(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Two concentric cylinders with radii a & b (b>a) with an infinitely long grounded strip along the z-direction are given potentials [tex]\phi_1[/tex] and [tex]\phi_2[/tex].

Find [tex]\Phi(r,\phi)[/tex] for a<r<b

Boundary conditions:

[tex]\Phi(r,2n\pi)=0[/tex]

[tex]\Phi(a,\phi)=\phi_1[/tex]

[tex]\Phi(b,\phi)=\phi_2

[/tex]

2. Relevant equations

The laplace equation in polar coordinates (as the solution is independent of z)

[tex]

\Delta^2\Phi=0[/tex]

Ansatz:

[tex]\Phi=R(r)Q(\phi)[/tex],

compute

[tex]\frac{r^2}{R(r)Q(\phi)}\Delta^2(R(r)Q(\phi))=0[/tex]

[tex]\frac{1}{r}\frac{d}{dr}(r\frac{dR}{dr})+\frac{1}{Q(\phi)}\frac{dQ(\phi}{d\phi^2}=0[/tex]

Assume that the first term equals n^2, which means that the second one equals -n^2

-> Solutions: (n=0):

[tex]R(r)=a_0 +b_0ln(r);[/tex]

[tex]Q(\phi)=A_0+B_0\phi[/tex]

and (n>1):

[tex]R(r)=a_nr^n +b_nr^{-n};[/tex]

[tex]Q(\phi)=A_ncos(n\phi)+B_nsin(n\phi) [/tex]

[tex]\Phi(r,\phi)=(a_0 +b_0ln(r))(A_0+B_0\phi)+\sum_{n=1}^\infty(a_nr^n +b_nr^{-n})(A_ncos(n\phi)+B_nsin(n\phi))[/tex]

3. The attempt at a solution

From the boundary conditions we see that the solution needs to be periodic in [tex]\phi=2n\pi; n=0,1,...[/tex] which leads to

[tex]A_n=A_0=B_0=0[/tex]

This gives the solution:

[tex]\Phi(r,\phi)=\sum_{n=1}^\infty(a_nr^n +b_nr^{-n})sin(n\phi)[/tex]

Where I have now put B_n coefficients into a_n and b_n.

Now I need to satisfy

[tex]\Phi(a,\phi)=\phi_1[/tex]

[tex]\Phi(b,\phi)=\phi_2[/tex]

[tex]\Phi(a,\phi)=\sum_{n=1}^\infty(a_na^n +b_na^{-n})sin(n\phi)=\phi_1[/tex]

[tex]\Phi(a,\phi)=\sum_{n=1}^\infty(a_nb^n +b_nb^{-n})sin(n\phi)=\phi_2[/tex]

The problem with this is that I would say that the solutions need to be symmetric for phi = [0,pi] and phi = [0,-pi] i.e. The solution should be mirrored in these two domains Phi(r,phi)=Phi(r,-phi), and thus it would be sufficient to solve the boundary conditions only in one of these domains. But only having sinus functions would lead to a discontinuity at phi=plusminus pi... which makes me question the legitimity of this solution.

Furthermore I don't know how to satisfy the boundary conditions on r=a,b; I guess that if I assume phi_1>phi_2 then a_n=0, likewise, phi_1<phi_2 gives b_n=0 and phi_1=phi_2 gives R(r) = constant.

But, to satisfy the finite constant potentials at r=a,b means that we can not have sin(n\phi) solutions i.e. n=0 Which also means that we can't satisfy the fact that the strip is grounded (Phi=0).

So all in all I would assume that the ansatz is wrong... can anyone give a helping hand??

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Find potential between 2 conc. cyl. with grounded strip

Have something to add?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**