91Ó°ÊÓ

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

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

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

$\stackrel{\mathbf{→}}{\mathbf{E}\mathbf{⃗}}\mathbf{=}\mathbf{-}\stackrel{\mathbf{→}}{\mathbf{∇}}\mathit{V}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

The continuity equation is expressed as

${\mathbf{∇}}^{\mathbf{2}}\mathit{V}\mathbf{=}\mathbf{-}\frac{\mathbf{ÒÏ}}{{\mathbf{Î\mu }}_{\mathbf{0}}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

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

Consider the integral

$I=∫\stackrel{→}{E_1}â‹\dots \stackrel{→}{E_2}d\mathrm{Ï„}$

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

role="math" localid="1658560798798" $$\begin{gathered} I=∫\stackrel{â‡\P Ä}{∇}{V}_1·\stackrel{â‡\P Ä}{∇}{V}_{2}d\mathrm{Ï„} \\ =∫\stackrel{â‡\P Ä}{∇}·\left(V_1\stackrel{â‡\P Ä}{∇}{V}_2\right)d\stackrel{→}{a-}∫V_1\stackrel{â‡\P Ä}{∇}{V}_{2}d\mathrm{Ï„} \end{gathered}$$

Invoke divergence theorem in the first term to get

role="math" localid="1658560641462" $l=âˆ\circledR \stackrel{â‡\P Ä}{∇}·\left(V_1\stackrel{â‡\P Ä}{∇}{V}_2\right)d\stackrel{→}{a-}∫V_1\stackrel{â‡\P Ä}{∇}{V}_{2}d\mathrm{Ï„}$

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

$l=-∫V_1\stackrel{â‡\P Ä}{∇}{V}_{2}d\mathrm{Ï„}$

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

$l=\frac{1}{{\mathrm{Î\mu }}_0}∫V_1\stackrel{â‡\P Ä}{∇}{V}_{2}d\mathrm{Ï„}$

Replace the second step with

$$\begin{gathered} I=∫\stackrel{â‡\P Ä}{∇}{V}_1·\stackrel{â‡\P Ä}{∇}{V}_{2}d\mathrm{Ï„} \\ =∫\stackrel{â‡\P Ä}{∇}·\left(V_1\stackrel{â‡\P Ä}{∇}{V}_2\right)d\mathrm{Ï„}-∫{V_1∇}^2{V}_{2}d\mathrm{Ï„} \end{gathered}$$

The final form of the integral will be

$l=\frac{1}{{\mathrm{Î\mu }}_0}∫V_2{\mathrm{ÒÏ}}_{1}d\mathrm{Ï„}$

Compare the two results to get

$∫V_1{\mathrm{ÒÏ}}_{2}d\mathrm{Ï„}=∫V_2{\mathrm{ÒÏ}}_{1}d\mathrm{Ï„}$

Thus, the Green's reciprocity theorem is proved

The charge on is given by

$∫{\mathrm{ÒÏ}}_{1}d\mathrm{Ï„}=Q$

The charge on is given by

$∫{\mathrm{ÒÏ}}_{2}d\mathrm{Ï„}=Q$

Use Green's reciprocity theorem to write

$∫V_{ba}{\mathrm{ÒÏ}}_{1}d\mathrm{Ï„}=∫V_{ab}{\mathrm{ÒÏ}}_{1}d\mathrm{Ï„}$

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

$$\begin{gathered} V_{ba}∫{\mathrm{ÒÏ}}_{1}d\mathrm{Ï„}=V_{ab}∫{\mathrm{ÒÏ}}_{2}d\mathrm{Ï„} \\ {V}_{ba}Q=V_{ab}Q \\ {V}_{ba}=V_{ab} \end{gathered}$$

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