In the classic image problem of electrostatics (an infinite conducting planar sheet and a point charge above the sheet), one calculates the potential and hence field in the region above the conducting sheet. Then the standard surface charge calculation applies the boundary condition
\begin{equation} \sigma = \epsilon_0 \left( \vec{\mathbf{E}}_{above} - \vec{\mathbf{E}}_{below} \right) \cdot \hat{n} \end{equation} at the conducting sheet, taking $\vec{\mathbf{E}}_{below}$ to be zero.

I cannot see why the field must vanish in the region below the sheet; at any rate, this region is not the interior of any conductor. Can anybody explain why (rigorously)?


1 Answer 1


Don't worry, I think I understand now. One simply applies a Uniqueness theorem to the region below the conducting sheet (which I should have mentioned in the question, is grounded) to conclude that $\vec{\mathbf{E}}_{below} = 0$.

  • 3
    $\begingroup$ That's exactly correct. $\endgroup$ Commented Aug 14, 2015 at 13:44

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