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Integral Over all Space for Charge Density - Exponential Fun

  1. Mar 12, 2015 #1
    1. The problem statement, all variables and given/known data

    The electric field is described by [itex] E = \frac{C e^{-br}}{r^2} [/itex],
    find the charge density and then integrate over all-space and show it's zero.



    2. Relevant equations

    [itex] E = \frac{C e^{-br}}{r^2} \\ \\ [/itex]
    [itex] \nabla \cdot E = \frac{\rho}{\epsilon_0} \\ \\ [/itex]
    [itex] \rho(r) = Q \delta^3 (r-r_0) \\ \\ [/itex]

    [itex]I_{AllSpace} = \int_{0}^{\infty}\:f(r,\phi,\theta) \: r^2 \: dr \: sin\theta \: d\theta \: d\phi \\ \\ [/itex]
    [itex]\nabla = \frac{d}{dr} \hat{r} \\ \\ [/itex]
    [itex] \int_{0}^{\infty}f(r)Q \delta^3(r-r_0)dr = f(r_0) \\ \\ [/itex]

    3. The attempt at a solution

    I tried to take [itex] \nabla E = \frac{\rho}{\epsilon_0} [/itex] and used the quotient solve for [itex]\rho(r)[/itex] and then integrated over all space and I do not get a zero.

    With [itex]\nabla \cdot E = (\frac{-e^{-ar}}{r} \:(ra+2))[/itex]
    [itex]\rho(r) = (\frac{-e^{-ar}}{r} \:(ra+2)) \: \epsilon_0[/itex]

    and integrating over all space gives me

    [itex]4\pi \int_{0}^{\infty}\rho(r) \: r^2 \: dr[/itex]

    and this gives me [itex](-1 + undefined) \: \epsilon_0[/itex]

    my only other idea here is to use the delta function, which in the case of the radial integral does give me a zero due to the [itex] r^2 [/itex] term being evaluated at zero.

    [itex]E_{AllSpace} = 4\pi \int_{0}^{\infty}\rho(r) \: r^2 \: dr = 4\pi Q \int_{0}^{\infty} \delta^3(r-r_0)r^2 \:dr = 0 [/itex]

    This second solution seems a bit short but I think it is still valid, I'm not sure if my first attempt if I made a mistake on the integral or if there is some other way of doing this? Any help would be appreciated.
     
  2. jcsd
  3. Mar 12, 2015 #2
    What's the direction of the electric field vector? You stated it as if it's a scalar.
     
  4. Mar 12, 2015 #3
    Ahh yes, it is in the [itex]\hat{r}[/itex] direction.
     
  5. Mar 13, 2015 #4
    OK, so I found an issue and corrected for it, I was using [itex] \nabla [/itex] in cartesian coordinates, and it should have been in spherical.

    My result after using spherical component for [itex] r [/itex] from [itex] \nabla [/itex], I get

    [itex] \rho(r)_{AllSpace} = 4\pi \int_{0}^{\infty} \frac{-bA}{\epsilon_0 } e^{-br} = A[/itex]

    but I noticed that if I still use the delta function I get the desired answer of zero, but I think it is Q that I am trying to find? Any help is appreciated.
     
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