Voltage due to electrostatic induction

AI Thread Summary
The discussion revolves around calculating the voltage difference between two parallel conducting plates when a point charge is placed at an arbitrary location. It is noted that intuitively, the voltage should be zero if the charge is equidistant from both plates or far away, due to equal induced charges or uniform electric field strength. However, there is uncertainty regarding the potential difference for other positions of the charge, suggesting that an analytic solution may not be feasible. The conversation proposes using Poisson's equation with appropriate boundary conditions for numerical simulation to find the voltage difference. Overall, the complexity of the problem highlights the challenges in electrostatics when dealing with arbitrary charge placements.
mcconohy
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Hello All,
I have an arcane curiosity that cannot be answered anywhere I look (including Griffiths' intro electrodynamics book and the internet). This question seems so simple but I may just be over-thinking it.

If there are two square, parallel, finite, conducting plates and a point change q placed at an arbitrary location in space, what will be the voltage difference between the two plates (i.e. hook a voltmeter up to the two plates, what will the reading be).

Intuitively I know that if the point charge is halfway between the plates, the voltage should be zero because the induced charge should be equal on each plate. Also, the voltage should be zero if the charge is placed far away from the plates because the electric field strength is practically the same on the two plates.
Any help would be great. Thanks in advance!
 
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In general, there will be some potential difference. I doubt that there is an analytic solution for arbitrary charge positions. Simulate it?
 
So would it be appropriate to use Poisson's equation with two boundary conditions that the voltage is constant for the positions of the conducting plates (or that the electric field is perpendicular to the conductor)?
 
For the numerical simulation? Right.
 
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