# Boundary conditions of 2 conductors

1. Nov 17, 2013

### Observer Two

1. The problem statement, all variables and given/known data

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

2. Relevant equations

What we need to do is solve the Laplace equation:

Δ$\phi$ = 0

for the boundary conditions which are implied by the picture.

3. The attempt at a solution

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

$\phi(r, \varphi) = R(r)\Psi(\varphi)$

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

$R(r) = c_1 e^{k \cdot ln(r)} + c_2 e^{-k \cdot ln(r)}$

$\Psi(\varphi) = c_3 sin(k \varphi) + c_4 cos(k \varphi)$

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

$\phi(r, 0) = 0$ and $\phi(r, \alpha) = 0$

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

2. Nov 17, 2013

### 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 $c_4$?

3. Nov 17, 2013

### Observer Two

I already did your second question. It tells me $c_4$ = 0.

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

This would still leave me with 2 unknown constants however. $c_2$ and $c_3$

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 ...

4. Nov 17, 2013

### 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

5. Nov 17, 2013

### WannabeNewton

6. Nov 17, 2013

### Observer Two

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

One constant remains though. So now I have:

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

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 $x_0$ = 0 for the very first summand $f(x_0)$ doesn't compute because of ln(0) ...