91影视

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

${\mathbf{V}}_{\mathbf{avg}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{蟺搁}}^{\mathbf{2}}}{\mathbf{鈭�}}_{\mathbf{s}}\mathbf{VDS}$ 鈥︹�� (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.

$\mathrm{dS}={\mathrm{R}}^2\sin \mathrm{胃诲胃诲蠒}$ 鈥︹�� (2)

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

Now, substitute role="math" localid="1657519318035" ${\mathrm{R}}^2\mathrm{蝉颈苍胃诲胃蠒}\mathrm{for}\mathrm{dS}\mathrm{in}\mathrm{equation}\left(1\right).$

$$\begin{gathered} {\mathrm{V}}_{\mathrm{avg}}=\frac{1}{4{\mathrm{蟺搁}}^2}{鈭�}_{\mathrm{s}}\mathrm{V}\left(\mathrm{R},\mathrm{胃},\mathrm{蠒}\right){\mathrm{R}}^2\mathrm{sin胃诲胃诲蠒} \\ =\frac{1}{4\mathrm{蟺}}{鈭�}_{\mathrm{s}}\mathrm{V}\left(\mathrm{R},\mathrm{胃},\mathrm{蠒}\right)\mathrm{sin胃诲胃诲蠒} \\ \mathrm{Take}\mathrm{the}\mathrm{derivation}\mathrm{of}\mathrm{the}\mathrm{equation}\left(1\right)\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{R}. \\ \frac{{\mathrm{dV}}_{\mathrm{avg}}}{\mathrm{dR}}=\frac{\mathrm{d}}{\mathrm{dR}}\left(\frac{1}{4{\mathrm{蟺搁}}^2}{鈭�}_{\mathrm{s}}\mathrm{V}\left(\mathrm{R},\mathrm{胃},\mathrm{蠒}\right){\mathrm{R}}^2\sin \mathrm{胃诲胃诲蠒}\right) \\ =\frac{1}{4{\mathrm{蟺搁}}^2}{鈭�}_{\mathrm{s}}\frac{鈭�\mathrm{V}}{鈭�\mathrm{R}}{\mathrm{R}}^2\sin \mathrm{胃诲胃诲蠒} \\ =\frac{1}{4{\mathrm{蟺搁}}^2}{鈭�}_{\mathrm{s}}\left(鈭�\mathrm{V}.\hat{\mathrm{r}}\right){\mathrm{R}}^2\sin \mathrm{胃诲胃诲蠒} \\ =\frac{1}{4{\mathrm{蟺搁}}^2}{鈭�}_{\mathrm{s}}\left(鈭�\mathrm{V}\right)\left({\mathrm{R}}^2\sin \mathrm{胃诲胃诲蠒}\hat{\mathrm{r}}\right) \\ \mathrm{Substitute}{\mathrm{R}}^2\mathrm{sin胃诲胃诲蠒}\mathrm{for}\mathrm{dS}\mathrm{in}\mathrm{above}\mathrm{equation} \\ \frac{{\mathrm{dV}}_{\mathrm{avg}}}{\mathrm{dR}}=\frac{1}{4{\mathrm{蟺搁}}^2}{鈭�}_{\mathrm{s}}\left(鈭�\mathrm{V}\right)\mathrm{dS}.......\left(3\right) \\ \mathrm{Hence},\mathrm{the}\mathrm{derivative}\mathrm{of}\mathrm{average}\mathrm{volume}\mathrm{is}\frac{{\mathrm{dV}}_{\mathrm{avg}}}{\mathrm{dR}}=\frac{1}{4{\mathrm{蟺搁}}^2}{鈭�}_{\mathrm{s}}\left(鈭�\mathrm{V}\right)\mathrm{dS}. \end{gathered}$$

Use the divergence theorem,

Write the expression foe divergence theorem.

$鈭�\left(鈭�.\mathrm{A}\right)\mathrm{dV}=鈭�\mathrm{A}.\mathrm{dS}$

Applying divergence theorem to equation (3),

$\frac{{\mathrm{dV}}_{\mathrm{avg}}}{\mathrm{dR}}=\frac{1}{4{\mathrm{蟺搁}}^2}{鈭�}_{\mathrm{v}}{鈭�}^2\mathrm{VdV}$

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

As ,$R鈫�0,$

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.