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

In summary, the problem involves two semi-infinite conducting planes at a 30º angle, one at ground potential and the other at V0. The potential between the conductors can be found using Laplace's equation in cylindrical coordinates. The boundary conditions are V=0 at phi=0 and V=V0 at phi=alpha. The justification for ignoring the radial variable is that the potential is independent of s for specific values of alpha, and the general form of the solution for V uses separation of variables.
  • #1
OhNoYaDidn't
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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]
IMG_1212.jpg


IMG_1216.jpg

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?
 
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  • #2
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##.
 
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  • #3
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?
 
  • #4
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)##.
 
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  • #5
TSny said:
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 .
 
  • #6
phyahmad said:
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.
 

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

1. What is a semi-infinite plane at an angle?

A semi-infinite plane at an angle is a geometric concept that describes an infinite plane that extends infinitely in one direction but is bounded at a particular angle. This means that the plane has no end in one direction, but is cut off at a specific angle in the other direction.

2. How are semi-infinite planes at an angle used in science?

Semi-infinite planes at an angle are used in various fields of science, including physics, mathematics, and engineering. They are often used to model physical systems and phenomena, such as heat transfer, fluid flow, and electromagnetic fields.

3. What is the significance of using semi-infinite planes at an angle in research?

Semi-infinite planes at an angle provide a simplified way to study and analyze complex systems. They allow scientists to make approximations and assumptions that make the calculations and analysis more manageable and easier to understand.

4. How do you calculate the properties of semi-infinite planes at an angle?

The properties of a semi-infinite plane at an angle can be calculated using mathematical equations and formulas. These calculations often involve concepts such as trigonometry, calculus, and differential equations, depending on the specific application.

5. Can semi-infinite planes at an angle be found in nature?

Yes, semi-infinite planes at an angle can be found in nature. For example, the Earth's surface can be approximated as a semi-infinite plane at an angle, with the Earth's curvature being negligible at a local scale. Other examples include the reflection of light off a flat surface and the flow of fluids over a flat surface at an angle.

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