Q29P - Introduction to Electrodynamics

Question

Check that Eq. 2.29 satisfies Poisson's equation, by applying the Laplacian and using Eq. 1.102.

Step-by-Step Solution

Answer

The equation $V\left(r\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\int \frac{P\left(r^{\text{'}}\right)}{r}d{x}^{\text{'}}$satisfies the Poisson’s equation.

Step 1: Determine Poisson’s equation.

Write the expression for Poisson’s.

${\mathbf{\nabla }}^{\mathbf{2}}\mathbf{V}\mathbf{=}\mathbf{-}\frac{\mathbf{P}}{{\mathbf{\epsilon }}_{\mathbf{0}}}$                 …… (1)

 

Here, ${\epsilon }_0$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 p is,

$\mathbf{V}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\tau \tau \epsilon }}_{\mathbf{0}}}\mathbf{\int }\frac{\mathbf{P}\mathbf{\left(}{\mathbf{r}}^{\mathbf{\text{'}}}\mathbf{\right)}}{\mathbf{r}}{\mathbf{d\tau }}^{\mathbf{\text{'}}}$

Step 2: Determine Laplacian equation

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

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

Now applying Laplacian to the equation (1).

${\mathbf{\nabla }}^{\mathbf{2}}\mathit{V}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\int }{\mathbf{\nabla }}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{r}}\mathbf{\right)}\mathit{p}\mathbf{\left(}{\mathit{r}}^{\mathbf{\text{'}}}\mathbf{\right)}\mathit{d}{\mathit{\tau }}^{\mathbf{\text{'}}}$
Substitute $\mathbf{-}\mathbf{4}{\mathbf{\pi \delta }}^{\mathbf{3}}\mathbf{(}\mathbf{r}\mathbf{-}{\mathbf{r}}^{\mathbf{\text{'}}}\mathbf{)}\mathbf{ }\mathbf{for}\mathbf{ }{\mathbf{\nabla }}^{\mathbf{2}}\frac{\mathbf{1}}{\mathbf{r}}$for in the equation.

$$\begin{gathered} {\mathbf{\nabla }}^{\mathbf{2}}\mathit{V}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\int }\left[-4{\mathrm{\pi \delta }}^3(r-r^{\text{'}})\right]\mathit{p}\left(r^{\text{'}}\right)\mathit{d}{\mathit{\tau }}^{\mathbf{\text{'}}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{-}\mathbf{4}\mathbf{\pi }}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{\int }{\mathbf{\delta }}^{\mathbf{3}}\mathbf{(}\mathbf{r}\mathbf{-}{\mathbf{r}}^{\mathbf{\text{'}}}\mathbf{)}\mathit{p}\left(r^{\text{'}}\right)\mathit{d}{\mathit{\tau }}^{\mathbf{\text{'}}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{-}\mathbf{4}\mathbf{\pi }}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{\int }\mathit{p}\left(r^{\text{'}}\right){\mathbf{\delta }}^{\mathbf{3}}\mathbf{(}\mathbf{r}\mathbf{-}{\mathbf{r}}^{\mathbf{\text{'}}}\mathbf{)}\mathit{d}{\mathit{\tau }}^{\mathbf{\text{'}}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{-}\mathbf{1}}{{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{\int }\mathit{p}\left(r^{\text{'}}\right){\mathbf{\delta }}^{\mathbf{3}}\mathbf{(}\mathbf{r}\mathbf{-}{\mathbf{r}}^{\mathbf{\text{'}}}\mathbf{)}\mathit{d}{\mathit{\tau }}^{\mathbf{\text{'}}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ } \end{gathered}$$

Step 3: Determine the Proof.

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

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

So the ${\mathbf{\nabla }}^{\mathbf{2}}\mathit{V}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{-}\mathbf{1}}{{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{\int }\mathit{p}\mathbf{\left(}{\mathit{r}}^{\mathbf{\text{'}}}\mathbf{\right)}{\mathit{\delta }}^{\mathbf{3}}\mathbf{(}\mathit{r}\mathbf{-}{\mathit{r}}^{\mathbf{\text{'}}}\mathbf{)}\mathit{d}{\mathit{\tau }}^{\mathbf{\text{'}}}$equation becomes,

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

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

Hence, the equation $\mathit{V}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\int }\frac{\mathbf{p}\mathbf{\left(}{\mathbf{r}}^{\mathbf{\text{'}}}\mathbf{\right)}}{\mathbf{r}}\mathit{d}{\mathit{\tau }}^{\mathbf{\text{'}}}$satisfies the Poisson’s equation.