91Ó°ÊÓ

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

Then,

${\mathrm{E}}_1=−∇{\mathrm{V}}_1$ …… (1)

$E_2=−∇V_2$ …… (2)

Now, write the expression for electric displacement.

$D_1=\mathrm{Î\mu }{E}_1$ …… (3)

$D_2=\mathrm{Î\mu }{E}_2$ …… (4)

Now, define$V_3=V_2−V_1$

Then,

$E_3=E_2−E_1$

$D_3=D_2−D_1$

Now compute,

${∫}_v∇â‹\dots \left(V_3{D}_3\right)d\mathrm{Ï„}$

By Gauss divergence theorem, then

${∫}_v∇â‹\dots \left(V_3{D}_3\right)d\mathrm{Ï„}={∫}_s\left(V_3{D}_3\right)da$

But $V_3=0$on$s$

Then,

$\begin{array}{l}{∫}_v∇â‹\dots \left(V_3{D}_3\right)d\mathrm{Ï„}\\ =0\end{array}$

Thus,

$∫(∇V_3)â‹\dots D_{3}d\mathrm{Ï„}+∫V_3(∇â‹\dots D_3)d\mathrm{Ï„}=0$

We know that,

$∇â‹\dots D_3=∇â‹\dots D_2−∇â‹\dots D_1$

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

Then,

$∫(∇V_3)â‹\dots D_{3}d\mathrm{Ï„}=0$

We know that,

$∇â‹\dots V_3=∇â‹\dots V_2−∇â‹\dots V_1$

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

$D_3=\mathrm{Î\mu }{E}_3$

Then,

$∫−\mathrm{Î\mu }{\left(E_3\right)}^{2}d\mathrm{Ï„}=0$

Here,$\mathrm{Î\mu }>0$

Then,$E_3=0$

$−∇V_3=0$

$V_3$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.