Infinite conducting plane Let $\pi$ be an infinite conducting plane laying in $z=0$.
the plane is kept in potential of 8 volts, ($\phi(z=0)=8[V]$).
Prove or disprove:
the surface charge density - $\sigma$ is well defined.(there is a single solution to the surface charge density.
What I thought:
You have to solve a Laplace Equation for the potential above the surface and below the surface.
The solution to laplace equation is unique, and therefore the electric field above and below is unique and well defined. According to the equation: $\hat{n}\cdot (\vec{D_2}-\vec{D_1}) = \rho_s$ the surface charge density is also unique and well defined.
 A: Without knowing what happens at any other point, the solution is indeed non-unique. The potential has to obey Laplace's equation below and above the plane, so
$$\phi''(z) = 0,$$
meaning that
$$\phi(z) = \begin{cases}
A_+ z + B_+ & z>0\\[5pt]
A_- z + B_- & z<0
\end{cases}$$
Note that since the system has translational invariance along the $x$ and $y$ directions, all quantities are only a function of $z$.
Now the only boundary condition given is that $\phi(0) = 8$ V. So $B_+ = B_- = 8$ V, i.e. all potentials of the form
$$\phi(z) = \begin{cases}
A_+ z + 8 \ \mathrm{V} & z\geq0\\[5pt]
A_- z + 8 \ \mathrm{V}& z<0
\end{cases}$$
are valid solutions. The corresponding surface charge density $\rho_s = \epsilon_0 \hat{\mathbf{z}}\cdot \big[\mathbf{E}(0^+) - \mathbf{E}(0^-)\big]$ is
\begin{align}
\rho_s &= -\epsilon_0 \Bigg[\frac{\partial \phi}{\partial z}(0^+)-\frac{\partial \phi}{\partial z}(0^-)\Bigg]\\[5pt]
&= -\epsilon_0 (A_+-A_-),
\end{align}
which is clearly non-unique, as it depends on $A_\pm$. I encourage you to look at the proof of the uniqueness theorem and see what assumption breaks down for this example.
