- #1
FaraDazed
- 347
- 2
Homework Statement
There is an electrostatic potential, given by ## \phi(r)=\phi_0 e^{-\alpha r}## where ##\phi_0## and ##\alpha## are some constants.
A: Find the electric charge denisty ##\rho(r)## which produces this aforementioned potential.
A: What is the electric field ##\vec{E}(\vec{r})## due to this charge denisty?
Homework Equations
##\nabla^2 \phi = \frac{- \rho(\vec{r})}{\epsilon_0}##
##\vec{E}(\vec{r}) = - \nabla \phi(\vec{r})##
The Attempt at a Solution
Very new to this material so still am a bit confused and when I can use certain equations and relations etc.
I have used the two equations in the section above, but am not sure if I can. For example the first equation the charge density is a function of the vector r whilst in my question it is just r; although I assume/hope that that just means that is is a constant charge density and the relation still holds in that case.
Also not sure if this is the most suitable section of the forum, so move if needed.
Anyway for A I have done
[tex]
\nabla^2 \phi = \frac{- \rho(r)}{\epsilon_0} \\
\nabla \cdot \nabla \phi = \frac{-\rho(r)}{\epsilon_0} \\
[/tex]
Then ## \nabla \phi## is ..
[tex]
\nabla \phi(r) = (\phi(r))' \frac{\vec{r}}{r} \\
\nabla \phi(r) = -\alpha \phi_0 e^{-\alpha r}\frac{\vec{r}}{r} \\
[/tex]
Then for ##\nabla^2 \phi =##...
[tex]
\nabla \cdot (-\alpha \phi_0 e^{-\alpha r}\frac{\vec{r}}{r})
[/tex]
Let ##a=-\alpha \phi_0 e^{-\alpha r}## and ##\vec{B}= \frac{\vec{r}}{r}##
Then ## \nabla \cdot a \vec{B} = \vec{B} \cdot \nabla a + a \nabla \cdot \vec{B} ##
Then just ##\vec{B} \cdot \nabla a##
[tex]
\vec{B} \cdot \nabla a = \frac{\vec{r}}{r} \cdot - \alpha^2 \phi_0 e^{-\alpha r} \frac{\vec{r}}{r} = -\alpha^2 \phi_0 e^{-\alpha r} \frac{r^2}{r^2} = -\alpha^2 e^{-\alpha r}
[/tex]
Then for ##a \nabla \cdot \vec{B}##
[tex]
\nabla \cdot \vec{B} = \nabla \cdot r^{-1} \vec{r} = \vec{r} \cdot \nabla r^{-1} + r^{-1} \nabla \cdot \vec{r} \\
\nabla \cdot \vec{B} = \vec{r} \cdot - r^{-2} \frac{\vec{r}}{r} + 3 r^{-1} = - \frac{r^2}{r^3} + r^{-1} = \frac{2}{r}
[/tex]
So that ##a \nabla \cdot \vec{B} = \frac{- 2 \alpha \phi_0 e^{-\alpha r}}{r} ##So then ##\nabla^2 \phi = \vec{B} \cdot \nabla a + a \nabla \cdot \vec{B} ##
[tex]
\nabla^2 \phi = - \alpha^2 \phi_0 e^{- \alpha r} - \frac{- 2 \alpha \phi_0 e^{-\alpha r}}{r} \\
\nabla^2 \phi = \phi_0 e^{-\alpha r}(- \alpha^2 - \frac{2 \alpha}{r}) = \frac{- \rho(r) }{\epsilon_0} \\
\rho(r) =- \epsilon_0 \phi_0 e^{-\alpha r}(\alpha^2 + \frac{2 \alpha}{r})
[/tex]And then for B I don't know if this is correct as it seemed to easy but...
[tex]
\vec{E} ( \vec{r} ) = - \nabla \phi ( \vec{r} ) \\
\vec{E} ( \vec{r} ) = \alpha^2 e^{-\alpha r} \frac{\vec{r}}{r}
[/tex]
Any help or advice is very much appreciated! Thank you :)