In Griffiths introduction to Electrodynamics, the classic image problem is presented: There is a charge q, above a grounded conducting plane.

The boundary conditions are therefore: 1.V=0, at plane 2.v=0 at infinity

My question is, since potentials are harmonic functions and the potential is zero both inside and on boundary, shouldn't the potential therefore be zero everywhere in R^3 ?


1 Answer 1


No, because the potential in this case in not an harmonic function.

Your statement was correct if the differential equation you have had to solve was the Laplace's equation $$\nabla^2V=0$$ which accounts for space without charges. In that case the potential was indeed an harmonic function, and in order to satisfy the maximum principle - it must have vanish identically.

However, in your particular case there are charges in the volume of interest, and thus one needs to fulfill the Poisson's equation $$\nabla^{2}V=-\frac{Q}{\varepsilon_{0}}\delta\left(x\right)\delta\left(y\right)\delta\left(z-d\right)$$ in the upper half part of $\mathbb{R}^{3}$. Here I assumed the charge $Q$ is placed at $\vec{r}=\left(0,0,d\right)$. Therefore, the function is not harmonic and the maximum principle is not applicable.

  • $\begingroup$ But in that case, the potential reaches a local maximum inside the boundary. $\endgroup$
    – hola
    Nov 11, 2017 at 15:24
  • $\begingroup$ In the case of the image problem? The potential is zero inside the boundary (conducting plate) as you stated. $\endgroup$
    – eranreches
    Nov 11, 2017 at 15:30
  • $\begingroup$ At the boundary at infinity, the potential is zero, so how can the potential be nonzero inside this boundary since potentials are harmonic functions $\endgroup$
    – hola
    Nov 11, 2017 at 15:41
  • $\begingroup$ Simply because the potential is not an harmonic function - due to charges in the region of interest. Thus its maximum doesn't need to be in the boundaries. I will edit my answer to stress that. $\endgroup$
    – eranreches
    Nov 11, 2017 at 15:45

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