1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Determine charge at origin, based on charge density function

  1. Sep 23, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-9-23_18-30-1.png

    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

    mfb

    User Avatar
    2016 Award

    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

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Set them up, then find useful bounds on them.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Determine charge at origin, based on charge density function
  1. Charge at origin (Replies: 2)

Loading...