Question

(a) Suppose a charge distribution ${\mathbf{p}}_{\mathbf{1}}\mathbf{\left(}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{r}}\mathbf{\right)}$produces a potential ${\mathit{V}}_{\mathbf{1}}\mathbf{\left(}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{r}}\mathbf{\right)}$, and some othercharge distribution ${\mathbf{p}}_2\mathbf{\left(}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{r}}\mathbf{\right)}$produces a potential ${\mathit{V}}_2\mathbf{\left(}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{r}}\mathbf{\right)}$. [The two situations mayhave nothing in common, for all I care-perhaps number 1 is a uniformlycharged sphere and number 2 is a parallel-plate capacitor. Please understand that ${\mathbf{p}}_{\mathbf{1}}$and ${\mathbf{p}}_2$ are not present at the same time;we are talking about two differentproblems,one in which only ${\mathbf{p}}_{\mathbf{1}}$is present, and another in which only ${\mathbf{p}}_2$ ispresent.] Prove Green's reciprocity theorem:

${\mathbf{\int }}_{\mathbf{a}\mathbf{l}\mathbf{l}\mathbf{}\mathbf{s}\mathbf{p}\mathbf{a}\mathbf{c}\mathbf{e}}{\mathit{p}}_{\mathbf{1}}{\mathit{V}}_2\mathit{d}\mathit{\tau }\mathbf{=}{\mathbf{\int }}_{\mathbf{a}\mathbf{l}\mathbf{l}\mathbf{}\mathbf{s}\mathbf{p}\mathbf{a}\mathbf{c}\mathbf{e}}{\mathit{p}}_{\mathbf{2}}{\mathit{V}}_{\mathbf{1}}\mathit{d}\mathit{\tau }$

[Hint:Evaluate $\mathbf{\int }{\overrightarrow{\mathbf{E}}}_{\mathbf{1}}\mathbf{}\mathit{X}{\overrightarrow{\mathbf{E}}}_{\mathbf{2}}\mathit{d}\mathit{\tau }$two ways, first writing ${\overrightarrow{\mathbf{E}}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\overrightarrow{\mathbf{\nabla }}{\mathit{V}}_{\mathbf{1}}$and using integrationby parts to transfer the derivative to ${\overrightarrow{\mathbf{E}}}_2$, then writing ${\overrightarrow{\mathbf{E}}}_2\mathbf{=}\mathbf{-}\overrightarrow{\mathbf{\nabla }}{\mathit{V}}_2$and transferring the derivative to ${\overrightarrow{\mathbf{E}}}_{\mathbf{1}}$.]

(b) Suppose now that you have two separated conductors (Fig. 3.41). If you chargeup conductor by amount $\mathit{Q}$(leaving uncharged), the resulting potential of bis, say,${\mathit{V}}_{\mathbf{a}\mathbf{b}}$.On the other hand, if you put that same charge on conductor (leaving uncharged), the potential of would be.Use Green's reciprocitytheorem to show that ${\mathit{V}}_{\mathbf{ab}}\mathbf{=}{\mathit{V}}_{\mathbf{b}\mathbf{a}}$(an astonishing result, since we assumed nothingabout the shapes or placement of the conductors).

Short answer

(a) The Green's reciprocity theorem, that is $\int V_1{\rho }_{2}d\tau =\int V_2{\rho }_{1}d\tau$has been proved.

(b) The potential due to a volume charge density on another volume is equal to the potential on the first volume if the same total charge is shifted to the second volume.

Step-by-step solution

Given data

The charge distributions are ${}_{1}and{\rho }_2.$

The potential due to the two charge distributions are $V_{1}and{V}_2.$

Expression for electric field and continuity equation

The electric field is expressed as a function of potential as follows

$\overrightarrow{\mathbf{E}\mathbf{⃗}}\mathbf{=}\mathbf{-}\overrightarrow{\mathbf{\nabla }}\mathit{V}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

The continuity equation is expressed as

${\mathbf{\nabla }}^{\mathbf{2}}\mathit{V}\mathbf{=}\mathbf{-}\frac{\mathbf{\rho }}{{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Here, ${\mathrm{\epsilon }}_0$ is the permittivity of free space.

Proof of Green's reciprocity theorem

Consider the integral

$I=\int \overrightarrow{E_1}\cdot \overrightarrow{E_2}d\tau$

Substitute the expression for electric field mentioned in equation (1)

role="math" localid="1658560798798" $$\begin{gathered} I=\int \stackrel{\rightharpoonup }{\nabla }{V}_1·\stackrel{\rightharpoonup }{\nabla }{V}_{2}d\tau \\ =\int \stackrel{\rightharpoonup }{\nabla }·\left(V_1\stackrel{\rightharpoonup }{\nabla }{V}_2\right)d\overrightarrow{a-}\int V_1\stackrel{\rightharpoonup }{\nabla }{V}_{2}d\tau \end{gathered}$$

Invoke divergence theorem in the first term to get

role="math" localid="1658560641462" $l=\oint \stackrel{\rightharpoonup }{\nabla }·\left(V_1\stackrel{\rightharpoonup }{\nabla }{V}_2\right)d\overrightarrow{a-}\int V_1\stackrel{\rightharpoonup }{\nabla }{V}_{2}d\tau$

Here represents the area vector of the surface enclosing the volume, but this surface can be kept at infinite distance from the charge where . Thus the first term vanishes and

$l=-\int V_1\stackrel{\rightharpoonup }{\nabla }{V}_{2}d\tau$

Use the continuity equation in equation (2) for charge density 2

$l=\frac{1}{{\epsilon }_0}\int V_1\stackrel{\rightharpoonup }{\nabla }{V}_{2}d\tau$

Replace the second step with

$$\begin{gathered} I=\int \stackrel{\rightharpoonup }{\nabla }{V}_1·\stackrel{\rightharpoonup }{\nabla }{V}_{2}d\tau \\ =\int \stackrel{\rightharpoonup }{\nabla }·\left(V_1\stackrel{\rightharpoonup }{\nabla }{V}_2\right)d\tau -\int {V_1\nabla }^2{V}_{2}d\tau \end{gathered}$$

The final form of the integral will be

$l=\frac{1}{{\epsilon }_0}\int V_2{\rho }_{1}d\tau$

Compare the two results to get

$\int V_1{\rho }_{2}d\tau =\int V_2{\rho }_{1}d\tau$

Thus, the Green's reciprocity theorem is proved

Proof of  Vab=Vba

The charge on is given by

$\int {\rho }_{1}d\tau =Q$

The charge on is given by

$\int {\rho }_{2}d\tau =Q$

Use Green's reciprocity theorem to write

$\int V_{ba}{\rho }_{1}d\tau =\int V_{ab}{\rho }_{1}d\tau$

Since the potential is measured outside the respective charge volumes, take them outside the integral and write

$$\begin{gathered} V_{ba}\int {\rho }_{1}d\tau =V_{ab}\int {\rho }_{2}d\tau \\ {V}_{ba}Q=V_{ab}Q \\ {V}_{ba}=V_{ab} \end{gathered}$$

Thus, it is proved that $V_{ba}=V_{ab}$.