Question

Prove that the field is uniquely determined when the charge density $\mathit{\rho }$

is given and either V or the normal derivative a $\mathbf{\partial }\mathit{V}\mathbf{/}\mathbf{\partial }\mathit{n}$ is specified on each boundary surface. Do not assume the boundaries are conductors, or that V is constant over any given surface.

Short answer

Answer:

The electric field is uniquely determined when the charge density $\mathit{\rho }$ is given and either V or the normal derivative a $\mathbf{\partial }\mathit{V}\mathbf{/}\mathbf{\partial }\mathit{n}$ is specified on each boundary surface.

Step-by-step solution

Given data

Either the potential V or the normal derivative of potential $\partial V/\partial n$is specified on each boundary surface.

Second uniqueness theorem

From the proof of the second uniqueness theorem

$\underset{\mathbf{S}}{\mathbf{\oint }}{\mathit{V}}_{\mathbf{3}}{\overrightarrow{\mathbf{E}}}_{\mathbf{3}}\mathbf{·}\mathit{d}\overrightarrow{\mathbf{a}}\mathbf{=}\mathbf{-}\underset{\mathbf{V}}{\mathbf{\int }}{\left(E_3\right)}^{\mathbf{2}}\mathit{d}\mathit{\tau }$

Here, ${\mathit{V}}_{\mathbf{3}}$is the difference in potential of two points on a surface S. ${\mathit{E}}_{\mathbf{3}}$ is the difference in electric field of two points in a volume V that S encloses.

Uniqueness of field in a volume

If the potential is specified on a boundary surface

${\mathit{V}}_{\mathbf{3}}\mathbf{=}\mathbf{0}$

If the normal derivative of potential (that is $\frac{\mathbf{\partial }\mathbf{V}}{\mathbf{\partial }\mathbf{n}}\mathbf{=}\mathbf{-}{\mathit{E}}_{\mathbf{\perp }}$) is specified on a boundary surface

${\mathit{E}}_{\mathbf{3}\mathbf{,}\mathbf{\perp }}\mathbf{=}\mathbf{0}$

Any of these conditions in equation (1) gives

${\mathit{E}}_{\mathbf{3}}\mathbf{=}\mathbf{0}$

Thus, the electric field is uniquely determined.