]
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 [tex]r^{-l}[/tex] 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
 
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  • #4
finalnothing said:
However that is something I already considered. For the case where u=1 and I have terms [tex](u-1)^n[/tex] 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 [itex]u^2[/itex] in that manner, you'd have to expand out every term in the infinite Taylor series (terms like [itex](u-1)^{677}[/itex] will contain multiples of every power of [itex]u[/itex] from 1 to 677, including [itex]u^2[/itex]... You would quite literally be calculating coefficients for eternity.
 
  • #5




Thank you for your question. It seems like you are on the right track in terms of using the boundary condition V(r, 0) to solve for the B coefficients. However, instead of expanding in terms of u, I would suggest using the definition of Legendre polynomials to rewrite the potential V(r, 0) in terms of Legendre polynomials. This will allow you to use the orthogonality property of Legendre polynomials to solve for the B coefficients.

As for your question about which point to expand the Taylor series at, it is important to note that the potential is only defined for r > R, so expanding around u = 0 (r >> R) would not be appropriate. Expanding around u = 1 (r = R) would give you a good approximation for the potential near the edge of the disk, but it may not be accurate for larger values of r. A better approach would be to take an average of the two expansions, as this would give you a better approximation for the potential over a larger range of r values.

I hope this helps guide you in the right direction. Keep in mind that this problem may require an approximate solution, as finding the exact solution for a uniformly charged disk off axis can be quite difficult. Good luck with your calculations!
 

Related to Potential of uniformly charged disk off axis

1. What is the potential of a uniformly charged disk off axis at a given point?

The potential of a uniformly charged disk off axis at a given point can be calculated using the equation V = kQ/r, where V is the potential, k is the Coulomb constant, Q is the charge of the disk, and r is the distance from the disk to the point.

2. How does the potential of a uniformly charged disk off axis differ from that of a uniformly charged disk on axis?

The potential of a uniformly charged disk off axis is different from that of a uniformly charged disk on axis because the distance from the point to the disk is not perpendicular to the disk, leading to a different value for r in the equation V = kQ/r.

3. What is the direction of the electric field produced by a uniformly charged disk off axis?

The direction of the electric field produced by a uniformly charged disk off axis is perpendicular to the disk and points away from the disk for points above the disk, and towards the disk for points below the disk.

4. How does the potential of a uniformly charged disk off axis change as the distance from the disk increases?

The potential of a uniformly charged disk off axis decreases as the distance from the disk increases, following the inverse square law. This means that the potential decreases at a faster rate as the distance increases.

5. Can the potential of a uniformly charged disk off axis ever be zero?

Yes, the potential of a uniformly charged disk off axis can be zero at a point directly above or below the center of the disk. This is because the distance from the disk to these points is infinite, making the potential at these points equal to 0 according to the equation V = kQ/r.

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