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thanks.

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- Thread starter eok20
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thanks.

- #2

Homework Helper

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- #4

Homework Helper

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Put the top side of the box above the plane.

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I may be having an answer about what you are asking for.

You may think about the problem in reverse. To be clear, if one takes two equal and sign opposite charges +q and -q. let r be the distance between them. Then from basic electrostatics, we know that each charge will create its own potential in the surrounding space. Equipotential surfaces will then be created around the given charge +q and -q respectively. Knowing that the expression of the potential created by a point charge Q at any point of the space is given by: U=Q/(4pi*epsilon0*R) with R being the distance between Q and the point of the space where the potential is evaluated. Then after the theorem of superposition (applicable to the electrostatic potential), at any point of the space around +q and -q, the total potential is the sum of U(+q) and U(+q). This lead to find a an equipotential surface where the total potential U=U(+q)+U(-q) is zero (especially for the particular case where +q and -q have identical magnitudes in Coulomb unit). It is rather clear that this equipotential will be placed at a distance r/2 from each one of the two charges. Let the equipotential surface where the total potential is zero be S. This mean that if one brings any metallic sheet and insure its potential to be zero (by grounding it for example), and fit it exactly to S, then this will change nothing of the electrostatics configuration of the two charges +q and -q. And here you got your answer. This means that solving the electrostatic interaction between a point charge +q and a grounded surface is exactly the same as if it was between two point charges +q and -q placed at equal distances from each part of the grounded surface. For more clear interpretation i recommand you to look to the First volume of Feynman lectures on electromagnetics, "the method of images" section. Its very well exposed.

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