Boundary conditions of 2 conductors

[Observer Two]

Homework Statement

Ignore the text in German. You just need to see the picture. 2 conductors both with potential 0 are given. [itex]\alpha[/itex] is the angle between the conductors. (r, [itex]\varphi[/itex]) are polar coordinates pointing to a point in the plane.

Homework Equations

What we need to do is solve the Laplace equation:

Δ[itex]\phi[/itex] = 0

for the boundary conditions which are implied by the picture.

The Attempt at a Solution

My attempt at a solution was to separate the variables in polar coordinates:

[itex]\phi(r, \varphi) = R(r)\Psi(\varphi) [/itex]

I think I did this correctly. It gives me the 2 ODEs:

[itex]R(r) = c_1 e^{k \cdot ln(r)} + c_2 e^{-k \cdot ln(r)}[/itex]

[itex]\Psi(\varphi) = c_3 sin(k \varphi) + c_4 cos(k \varphi)[/itex]



Now my problem is that I can only see 2 boundary conditions in the above picture:

[itex]\phi(r, 0) = 0[/itex] and [itex]\phi(r, \alpha) = 0[/itex]

But this only gives me the constants [itex]c_4[/itex] and k. How do I get the other constants? Are there any more boundary conditions which I'm blind to see?

Thanks in advance.

 

[WannabeNewton]

Should the potential blow up as ##r\rightarrow \infty##? What does this tell you about ##c_1##?

Sweeping out ##\varphi## takes you from one conducting plate to the other and both plates are grounded. What does that tell you about ##c_4##?

 

[Observer Two]

I already did your second question. It tells me [itex]c_4[/itex] = 0.

The first one I didn't see however. The potential should disappear of course in infinity. This tells me [itex]c_1 = 0[/itex].

This would still leave me with 2 unknown constants however. [itex]c_2[/itex] and [itex]c_3[/itex]

What about r → 0? Can I say that the potential has to disappear there as well? Probably not otherwise the entire R(r) term would be 0.

Hmmm ...

 

[Observer Two]

The next question would be to approximate the potential for small r. So from that question alone I'd say that it should depend on r? lol

 

[WannabeNewton]

From the way you've stated it in post #1, I honestly don't know what else it could be. Take look at example 7.3 on page 204 of this pdf and the figure at the bottom of said page: http://www.ie.itcr.ac.cr/acotoc/Mae..._Comunicacion_II/Material/Biblio2/chapt07.pdf

Maybe I'm missing something obvious that someone else can point out.

 

[Observer Two]

I found out that another constant can be determined by the equation Δ [itex](R(r) \Psi(\varphi)) = 0[/itex] itself. It gives [itex]c_3 = -1[/itex].

One constant remains though. So now I have:

[itex]\Phi(r, \varphi) = -C e^{-\frac{n \pi}{\alpha} \cdot ln(r)} sin(\frac{n \pi}{\alpha} \varphi)[/itex]

Oh well ... Close enough I guess.

Could you help me with the approximation for little r? I guess it's supposed to be a Taylor Series approximation but if I want to develop around [itex]x_0[/itex] = 0 for the very first summand [itex]f(x_0)[/itex] doesn't compute because of ln(0) ...