Chapter 4: Q4.38P (page 207)

Prove the following uniqueness theorem: A volume $V$ 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 $S$ of $V$ ( $V = 0$ at infinity would be suitable) then the potential throughout is uniquely determined.

Short Answer

Expert verified

Here, the uniqueness theorem is proved.

Step by step solution

Define function

Consider the two solutions having potentials $V_{1}$ and $V_{2}$ .

Then,

$E_{1} = - \nabla V_{1}$ …… (1)

$E_{2} = - \nabla V_{2}$ …… (2)

Now, write the expression for electric displacement.

$D_{1} = ε E_{1}$ …… (3)

$D_{2} = ε E_{2}$ …… (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,

$\int_{v} \nabla \cdot (V_{3} D_{3}) d τ$

By Gauss divergence theorem, then

$\int_{v} \nabla \cdot (V_{3} D_{3}) d τ = \int_{s} (V_{3} D_{3}) d a$

But $V_{3} = 0$ on $s$

Then,

$\begin{array}{l} \int_{v} \nabla \cdot (V_{3} D_{3}) d τ \\ = 0 \end{array}$

Thus,

$\int (\nabla V_{3}) \cdot D_{3} d τ + \int V_{3} (\nabla \cdot D_{3}) d τ = 0$

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,

$\int (\nabla V_{3}) \cdot D_{3} d τ = 0$

We know that,

$\nabla \cdot V_{3} = \nabla \cdot V_{2} - \nabla \cdot V_{1}$

$\begin{array}{l} = - E_{2} + E_{1} \\ = - E_{3} \end{array}$

$D_{3} = ε E_{3}$

Then,

$\int - ε {(E_{3})}^{2} d τ = 0$

Here, $ε > 0$

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.

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Short Answer

Step by step solution

Define function

Determine theorem

Determine theorem

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