Point charge with grounded conducting planes angled 120

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SUMMARY

The discussion focuses on solving the electric potential of a point charge located on the internal angle bisector of a 120-degree dihedral angle formed by two grounded conducting planes. The method of images is deemed ineffective for this configuration, necessitating the use of a surface charge distribution and cylindrical coordinates. Participants suggest employing Green's functions and separation of variables to derive the potential, emphasizing the need for a piece-wise solution that adheres to the boundary conditions of V=0 at the planes and at infinity.

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  • Understanding of electric potential and boundary conditions in electrostatics
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  • Knowledge of Green's functions and their application in solving differential equations
  • Proficiency in cylindrical coordinates and Laplace's equation
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Homework Statement


The problem states:

"A point charge q is located at a fixed point P on the internal angle bisector of a 120 degree dihedral angle between two grounded conducting planes. Find the electric potential along the bisector."

Homework Equations



ΔV = 0
with Dirichlet boundary condition: V = 0 at the two planes and also at infinity.

The Attempt at a Solution


My attempt: Apparently, the method of images does not work. The method of images only works when the given angle is 360/(even number).

Alternatively, if we assume a surface charge distribution σ, we can assume by symmetry that σ is symmetrical for the two half planes, so we can use cylindrical coordinates, assuming σ = σ(z,θ,φ) for φ = 0 and 120°, and P = (0, π/2, 60°). We will have to solve an integral equation, but I am a novice in the topic of integral equation. Did anyone solve it analytically or numerically? Thanks.
 
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If the method of images does not work then you may need to explicitly find the Green's function for the problem. To begin you will need the solution to the Laplace equation for two intersecting grounded planes. Do you know this formula?
 
I know the cylindrical form of the Laplace equations. Is it possible to solve using the separation of variables trick when the boundary condition V=0?
 
You can use separation of variables to find the general solution, although it is easy to just look it up. It's probably solved for you in your textbook. I should ask, are you familiar with Green's functions?

Once you have the general solution, you can assume the charge is located at the position ##(r_{0},\phi_{0})## so the charge density can be written as
$$\rho=\frac{q}{r}\delta(r-r_{0})\delta(\phi-\phi_{0})$$
The solution can be constructed piece-wise for the region ##r<r_{0}## and ##r>r_{0}##. The potential should be continuous everywhere and satisfy the given boundary conditions which will set all of the integration constants but one. The final constant is determined by the discontinuity in the first derivative at the location of the charge ##(r_{0},\phi_{0})##.

For now, can you construct the piece-wise potential valid in the regions ##r<r_{0}## and ##r>r_{0}##?
 
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