In Section 3.1.4, I proved that the electrostatic potential at any point

in a charge-free region is equal to its average value over any spherical surface

(radius R )centered at .Here's an alternative argument that does not rely on Coulomb's law, only on Laplace's equation. We might as well set the origin at P .Let ${\mathbf{V}}_{\mathbf{ave}}\left(R\right)$be the average; first show that

$\frac{{\mathbf{dV}}_{\mathbf{ave}}}{\mathbf{dR}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi R}}^{\mathbf{2}}}\mathbf{\int }\mathbf{\nabla }\mathbf{V}\mathbf{.}\mathbf{da}$

(note that the ${\mathbf{R}}_{\mathbf{2}}$in da cancels the $\mathbf{1}\mathbf{/}{\mathbf{R}}^{\mathbf{2}}$out front, so the only dependence on R

is in itself). Now use the divergence theorem, and conclude that if Vsatisfies

Laplace's equation, then,${\mathbf{V}}_{\mathbf{ave}}\left(0\right)\mathbf{=}\mathbf{V}\left(P\right)\mathbf{,}\mathbf{}\mathbf{for}\mathbf{}\mathbf{all}\mathbf{}{\mathbf{R}}^{\mathbf{18}}$.

The electrostatic potential at any point P is equal to the average over any spherical surface centered at point P.

${\mathbf{V}}_{\mathbf{avg}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi R}}^{\mathbf{2}}}{\mathbf{\int }}_{\mathbf{s}}\mathbf{VDS}$ …… (1)

Here, ${\mathrm{V}}_{\mathrm{avg}}$is the average potential, R is the radius of the sphere, dS is the small elemental surface of the sphere and V is the potential through the elemental surface area.

Write the expression for the three dimensional for the element surface area of the sphere.

$\mathrm{dS}={\mathrm{R}}^2\sin \mathrm{\theta d\theta d\varphi }$ …… (2)

Here, $d\theta$and $d\varphi$are the angles of space length in xy (horizontal) and yz (vertical) direction.

Now, substitute role="math" localid="1657519318035" ${\mathrm{R}}^2\mathrm{sin\theta d\theta \varphi }\mathrm{for}\mathrm{dS}\mathrm{in}\mathrm{equation}\left(1\right).$

$$\begin{gathered} {\mathrm{V}}_{\mathrm{avg}}=\frac{1}{4{\mathrm{\pi R}}^2}{\int }_{\mathrm{s}}\mathrm{V}\left(\mathrm{R},\mathrm{\theta },\mathrm{\varphi }\right){\mathrm{R}}^2\mathrm{sin\theta d\theta d\varphi } \\ =\frac{1}{4\mathrm{\pi }}{\int }_{\mathrm{s}}\mathrm{V}\left(\mathrm{R},\mathrm{\theta },\mathrm{\varphi }\right)\mathrm{sin\theta d\theta d\varphi } \\ \mathrm{Take}\mathrm{the}\mathrm{derivation}\mathrm{of}\mathrm{the}\mathrm{equation}\left(1\right)\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{R}. \\ \frac{{\mathrm{dV}}_{\mathrm{avg}}}{\mathrm{dR}}=\frac{\mathrm{d}}{\mathrm{dR}}\left(\frac{1}{4{\mathrm{\pi R}}^2}{\int }_{\mathrm{s}}\mathrm{V}\left(\mathrm{R},\mathrm{\theta },\mathrm{\varphi }\right){\mathrm{R}}^2\sin \mathrm{\theta d\theta d\varphi }\right) \\ =\frac{1}{4{\mathrm{\pi R}}^2}{\int }_{\mathrm{s}}\frac{\partial \mathrm{V}}{\partial \mathrm{R}}{\mathrm{R}}^2\sin \mathrm{\theta d\theta d\varphi } \\ =\frac{1}{4{\mathrm{\pi R}}^2}{\int }_{\mathrm{s}}\left(\nabla \mathrm{V}.\hat{\mathrm{r}}\right){\mathrm{R}}^2\sin \mathrm{\theta d\theta d\varphi } \\ =\frac{1}{4{\mathrm{\pi R}}^2}{\int }_{\mathrm{s}}\left(\nabla \mathrm{V}\right)\left({\mathrm{R}}^2\sin \mathrm{\theta d\theta d\varphi }\hat{\mathrm{r}}\right) \\ \mathrm{Substitute}{\mathrm{R}}^2\mathrm{sin\theta d\theta d\varphi }\mathrm{for}\mathrm{dS}\mathrm{in}\mathrm{above}\mathrm{equation} \\ \frac{{\mathrm{dV}}_{\mathrm{avg}}}{\mathrm{dR}}=\frac{1}{4{\mathrm{\pi R}}^2}{\int }_{\mathrm{s}}\left(\nabla \mathrm{V}\right)\mathrm{dS}.......\left(3\right) \\ \mathrm{Hence},\mathrm{the}\mathrm{derivative}\mathrm{of}\mathrm{average}\mathrm{volume}\mathrm{is}\frac{{\mathrm{dV}}_{\mathrm{avg}}}{\mathrm{dR}}=\frac{1}{4{\mathrm{\pi R}}^2}{\int }_{\mathrm{s}}\left(\nabla \mathrm{V}\right)\mathrm{dS}. \end{gathered}$$

Use the divergence theorem,

Write the expression foe divergence theorem.

$\iiint \left(\nabla .\mathrm{A}\right)\mathrm{dV}=\iint \mathrm{A}.\mathrm{dS}$

Applying divergence theorem to equation (3),

$\frac{{\mathrm{dV}}_{\mathrm{avg}}}{\mathrm{dR}}=\frac{1}{4{\mathrm{\pi R}}^2}{\int }_{\mathrm{v}}{\nabla }^2\mathrm{VdV}$

Through R increases, if the average potential ${\mathrm{V}}_{\mathrm{avg}}$kept constant, then it remains as constant for increasing R.

As ,$R\to 0,$

It becomes true and satisfies Laplace equation and gives the result as,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{avg}}\left(\mathrm{R}\right)={\mathrm{V}}_{\mathrm{avg}}\left(0\right) \\ =\mathrm{V}\left(\mathrm{P}\right) \end{gathered}$$

Hence proved.