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Determine charge at origin, based on charge density function

  1. Sep 23, 2015 #1
    1. The problem statement, all variables and given/known data

    a) and b) are no problem.

    I need help to solve c) and d)

    2. Relevant equations
    c) Delta dirac function
    Gauss' law

    d) Gauss' law
    ## \int_V {\rho \, d\tau} = Q_{enclosed} ##

    3. The attempt at a solution
    By taking laplace on the potential I get:

    ## \rho(\mathbf{r}) = \frac{q_0}{4 \, \pi \, r} \, e^{-r/\lambda} \, \left( \frac{cos^2(\theta)}{\lambda^2} + \frac{2}{r^2} (1-3 \, cos^2(\theta)) \right)##

    c) I got a hint that it was a good idea to use the dirac delta function along with the charge distribution.

    But I'm not exactly sure why. As I understand it the dirac delta function "picks out" the value of a function at zero. So I'd get:

    ## \int {\rho(\mathbf{r}) \, \delta(r) \, dr} = \rho(0) ##

    I realize that there must be a dimensional problem here, but I'm not sure how to use a delta function in 3D and spherical coordinates.
    Also how will it help me to find the density at the origin? Can I apply Gauss' law here and let the radius go towards zero to get the charge in the origin?

    d) I want to solve the integral

    ## Q = \int_0^\pi \int_0^{2 \, \pi} \int_a^{\infty} \, \rho(\mathbf{r}) \, r^2 \, \sin(\theta) \, dr \, d\theta \, d\phi ##

    I tried evaluating this with Maple.
    By assuming a>0 I get a complex function multiplied by infinity, which is not of much use.
    If I also assume lambda>0 (as it says in the problem) I get rid of the infinity, but get exponential integrals instead.
    I'm not sure how to move on from here. I suspect I need to modify my function for charge distribution by assumptions, to make it simpler.
  2. jcsd
  3. Sep 23, 2015 #2


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    Staff: Mentor

    This assumes your charge density formula doesn't have problems with r=0. That might be, but I wouldn't rely on it.

    Using the electric field, you can determine the total charge up to a radius a and then let a go to zero.
    The opposite limit also works for (d) and does not need evaluating any actual integrals.
  4. Sep 24, 2015 #3
    By best idea was also to use Gauss' law. ## \int_V \mathbf{\nabla} \cdot \mathbf{E} \, d\tau = \frac{1}{\epsilon_0} \, Q_{enclosed} ##
    How can I avoid evaluating any actual integrals?
    The charge density/ elctric field depends on both r and theta.
  5. Sep 24, 2015 #4


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    Staff: Mentor

    Set them up, then find useful bounds on them.
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