1) The first case we consider is a charge q placed a distance d along the z axis in front of an infinite flat grounded plane (=0 on the plane). We want V²= -4πq(x)(y)(z — d), and = 0 for z = 0. If we find a solution to the above problem, we know it is the unique solution. q 7=0 Solution: put a fictitious image charge -q at z = -d. The potential is then the Coulomb potential from the real charge +q and the image charge -q. ø(r) = q -9 √ x² + y² + (z − d)² - x²+y+2+(z+d)² (2.2.1) Summary: What is going on in this problem? If there was no charge q, then the grounded conducting plane would be charge free with σ = 0. But now we put the charge q in front of the plane. If the plane stayed charge free, then the only electric field in the system would be that produced by q. But the field from q would not be perpendicular to the surface of the plane as it must be, because the plane is the surface of a conductor. The presence of the charge q in front of the plane therefore induces charge σ to appear on the plane. This σ arranges itself non-uniformly over the surface of the plane so that total electric field from q and from σ is then perpendicular to the plane. Where does this σ come from? It comes from the source that is causing the plane to stay grounded with = 0. The image charge is a trick that lets us compute σ and hence the total E to the right of the plane where the charge q is. Note, the image charge must lie to the left of the plane, where we are not trying to find E. Discussion Question 2.2.1 For the above problem, what is the total true physical electric field E to the left of the conducting plane? What is the E field on the left side of the plane that is produced by the induced charge σ? Suppose now that the conductor did not fill the half space z < 0, but was only a thin conducting plane of thickness w, with the right hand surface at z = 0, and the left hand surface at z=-w. What would the electric field be to the right, to the left, and inside the conducting plane?

1) The first case we consider is a charge q placed a distance d along the z axis in front of an infinite flat grounded plane (=0 on the plane). We want V²= -4πq(x)(y)(z — d), and = 0 for z = 0. If we find a solution to the above problem, we know it is the unique solution. q 7=0 Solution: put a fictitious image charge -q at z = -d. The potential is then the Coulomb potential from the real charge +q and the image charge -q. ø(r) = q -9 √ x² + y² + (z − d)² - x²+y+2+(z+d)² (2.2.1) Summary: What is going on in this problem? If there was no charge q, then the grounded conducting plane would be charge free with σ = 0. But now we put the charge q in front of the plane. If the plane stayed charge free, then the only electric field in the system would be that produced by q. But the field from q would not be perpendicular to the surface of the plane as it must be, because the plane is the surface of a conductor. The presence of the charge q in front of the plane therefore induces charge σ to appear on the plane. This σ arranges itself non-uniformly over the surface of the plane so that total electric field from q and from σ is then perpendicular to the plane. Where does this σ come from? It comes from the source that is causing the plane to stay grounded with = 0. The image charge is a trick that lets us compute σ and hence the total E to the right of the plane where the charge q is. Note, the image charge must lie to the left of the plane, where we are not trying to find E. Discussion Question 2.2.1 For the above problem, what is the total true physical electric field E to the left of the conducting plane? What is the E field on the left side of the plane that is produced by the induced charge σ? Suppose now that the conductor did not fill the half space z < 0, but was only a thin conducting plane of thickness w, with the right hand surface at z = 0, and the left hand surface at z=-w. What would the electric field be to the right, to the left, and inside the conducting plane?