Find the general solution to Laplace's equation in spherical coordinates, for the case where V depends only on r. Do the same for cylindrical coordinates, assuming v depends only on s.

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

${\mathbf{∆}}^{\mathbf{2}}\mathit{V}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\left(r^2\frac{\partial V}{\partial r}\right)\mathbf{+}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{\theta }}\left(sin\theta \frac{\partial V}{\partial \theta }\right)\mathbf{+}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{\theta }}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{V}}{\mathbf{\partial }{\mathbf{\varphi }}^{\mathbf{2}}}$ …… (1)

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

Then,

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

Rearrange the equation (2),

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

Thus,

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

${\mathbf{\nabla }}^{\mathbf{2}}\mathit{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{s}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{s}}\left(s\frac{\partial V}{\partial s}\right)\mathbf{+}\frac{\mathbf{1}}{{\mathbf{s}}^{\mathbf{2}}}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{V}}{\mathbf{\partial }{\mathbf{\varphi }}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{V}}{\mathbf{\partial }{\mathbf{Z}}^{\mathbf{2}}}$

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

Then,

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

The above equation in cylindrical coordinates is,

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

Thus,

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

Integrating both the sides

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

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

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.