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Singularity of potential

  1. Oct 19, 2016 #1
    1. The problem statement, all variables and given/known data

    Determine the electric field and the charge density associated with the potential ##\displaystyle{V(r)=A\frac{\exp{-\lambda r}}{r}}##, where ##A## and ##\lambda## are constants.

    2. Relevant equations

    3. The attempt at a solution

    The electric field is easy to determine:

    ##\displaystyle{{\bf{E}} = -\nabla V}##
    ##\displaystyle{= -A\ \nabla\left(\frac{\exp\left(-\lambda r\right)}{r}\right)}##
    ##\displaystyle{= -A\ \left(\frac{\partial}{\partial r}\left(\frac{\exp\left(-\lambda r\right)}{r}\right),0,0\right)}##
    ##\displaystyle{= A\ \left(\frac{(1+\lambda r)\exp\left(-\lambda r\right)}{r},0,0\right)}##

    But the electric field is singular at the origin.

    How does this affect the calculation of the charge density?
     
  2. jcsd
  3. Oct 19, 2016 #2
    Well, that means you have a point charge at the origin - recall that ##\nabla \cdot \left(\frac{\vec{r}}{r^2}\right) = 4 \pi \delta(r)##.
     
  4. Oct 19, 2016 #3
    In which step have I made the mistake?

    ##\rho = \epsilon_{0}\ \nabla\cdot{{\bf{E}}}##

    ##= \epsilon_{0}\ A\ \nabla\cdot{\left((1+\lambda r)\frac{\exp\left(-\lambda r\right)}{r},0,0\right)}##

    ##= \epsilon_{0}\ \frac{A}{r^{2}}\ \frac{\partial}{\partial r} \left(r^{2}(1+\lambda r)\left(\frac{\exp\left(-\lambda r\right)}{r}\right)\right)##

    ##= \epsilon_{0}\ \frac{A}{r^{2}}\ \frac{\partial}{\partial r} \left[(r+\lambda r^{2})\exp\left(-\lambda r\right)\right]##

    ##= \epsilon_{0}\ \frac{A}{r^{2}}\ \left[(1+2\lambda r)\exp\left(-\lambda r\right)-\lambda(r+\lambda r^{2})\exp\left(-\lambda r\right)\right]##

    ##= \epsilon_{0}\ \frac{A}{r^{2}}\ \left[(1+2\lambda r)-\lambda(r+\lambda r^{2})\right]\exp\left(-\lambda r\right)##

    ##= \epsilon_{0}\ A\ \left[1+\lambda r-(\lambda r)^{2})\right]\frac{\exp\left(-\lambda r\right)}{r^{2}}##
     
  5. Oct 19, 2016 #4
    Here:
     
  6. Oct 20, 2016 #5
    I only missed a negative sign in the electric field, right?
     
  7. Oct 20, 2016 #6
    No, the signs are right; its the denominator that should be ##r^2##, no?
     
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