For the potential of the charge over a conducting grounded plane
Physics

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For the potential of the charge over a conducting grounded plane : (a) Find the Electric Field vector. (b) Find the Electric field for x = 0, z = 0. Is it as you expected?
The trick for this is to use the method of images which is based on the uniqueness theorem. The uniqueness theorem states that the Electric field in a region is completely determined by the amount of charge in the region and the value of the electric potential at the boundaries of the region. In this case we know that the field from the plane goes to 0 at an infinite distance away. It is also 0 at the plane since the plane is grounded. But imagine that instead of a plane we had a charge of opposite sign but equal magnitude the same distance below the plane as the original charge is above the plane
* +q

* q
In that case the two fields from both charges would cancel out exactly halfway between them at the point where the plane was and the field for both charges would still go to 0 at infinity. Since in both cases the electric potential is the same and the charge density above the plane is the same the fields must be the same in both cases since by the uniqueness theorem the fields for the two situations are the same since their charge densities and boundary conditions are the same so we simply replace the plane by an opposite charge opposite of where the original charge is located. To find the field simply add the field vectors for each charge
E = kq/r1^2 + k(q)/r2^2
where k is Coulombs constant, q is the amount of charge, r1 is the distance from the point of interest to q, r2 is the distance from the point of interest to q and both terms in the sum are vectors pointing from or away from the charges.
On the plane at each point the E field from the top charge will cancel the E field from the bottom charge since the vector sum there will be 0 so the field there will be 0, which is consistent with the fact that the field inside conductors is always 0.
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