What Is the Potential Between Semi-Infinite Conducting Planes at a 30º Angle?

[OhNoYaDidn't]

This problem was on my exam last week, and I've been having some troubles coming up with a solution.

1. Homework Statement
Two semi-infinite conducting planes are connected by insulating glue, they make a 30º angle with each other. One of them is at ground potential, and the other one at V0. Find the potential between the conductors.
Note: There might be solutions, in which the potential has logarithmic dependence on the distance to the axis, or even independent of this distance.

Homework Equations

Laplace's equation in cylindrical coordinates.
3. The Attempt at a Solution [/B]

The boundary conditions i came up with are:
1- V=0 at phi=0
2- V=V0 at phi=alpha

But I'm note sure if this is right.
Can you guys please help me out?

 

[TSny]

Your work looks correct to me. But you might be required to justify ignoring the parts of the Laplacian that involve derivatives of the radial variable $s$ and derivatives of the variable $z$.

Of course, you can make your answer more specific by substituting the given value for $\alpha$.

 

[OhNoYaDidn't]

I'm assuming i can justify ignoring the z part by saying we have cylindrical symmetry. But what would the justification be for the s radial variable?

 

[TSny]

Use the fact that the potential for $\alpha = 0$ or $\alpha = 30^0$ is independent of $s$ along with the general form of the solution for $V$ using separation of variables: $V(s, \alpha, z) = f(s)g(\alpha)h(z)$.

 

[member 743765]

Use the fact that the potential for $\alpha = 0$ or $\alpha = 30^0$ is independent of $s$ along with the general form of the solution for $V$ using separation of variables: $V(s, \alpha, z) = f(s)g(\alpha)h(z)$.

I understand that but i have a question : if you moved radially outward the distance between you and the planes changes differently i mean if the angle between the planes is 30 degrees and you started to move at an angle of 10 degrees your distance from each plane will increase but at a different rate so why it doesnt depend on r .

 

[PhDeezNutz]

I understand that but i have a question : if you moved radially outward the distance between you and the planes changes differently i mean if the angle between the planes is 30 degrees and you started to move at an angle of 10 degrees your distance from each plane will increase but at a different rate so why it doesnt depend on r .

Instead of relying on intuition I think you should appeal to the "First Uniqueness Theorem" of Laplace's Equation which states

"The solution to Laplace's equation in some volume $\mathcal{V}$ is uniquely determined if $V$ is specified on the boundary surface $S$"

We have $V = \frac{V_0}{\alpha} \phi$

Clearly
$V \left( 0 \right) = 0$

and

$V\left( \alpha \right) = V_0$

So the boundary conditions are satisfied

and since the Laplacian in Spherical coordinates is

$\nabla^2 V = \frac{1}{r} \frac{\partial }{\partial r} \left(r \frac{\partial V}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 V}{\partial \phi^2 } + \frac{\partial^2 V}{\partial z^2}$

$\nabla^2 \left(\frac{V_0}{\alpha} \phi \right) = 0$

and it satisfies laplaces equation in the interior

Because it is satisfied on the boundary and in the interior by the Uniqueness Theorem it is the ONLY solution.