Question

Prove the following uniqueness theorem: A volume $\mathbf{V}$ contains a specified free charge distribution, and various pieces of linear dielectric material, with the susceptibility of each one given. If the potential is specified on the boundaries$\mathbf{S}$ of $\mathit{V}$($\mathit{V}\mathbf{=}\mathbf{0}$ at infinity would be suitable) then the potential throughout is uniquely determined.

Short answer

Here, the uniqueness theorem is proved.

Step-by-step solution

Define function

Consider the two solutions having potentials ${\mathrm{V}}_1$and $V_2$.

Then,

${\mathrm{E}}_1=-\nabla {\mathrm{V}}_1$ …… (1)

$E_2=-\nabla V_2$ …… (2)

Now, write the expression for electric displacement.

$D_1=\epsilon E_1$ …… (3)

$D_2=\epsilon E_2$ …… (4)

Now, define$V_3=V_2-V_1$

Then,

$E_3=E_2-E_1$

$D_3=D_2-D_1$

Determine theorem

Now compute,

${\int }_v\nabla \cdot \left(V_3{D}_3\right)d\tau$

By Gauss divergence theorem, then

${\int }_v\nabla \cdot \left(V_3{D}_3\right)d\tau ={\int }_s\left(V_3{D}_3\right)da$

But $V_3=0$on$s$

Then,

$\begin{array}{l}{\int }_v\nabla \cdot \left(V_3{D}_3\right)d\tau \\ =0\end{array}$

Thus,

$\int (\nabla V_3)\cdot D_{3}d\tau +\int V_3(\nabla \cdot D_3)d\tau =0$

We know that,

$\nabla \cdot D_3=\nabla \cdot D_2-\nabla \cdot D_1$

Determine theorem

A volume$V$contains a free charge distribution then$\nabla .D_3=0$

Then,

$\int (\nabla V_3)\cdot D_{3}d\tau =0$

We know that,

$\nabla \cdot V_3=\nabla \cdot V_2-\nabla \cdot V_1$

$\begin{array}{l}=-E_2+E_1\\ =-E_3\end{array}$

$D_3=\epsilon E_3$

Then,

$\int -\epsilon {\left(E_3\right)}^{2}d\tau =0$

Here,$\epsilon >0$

Then,$E_3=0$

$-\nabla V_3=0$

$V_3$is constant.

Then,$V_2-V_1$ is constant

As, $V_3=0$at the surface then$V_2-V_1=0$

Then,$V_2-V_1$ is everywhere.