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I know how to solve Laplace's equation for a point charge in front of a grounded conducting infinite plane. But I want to know what happens (both physics and math) when the infinite conducting plane isn't grounded, or is connected to a potential V.

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  • $\begingroup$ @sammygerbil Yes I have and also I looked it up in several books, Jackson's Classical Electrodynamics, Intro. to electrodynamics (Griffiths), Foundations of Electromagnetic Theory (Reitz, Milford, Christy) the examples in all these books limit to "Grounded" planes, $\endgroup$ – Alireza Feb 26 '17 at 17:51
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If the problem you're trying to solve only contains one point charge and the conducting infinite plane at potential V, then there is no physical difference between the plane's potential being 0 (grounded) or +V, because the electric potential may be globally shifted by a constant value everywhere...

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  • $\begingroup$ I don't feel quite right about this. Because in a grounded version we can assume electric field is zero on the other side of the plane but in this case one side of the plane has a positive charge and the other side has negative charge therefore there are electric fields on both sides due to this charges. I'm not sure about this either though. Just a beginner in image charges method. $\endgroup$ – Alireza Feb 26 '17 at 20:44
  • $\begingroup$ We actually know nothing about what happens on the other side of the plane. The image charge is a construction to help us write down an electric field that is perpendicular at the plane. $\endgroup$ – user3265043 Feb 26 '17 at 20:51
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If you would like to define V at infinity as zero and the conducting plane as not grounded, you can also think of the solution as a superposition of two different elctrostatic cases: Take the fields expression of a single charge and a grounded plane, and sum this with the fields given off by a plane of fixed potential.

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  • $\begingroup$ This sounds logical. But what are the boundary conditions in this problem so we can check the answer? For example in the grounded version we could say at x=0 (where the inf. plane is located) electric potential has to be zero. $\endgroup$ – Alireza Feb 27 '17 at 11:14
  • $\begingroup$ The important requirement for the boundary condition of a conductor is that the surface be equipotential. For example, say you have a a conducting plane fixed at V0 on the xy plane at z = 0 and a charged particle at z = d above the plane. I would set the boundary conditions to be V = V0 for z = 0 and V = 0 for z = inf. $\endgroup$ – Kthaxt Feb 28 '17 at 6:59
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    $\begingroup$ I don't understand why should V be equal to zero at z=inf. Because the electric potential due to an infinite charges plane increases by distance (since the electric field generated by it doesn't depend on distance) $\endgroup$ – Alireza Mar 5 '17 at 18:17
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I also think that there would be a difference between the two situations. Lets take a non-grounded PEC plane and say a positive charge lies above it. The field created by the charge will induce negative charges to be stacked on the top (towards the positive charge) of the PEC, and since there is no creation nor destruction of charge, the same and opposite charge will be induced on the bottom surface of the PEC. In this way, treating the thickness of the PEC as infinitely small, both induced surface charges cancel each other over large distances, as seen by the positive charge. In this way, nothing changes in respect to the field generated around the positive charge (since the PEC as no total charge). Though, if the PEC is grounded, it acts like an infinite source of charges that can be arranged in any way, dictated by the surrounding charges. So the positive charge induces an electric field on the upper surface of the grounded PEC, BUT no charge on the lower surface since charge must not be conserved in this situation (that is what ground means!). We are then left with a charge that as electric field lines pointing towards the PEC, all coming at right angles with the latter. This situation is exactly equal to a dipole made of the same positive charge, and a equal negative charge situated an equal distance normal to the PEC.

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