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Charge density for a given potential

  1. Feb 18, 2017 #1
    1. The problem statement, all variables and given/known data
    Given the electric potential ##V(r)=A\frac{e^{-\lambda r}}{r}## calculate the charge density ##\rho(r)## and the electric field ##E(r)##.
    They specify the answer for charge density should be: ##\rho = \epsilon_0 A(4\pi \delta^3(r)-\lambda^2e^{-\lambda r}/r)##

    2. Relevant equations
    ##E=-\nabla V##
    ##\nabla E = \frac{\rho}{\epsilon_o}##

    3. The attempt at a solution
    For the electric field I got: ##E = Ae^{-\lambda r}(1+\lambda r)/r^2##, pointing in the r direction and this is the same as their answer. Then I tried to take the divergence of E, and in spherical coordinates this would be ##\nabla E = \frac{1}{r^2}\frac{\partial r^2 E}{\partial r} = \frac{1}{r^2}\frac{\partial (Ae^{-\lambda r}(1+\lambda r))}{\partial r} ## . However this doesn't give me a delta function so I am not sure where I did something wrong.
     
  2. jcsd
  3. Feb 18, 2017 #2

    vela

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    Your expression for the divergence is only valid for ##r\ne 0##. You need to consider what's happening at ##r=0## in another way.
     
  4. Feb 19, 2017 #3
    Thank you! So I found a solution online to this problem but I am still a bit confused. I attached their solution. So i understand how they obtain the delta function but when they apply the divergence to the other term, they use ##\frac{\partial}{\partial r}##. Shouldn't they use the divergence in spherical coordinates which is ##\frac{1}{r^2}\frac{\partial (r^2 E')}{\partial r}##, with E' being the term that remained to be acted on?
     

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  5. Feb 19, 2017 #4

    TSny

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    They are using the identity
    upload_2017-2-19_16-56-10.png
    where ##\psi## is a scalar function.

    (See https://en.wikipedia.org/wiki/Vector_calculus_identities#Product_of_a_scalar_and_a_vector)

    The second term on the right does not involve a divergence. It involves a gradient.
     
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