causing ane stay gr 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?

causing ane stay gr 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?