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Classic electrostatics image problem surface charge

Physics Asked by Quaternion on April 14, 2021

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)?

2 Answers

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 $vec{mathbf{E}}_{below} = 0$.

Answered by Quaternion on April 14, 2021

My Answer

The problem here is to solve Poisson's equation with Dirichlet boundary conditions. The general integral solution [1] of Poisson's equation with Dirichlet boundary conditions is begin{align} V{(mathbf{x})} &= int_U G_D{(mathbf{x} , mathbf{x}')},rho{(mathbf{x}')} , left|dmathbf{x}'^3right| &- frac{1}{ 4,pi },oint_{partial U} left[G_D{(mathbf{x} , mathbf{x}')} V{(mathbf{x}')} , frac{partial G_D{(mathbf{x} ,mathbf{x}')}}{partial n'} right], dS'. end{align}

In the problem here, the region $U$ is defined as $$U = left{mathbb{R}^3 mid z < 0right}.$$ Additionally, the boundary $partial{U}$ is given by the union of the infinite plane $z = 0$, and the infinitely large hemisphere below $z=0$.

According to the premise of the problem, $V(mathbf{x}')=0$ for all $mathbf{x}' in partial{U}$. Thus, the general solution reduces to $$ V{(mathbf{x})} = int_U G{(mathbf{x} , mathbf{x}')},rho{(mathbf{x}')} , left|dmathbf{x}'^3right|,~text{for all}~mathbf{x} in U.$$ Further, according to the premise of the problem, $rho(mathbf{x}') = 0$ for all $mathbf{x}' in U$. Thus, the general solution is further reduced to $$V{(mathbf{x})} =0,~text{for all}~mathbf{x} in U.$$ Finally, since $mathbf{E} = -boldsymbol{nabla}V$, therefore $$mathbf{E}{(mathbf{x})} = mathbf{0},~text{for all}~mathbf{x} in U.$$

Discussion

I disagree that the answer, "One simply applies a uniqueness theorem to the region below the conducting sheet," is correct. The uniqueness theorem [2] does not give the solution---all the uniqueness theorem does is say that the solution for the electric field will be a unique solution. In my estimation, the existence theorem is what is actually required to determine the field below the conductor.

Bibliography

[1] John David Jackson, ``Classical Electrodynamics.''

[2] Wikipedia contributors. Uniqueness theorem for Poisson's equation [Internet]. Wikipedia, The Free Encyclopedia; 2021 Jan 7, 00:22 UTC [cited 2021 Jan 9]. Available from: https://en.wikipedia.org/w/index.php?title=Uniqueness_theorem_for_Poisson%27s_equation&oldid=998777694.

Answered by Michael Levy on April 14, 2021

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