I am making my way through Griffiths' Introduction to Electrodynamics and ran across the following passage:

Suppose a point charge $q$ is held a distance $d$ above an infinite grounded conducting plane. What is the potential in the region above the plane? It's not just $(1/4\pi\epsilon_0)q/r$, for $q$ will induce a certain amount of negative charge on the nearby surface of the conductor...

Later on, Griffiths states the boundary conditions for which we must solve Laplace equation:

  1. $V=0$ when $z=0$ (since the conducting plane is grounded), and 2
  2. $V \rightarrow 0$ far from the charge.

I understand that the electric field inside a conductor is zero, and that specifying that the conductor is grounded is synonymous with stating $V=0$, and that any induced charge will be on the boundary of that conductor. But in the case of a grounded plane, would the boundary not be the plane itself, such that $V\neq 0$? Where else would the induced charge that Griffiths refers to reside?


1 Answer 1


It is the combination of your original point charge above the plane with the induced negative charge on the plane that makes the potential zero. If the induced charge were zero, the plane would have a non-zero potential due to the given point charge.

You should look up method of images for a conducting plane.


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