Point charge with grounded conducting planes angled 120

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Homework Help Overview

The problem involves a point charge located on the internal angle bisector of a 120-degree dihedral angle formed by two grounded conducting planes. The objective is to find the electric potential along this bisector, considering the boundary conditions that the potential is zero at the planes and at infinity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the limitations of the method of images for this configuration and explore alternative approaches, such as using a surface charge distribution and cylindrical coordinates. Questions arise about the applicability of Green's functions and separation of variables in solving the Laplace equation under the given boundary conditions.

Discussion Status

The discussion is active, with participants offering various approaches to tackle the problem. Some suggest finding the Green's function explicitly, while others consider the separation of variables method. There is no explicit consensus on a single approach, but multiple lines of reasoning are being explored.

Contextual Notes

Participants note the challenge posed by the angle of the dihedral and the implications for the method of images. There is also mention of the need for familiarity with integral equations and Green's functions, indicating a potential gap in knowledge among some participants.

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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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