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Finding E field and charge density due to a potential given

  1. Nov 8, 2015 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations
    ##\nabla^2 \phi = \frac{- \rho(\vec{r})}{\epsilon_0}##
    ##\vec{E}(\vec{r}) = - \nabla \phi(\vec{r})##

    3. 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 dont 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 :)
     
  2. jcsd
  3. Nov 8, 2015 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    It would be much easier if you used the known expressions for the divergence and/or Laplacian in spherical coordinates. See https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates

    However, your method will work. But it looks like you might have made a couple of minor errors.

    Did you make a sign error on the right?

    Another sign error going from the first to second equation shown above?


    Check whether the factor of ##\alpha^2## is correct in the last equation.
     
  4. Nov 8, 2015 #3
    Ah, yes it looks like I did, thanks! Also kept forgetting to type the phi zero factor before all the exp's.

    Yes, so that and with the above error corrected aswell the last part shold be..

    [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]



    Oops yeah should only be alpha, and again I forgot the phi zero,

    So that should be
    [tex]
    \vec{E} ( \vec{r} ) = - \nabla \phi ( \vec{r} ) \\
    \vec{E} ( \vec{r} ) = \alpha \phi_0 e^{-\alpha r} \frac{\vec{r}}{r}
    [/tex]
     
  5. Nov 8, 2015 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I believe that's the correct expression for the electric field.
     
  6. Nov 8, 2015 #5
    Thank you. I just edited my reply, when I first posted it half of my reply got caught in the quote tags.
     
  7. Nov 8, 2015 #6

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Looks good except for a sign error in the exponent of ##e##.
     
  8. Nov 8, 2015 #7
    Ah yeah, I think that one was just a typo, the others were proper sign errors haha. Thanks for taking a look! :)
     
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