# Find potential between 2 conc. cyl. with grounded strip

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1. Nov 19, 2017

### vemarli

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 $$\phi_1$$ and $$\phi_2$$.

Find $$\Phi(r,\phi)$$ for a<r<b

Boundary conditions:
$$\Phi(r,2n\pi)=0$$
$$\Phi(a,\phi)=\phi_1$$
$$\Phi(b,\phi)=\phi_2$$
2. Relevant equations
The laplace equation in polar coordinates (as the solution is independent of z)
$$\Delta^2\Phi=0$$
Ansatz:
$$\Phi=R(r)Q(\phi)$$,
compute
$$\frac{r^2}{R(r)Q(\phi)}\Delta^2(R(r)Q(\phi))=0$$
$$\frac{1}{r}\frac{d}{dr}(r\frac{dR}{dr})+\frac{1}{Q(\phi)}\frac{dQ(\phi}{d\phi^2}=0$$

Assume that the first term equals n^2, which means that the second one equals -n^2
-> Solutions: (n=0):
$$R(r)=a_0 +b_0ln(r);$$
$$Q(\phi)=A_0+B_0\phi$$

and (n>1):
$$R(r)=a_nr^n +b_nr^{-n};$$
$$Q(\phi)=A_ncos(n\phi)+B_nsin(n\phi)$$

$$\Phi(r,\phi)=(a_0 +b_0ln(r))(A_0+B_0\phi)+\sum_{n=1}^\infty(a_nr^n +b_nr^{-n})(A_ncos(n\phi)+B_nsin(n\phi))$$
3. The attempt at a solution

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

$$A_n=A_0=B_0=0$$
This gives the solution:
$$\Phi(r,\phi)=\sum_{n=1}^\infty(a_nr^n +b_nr^{-n})sin(n\phi)$$
Where I have now put B_n coefficients into a_n and b_n.
Now I need to satisfy
$$\Phi(a,\phi)=\phi_1$$
$$\Phi(b,\phi)=\phi_2$$

$$\Phi(a,\phi)=\sum_{n=1}^\infty(a_na^n +b_na^{-n})sin(n\phi)=\phi_1$$
$$\Phi(a,\phi)=\sum_{n=1}^\infty(a_nb^n +b_nb^{-n})sin(n\phi)=\phi_2$$

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??

2. Nov 19, 2017

### vemarli

Obviously, if I may comment, the strip is situated at a<r<b and thus not in contact with the two cylinders (which would otherwise mean that they would all be equipotential and grounded). I wonder if the solution is a combination of two cases, r=a,b and a<r<b? What if I made the strip such that it was between a+d<r<b-d and then let d->0 (but never reach d=0?). Or is the ansatz simply wrong?

3. Nov 19, 2017

### TSny

Consider whether or not the boundary conditions can be satisfied with $n$ not being an integer.