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Finding Total Charge from E-field

  1. Jan 1, 2018 #1
    1. The problem statement, all variables and given/known data
    A static charge distribution has a radial electric field of magnitude
    ##E = \alpha \frac{e^{-\lambda r}}{r} ##
    where λ and α are positive constants. Calculate the total charge of the distribution.

    2. Relevant equations
    Gauss's law ##Q/\epsilon_0 = \int \vec{E} \cdot d\vec{S}##
    ##\rho/\epsilon_0 = \nabla \cdot \vec{E} ##

    3. The attempt at a solution
    I have tried two ways to go about this problem, without success. First I tried using Gauss's law, placing a Gaussian sphere of radius a on the distribution:
    ##Q/\epsilon_0 = \int \vec{E} \cdot d\vec{S} = 4\pi\alpha (ae^{-\lambda a}) ##
    We should now find the total charge by letting ## a \rightarrow \infty ##, but ##\lim_{a \rightarrow \infty} ae^{-\lambda a} = 0##, giving ##Q = 0## which i thought cannot be right.*

    My second method involved finding the charge distribution
    ##\rho/\epsilon_0 = \nabla \cdot \vec{E} = \alpha \frac{e^{-\lambda r}(1-\lambda r)}{r^2} ## but this gives an integral which is divergent!

    *I have done some further thinking and realised that ##Q_{tot} = 0## might be the answer, because the charge density changes sign at ##r = 1/\lambda## so the total net charge could as well be zero.

    However, how is it possible that the volume integral of #\rho# is divergent then? Or is it not?
     
  2. jcsd
  3. Jan 1, 2018 #2

    TSny

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    Make sure you are setting up the integral correctly. Are you using the correct form for the volume element of the integral?
     
  4. Jan 2, 2018 #3
    Isn't E = k Q / R^2 (spherical charge distribution)?
    Then 2 E R dR = k dQ.
    The integral doesn't diverge unless (1 / R) appears in the result at R = 0.
     
  5. Jan 2, 2018 #4

    TSny

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    Yes.
    This doesn't look right. From E = k Q / R2 you have k Q = R2 E. If you take the differential of this, you need to note that E is a function of R.

    But I don't see the point of doing this. kQ = R2E already tells you how much charge is contained in a sphere of radius R. So, taking the limit as R goes to infinity will give you the total charge of the system. This is essentially what @Marcus95 did in his first method of finding Qtot.
     
  6. Jan 2, 2018 #5
    I tend to agree, but why does R need to go to infinity?
    Was this implied in the problem?
     
  7. Jan 2, 2018 #6
    Also, doesn't L'Hopital's give a finite value for E as R goes to infinity?
    (It does appear to be zero)
     
  8. Jan 2, 2018 #7

    TSny

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    The charge density extends to infinity. So, to get the total charge you have to include all the charge out to infinity.
     
  9. Jan 2, 2018 #8

    TSny

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    ##E = \alpha \frac{e^{-\lambda r}}{r} ##

    L'Hopital gives zero for the limit, but you don't really need L'Hopital. The numerator goes to zero while the denominator goes to infinity as ##r## goes to infinity.
     
  10. Jan 7, 2018 #9
    My integral is:
    ##Q = \int_{r=0}^\infty \int_{\theta=0}^{\pi} \int_{\phi=0}^{2\pi} \alpha\epsilon_0 \frac{e^{-\lambda r}(1-\lambda r)}{r^2} \times r^2sin\theta d\theta dr d\phi ##
    which gives:
    ##Q = 4\phi \alpha \epsilon_0 \int_{r=0}^\infty e^{-\lambda r}(1-\lambda r) dr =4\phi \alpha \epsilon_0 [re^{-\lambda r}]_0^\infty = 0 ##
    so it all works out! Thanks a lot!
     
  11. Jan 7, 2018 #10

    TSny

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    OK. In the second equation, the ##\phi## should be ##\pi##??

    Otherwise, looks good.
     
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