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Homework Help: Total Charge what? potential by line inter

  1. Apr 25, 2005 #1
    alright, so i got this potentional equation [tex] \Phi = k_e Q e^{-\alpha r}/r[/tex]
    it askes me to find the total charge after calcing the charge density, so
    anyway...lets get the field [tex]\vec{E}=-\nabla \Phi[/tex]
    so yea then take the divergence fot the charge density
    [tex] \nabla \bullet \vec{E} = 4 \pi k_e p [/tex]
    so then im assuming to figure out the "total charge" im gonna use the density in a volume intergral and equate that to [tex] 4 \pi k_e Q_{enclosed} [/tex] but what is my limit? is it is a sphere i can define by radius a as my surface? or what?

    next question

    finding the potential associated with a Vector field A by line intergration in polar cords (or any cords for that matter) whats that mean
    [tex]\oint \vec{A} \bullet d\vec{r} = \Phi (b) - \Phi (a)[/tex] is that what they're talking about let me know
  2. jcsd
  3. Apr 25, 2005 #2


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    I can't exactly follow what you are talking about in the first case. Perpaps you could state the problem more specifically. The potential certainly suggests a spherically symmetric charge distribution. Is that potential good for all r?

    Your second question is more or less right, but that integration symbol is for closed path integrals where the resulting potential difference would be zero. You want an integral from point a to point b.
  4. Apr 25, 2005 #3
    alright arlight,

    the first one the total charge is what im asking what is the "total charge" mean

    [tex] \int \int^{2\pi}_0 \int^{\pi}_0 \nabla \bullet (-\nabla \Phi) dV = 4 \pi k_e Q_e[/tex]
    whats the last limits to solve for [tex]Q_e[/tex]
  5. Apr 25, 2005 #4
    Have you calculated divE yet? That gives you the charge density anywhere in space. So...how far out does your integral have to go?
  6. Apr 26, 2005 #5
    charge density is, [tex] p = Q \alpha^2 e^{-\alpha r}}/(4 \pi r) [/tex] so im thinking infinity which the intergral is undefined there
  7. Apr 26, 2005 #6


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    Assuming your charge density is correct (I think it is), the problem with the integral is not the limit at infinity, it's the limit at zero. I believe that if you start the integral at some radius b, you get a gamma function for the result. That's why I was initially wondering if the potential was good for all values of r.

    I'm not suggesting that you don't learn to integrate, but I was never that great at it, so I'm glad things like this are on the internet.

    http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=advanced [Broken]
    Last edited by a moderator: May 2, 2017
  8. Apr 26, 2005 #7

    But won't there be a differential volume element r^2dr?
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