91Ó°ÊÓ

Write the expression for Poisson’s.

${\mathbf{∇}}^{\mathbf{2}}\mathbf{V}\mathbf{=}\mathbf{-}\frac{\mathbf{P}}{{\mathbf{Î\mu }}_{\mathbf{0}}}$ …… (1)

Here, ${\mathrm{Î\mu }}_0$is the permittivity for the free space and p is the charge density.

Consider the formula for the potential due to volume charge of charge density pis,

$\mathbf{V}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï„Ï„Î\mu }}_{\mathbf{0}}}\mathbf{∫}\frac{\mathbf{P}\mathbf{\left(}{\mathbf{r}}^{\mathbf{\text{'}}}\mathbf{\right)}}{\mathbf{r}}{\mathbf{»åÏ„}}^{\mathbf{\text{'}}}$

Consider the expression from the properties of the 3-dimentional delta function.

$$\begin{gathered} {\mathbf{∇}}^{\mathbf{2}}\frac{\mathbf{1}}{\mathbf{r}}\mathbf{=}\mathbf{-}\mathbf{4}{\mathbf{πδ}}^{\mathbf{3}}\left(r\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{4}{\mathbf{πδ}}^{\mathbf{3}}\left(r-r^{\text{'}}\right) \end{gathered}$$

Now applying Laplacian to the equation (1).

${\mathbf{∇}}^{\mathbf{2}}\mathit{V}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}\mathbf{∫}{\mathbf{∇}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{r}}\mathbf{\right)}\mathit{p}\mathbf{\left(}{\mathit{r}}^{\mathbf{\text{'}}}\mathbf{\right)}\mathit{d}{\mathit{Ï„}}^{\mathbf{\text{'}}}$
Substitute $\mathbf{-}\mathbf{4}{\mathbf{πδ}}^{\mathbf{3}}\mathbf{(}\mathbf{r}\mathbf{-}{\mathbf{r}}^{\mathbf{\text{'}}}\mathbf{)}\mathbf{}\mathbf{for}\mathbf{}{\mathbf{∇}}^{\mathbf{2}}\frac{\mathbf{1}}{\mathbf{r}}$for in the equation.

$$\begin{gathered} {\mathbf{∇}}^{\mathbf{2}}\mathit{V}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}\mathbf{∫}\left[-4{\mathrm{πδ}}^3(r-r^{\text{'}})\right]\mathit{p}\left(r^{\text{'}}\right)\mathit{d}{\mathit{Ï„}}^{\mathbf{\text{'}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{-}\mathbf{4}\mathbf{Ï€}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\mathbf{∫}{\mathbf{δ}}^{\mathbf{3}}\mathbf{(}\mathbf{r}\mathbf{-}{\mathbf{r}}^{\mathbf{\text{'}}}\mathbf{)}\mathit{p}\left(r^{\text{'}}\right)\mathit{d}{\mathit{Ï„}}^{\mathbf{\text{'}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{-}\mathbf{4}\mathbf{Ï€}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\mathbf{∫}\mathit{p}\left(r^{\text{'}}\right){\mathbf{δ}}^{\mathbf{3}}\mathbf{(}\mathbf{r}\mathbf{-}{\mathbf{r}}^{\mathbf{\text{'}}}\mathbf{)}\mathit{d}{\mathit{Ï„}}^{\mathbf{\text{'}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{-}\mathbf{1}}{{\mathbf{Î\mu }}_{\mathbf{0}}}\mathbf{∫}\mathit{p}\left(r^{\text{'}}\right){\mathbf{δ}}^{\mathbf{3}}\mathbf{(}\mathbf{r}\mathbf{-}{\mathbf{r}}^{\mathbf{\text{'}}}\mathbf{)}\mathit{d}{\mathit{Ï„}}^{\mathbf{\text{'}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \end{gathered}$$

The generalized formula for the 3-diamentional delta function is,

${\mathbf{∫}}_{\mathbf{a}\mathbf{l}\mathbf{l}\mathbf{}\mathbf{s}\mathbf{p}\mathbf{a}\mathbf{c}\mathbf{e}}\mathit{p}\left(r^{\text{'}}\right){\mathit{δ}}^{\mathbf{3}}\mathbf{(}\mathit{r}\mathbf{-}{\mathit{r}}^{\mathbf{\text{'}}}\mathbf{)}\mathit{d}{\mathit{τ}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{p}\mathbf{\left(}\mathit{r}\mathbf{\right)}$

So the ${\mathbf{∇}}^{\mathbf{2}}\mathit{V}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{-}\mathbf{1}}{{\mathbf{Î\mu }}_{\mathbf{0}}}\mathbf{∫}\mathit{p}\mathbf{\left(}{\mathit{r}}^{\mathbf{\text{'}}}\mathbf{\right)}{\mathit{δ}}^{\mathbf{3}}\mathbf{(}\mathit{r}\mathbf{-}{\mathit{r}}^{\mathbf{\text{'}}}\mathbf{)}\mathit{d}{\mathit{Ï„}}^{\mathbf{\text{'}}}$equation becomes,

$$\begin{gathered} {\mathbf{∇}}^{\mathbf{2}}\mathit{V}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{-}\mathbf{1}}{{\mathbf{Î\mu }}_{\mathbf{0}}}\mathit{p}\mathbf{\left(}\mathit{r}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\frac{\mathbf{p}\mathbf{\left(}\mathbf{r}\mathbf{\right)}}{{\mathbf{Î\mu }}_{\mathbf{0}}} \end{gathered}$$

From the above simplification it is clear that, the above equation is same that as equation (1).

Hence, the equation $\mathit{V}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}\mathbf{∫}\frac{\mathbf{p}\mathbf{\left(}{\mathbf{r}}^{\mathbf{\text{'}}}\mathbf{\right)}}{\mathbf{r}}\mathit{d}{\mathit{Ï„}}^{\mathbf{\text{'}}}$satisfies the Poisson’s equation.