How to Solve Laplace's Equation for a Point Charge Near a Grounded Plane?

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This discussion focuses on solving Laplace's equation for a point charge located at a distance d from a grounded conducting plane. The user explores potential solutions using zonal harmonics and azimuthal symmetry, leading to a potential expression involving terms like Pn(theta)r^{-(n+1)}. The conversation highlights the complexity of the problem, suggesting that Cartesian or cylindrical coordinates may provide a more effective approach than the image charge method. Jason shares his experience using integral transform techniques and contour integration, emphasizing that the solution will likely yield a continuous sum rather than a discrete one.

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RickRazor
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A point charge q is situated at a distance d from a grounded conducting plane of infinite extent. Find the potential at different points in space.

I want to solve this problem without using the image charge idea.

I assumed azimuthal symmetry and took the zonal harmonics. And we know that as r tends to infinite the potential in the opposite side of the conducting plane goes to a constant. So we are left with the Pn(theta)r^{-(n+1)} terms.

$$V(r,\theta)=A_1 + \frac{C_1}{r} + \frac{C_2}{r^2}\cos(\theta)) + \frac{C_3}{2}(3\cos^2(\theta) - 1) + ...$$

When cos(theta)=d/r, the potential vanishes. So we get the condition:

$$A_1 + \frac{C_1}{r} + \frac{C_2}{r^2}\frac{d}{r} + \frac{C_3}{2}(3\frac{d^2}{r^2} - 1) + ... =0$$

for all r

Now how to progress further? How do I get the coefficients
 

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I don't think you are answering the same question in the image - it just asks for the total charge on the plane (which you can compute from the image solution). What is the real question you are trying to answer?

Jason
 
By the way, if you want to solve without images, you will have better luck I think if you use Cartesian or cylindrical coordinates since your boundary is a plane. It will be fairly involved - more difficult than the kinds of problems in a typical undergrad (in US) textbook like Griffiths. For fun I solved it using integral transform techniques and contour integration, but other approaches could be used. I'm pretty sure that whatever way you solve it, instead of a discrete sum of functions you will find a continuous sum (an integral). EDIT: the integral can be evaluated to yield the image solution, of course.

Jason
 
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