91Ó°ÊÓ

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

${\mathbf{V}}_{\mathbf{avg}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï€¸é}}^{\mathbf{2}}}{\mathbf{∫}}_{\mathbf{s}}\mathbf{VDS}$ …… (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.

$\mathrm{dS}={\mathrm{R}}^2\sin \mathrm{θ»åθ»åÏ•}$ …… (2)

Here, $d\mathrm{θ}$and $d\mathrm{ϕ}$are the angles of space length in xy (horizontal) and yz (vertical) direction.

Now, substitute role="math" localid="1657519318035" ${\mathrm{R}}^2\mathrm{²õ¾\pm ²Ôθ»åθϕ}\mathrm{for}\mathrm{dS}\mathrm{in}\mathrm{equation}\left(1\right).$

$$\begin{gathered} {\mathrm{V}}_{\mathrm{avg}}=\frac{1}{4{\mathrm{Ï€¸é}}^2}{∫}_{\mathrm{s}}\mathrm{V}\left(\mathrm{R},\mathrm{θ},\mathrm{Ï•}\right){\mathrm{R}}^2\mathrm{sinθ»åθ»åÏ•} \\ =\frac{1}{4\mathrm{Ï€}}{∫}_{\mathrm{s}}\mathrm{V}\left(\mathrm{R},\mathrm{θ},\mathrm{Ï•}\right)\mathrm{sinθ»åθ»åÏ•} \\ \mathrm{Take}\mathrm{the}\mathrm{derivation}\mathrm{of}\mathrm{the}\mathrm{equation}\left(1\right)\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{R}. \\ \frac{{\mathrm{dV}}_{\mathrm{avg}}}{\mathrm{dR}}=\frac{\mathrm{d}}{\mathrm{dR}}\left(\frac{1}{4{\mathrm{Ï€¸é}}^2}{∫}_{\mathrm{s}}\mathrm{V}\left(\mathrm{R},\mathrm{θ},\mathrm{Ï•}\right){\mathrm{R}}^2\sin \mathrm{θ»åθ»åÏ•}\right) \\ =\frac{1}{4{\mathrm{Ï€¸é}}^2}{∫}_{\mathrm{s}}\frac{∂\mathrm{V}}{∂\mathrm{R}}{\mathrm{R}}^2\sin \mathrm{θ»åθ»åÏ•} \\ =\frac{1}{4{\mathrm{Ï€¸é}}^2}{∫}_{\mathrm{s}}\left(∇\mathrm{V}.\hat{\mathrm{r}}\right){\mathrm{R}}^2\sin \mathrm{θ»åθ»åÏ•} \\ =\frac{1}{4{\mathrm{Ï€¸é}}^2}{∫}_{\mathrm{s}}\left(∇\mathrm{V}\right)\left({\mathrm{R}}^2\sin \mathrm{θ»åθ»åÏ•}\hat{\mathrm{r}}\right) \\ \mathrm{Substitute}{\mathrm{R}}^2\mathrm{sinθ»åθ»åÏ•}\mathrm{for}\mathrm{dS}\mathrm{in}\mathrm{above}\mathrm{equation} \\ \frac{{\mathrm{dV}}_{\mathrm{avg}}}{\mathrm{dR}}=\frac{1}{4{\mathrm{Ï€¸é}}^2}{∫}_{\mathrm{s}}\left(∇\mathrm{V}\right)\mathrm{dS}.......\left(3\right) \\ \mathrm{Hence},\mathrm{the}\mathrm{derivative}\mathrm{of}\mathrm{average}\mathrm{volume}\mathrm{is}\frac{{\mathrm{dV}}_{\mathrm{avg}}}{\mathrm{dR}}=\frac{1}{4{\mathrm{Ï€¸é}}^2}{∫}_{\mathrm{s}}\left(∇\mathrm{V}\right)\mathrm{dS}. \end{gathered}$$

Use the divergence theorem,

Write the expression foe divergence theorem.

$∭\left(∇.\mathrm{A}\right)\mathrm{dV}=âˆ\neg \mathrm{A}.\mathrm{dS}$

Applying divergence theorem to equation (3),

$\frac{{\mathrm{dV}}_{\mathrm{avg}}}{\mathrm{dR}}=\frac{1}{4{\mathrm{Ï€¸é}}^2}{∫}_{\mathrm{v}}{∇}^2\mathrm{VdV}$

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

As ,$R→0,$

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.