91Ó°ÊÓ

Write the value of ${∇}^{2}V$in spherical coordinates.

${\mathbf{∆}}^{\mathbf{2}}\mathit{V}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{r}}\left(r^2\frac{∂V}{∂r}\right)\mathbf{+}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{θ}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{θ}}\left(sin\mathrm{θ}\frac{∂V}{∂\mathrm{θ}}\right)\mathbf{+}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{θ}}\frac{{\mathbf{∂}}^{\mathbf{2}}\mathbf{V}}{\mathbf{∂}{\mathbf{ϕ}}^{\mathbf{2}}}$ …… (1)

Here, V is only depends only on r. Vis the potential, r is the variable.

Then,

${\mathbf{∇}}^{\mathbf{2}}\mathit{V}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{∂}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{r}}\left(r^2\frac{∂V}{∂r}\right)$

Rearrange the equation (2),

$$\begin{gathered} \frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{r}}\left(r^2\frac{∂V}{∂r}\right)\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{r}}\left(r^2\frac{∂V}{∂r}\right)\mathbf{=}\mathbf{0} \end{gathered}$$

Thus,

$$\begin{gathered} r^2\frac{∂V}{∂r}=Cons\tan t \\ {r}^2\frac{∂V}{∂r}=C \\ ∂V=\frac{C}{r^2}∂r \end{gathered}$$ ……. (3)

Integrate both the sides,

$\mathit{V}\mathbf{=}\frac{\mathbf{-}\mathbf{C}}{\mathbf{r}}\mathbf{+}\mathit{B}$ …… (4)

Here, B is constant.

Hence, the potential V is only depend on r only.

Write the equation,

${\mathbf{∇}}^{\mathbf{2}}\mathit{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{s}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{s}}\left(s\frac{∂V}{∂s}\right)\mathbf{+}\frac{\mathbf{1}}{{\mathbf{s}}^{\mathbf{2}}}\frac{{\mathbf{∂}}^{\mathbf{2}}\mathbf{V}}{\mathbf{∂}{\mathbf{ϕ}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{∂}}^{\mathbf{2}}\mathbf{V}}{\mathbf{∂}{\mathbf{Z}}^{\mathbf{2}}}$

Here, V is only depends only on s. Vis the potential, s is the variable.

Then,

${\mathbf{∇}}^{\mathbf{2}}\mathit{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{s}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{s}}\left(s\frac{∂V}{∂s}\right)$

The above equation in cylindrical coordinates is,

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{∇}}^{\mathbf{2}}\mathbf{=}\mathbf{0} \\ \frac{\mathbf{1}}{\mathbf{s}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{s}}\left(s\frac{∂V}{∂s}\right)\mathbf{=}\mathbf{0} \\ \frac{\mathbf{1}}{\mathbf{s}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{s}}\left(s\frac{∂V}{∂s}\right)\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{s}}\left(s\frac{∂V}{∂s}\right)\mathbf{=}\mathbf{0} \end{gathered}$$

Thus,

$$\begin{gathered} \mathit{s}\frac{\mathbf{∂}\mathbf{V}}{\mathbf{∂}\mathbf{s}}\mathbf{=}\mathit{C}\mathit{o}\mathit{n}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{t} \\ \mathit{s}\frac{\mathbf{∂}\mathbf{V}}{\mathbf{∂}\mathbf{s}}\mathbf{=}\mathit{C} \\ \mathbf{}\mathbf{}\frac{\mathbf{∂}\mathbf{V}}{\mathbf{∂}\mathbf{s}}\mathbf{=}\frac{\mathbf{C}}{\mathbf{s}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{∂}\mathit{V}\mathbf{=}\frac{\mathbf{C}}{\mathbf{s}}\mathbf{∂}\mathit{s} \end{gathered}$$ ….. (5)

Integrating both the sides

The integral of polynomial of $\frac{\mathbf{1}}{\mathbf{x}}$gives natural algorithm.

$\mathbf{∫}\frac{\mathbf{a}}{\mathbf{x}}\mathit{d}\mathit{x}\mathbf{=}\mathit{a}\mathbf{}\mathit{I}\mathit{n}\mathbf{}\mathit{x}\mathbf{+}\mathit{C}$

Then the above equation becomes,

$\mathit{V}\mathbf{=}\mathit{C}\mathbf{}\mathit{I}\mathit{n}\mathbf{}\mathit{s}\mathbf{+}\mathit{B}$ …… (6)

Here, B is constant.

Hence, the potential V depends on s only.