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Differential form of Gauss's Law

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data

    In a model of an atomic nucleus, the electric field is given by:

    E = α r for r < a

    where α is a constant and a is the radius of the nucleus.


    Use the differential form of Gauss's Law to calculate the charge density ρ inside the nucleus.


    2. The attempt at a solution

    Using the simple version of Gauss's law :

    [tex]\int_{S} \underline{E}.\underline{dS} = \int_{V} \frac{\rho}{\epsilon_{0}} dV [/tex]

    Yields a result [tex]\rho = \frac{3E\epsilon_{0}}{r} [/tex] for 0<r<=a

    1. The problem statement, all variables and given/known data

    However when using the differential form:

    [tex]\nabla . \underline{E} = \frac{\rho}{\epsilon_{0}}[/tex]

    [tex]\frac{1}{r} \frac{\partial (r E_{r})}{\partial r} = \frac{\rho}{\epsilon_{0}} [/tex]

    and when integrating with respect to r from 0 to a,

    [tex] \rho = {\epsilon_{0}} E (1 + ln a ) [/tex]


    Any helpful advice would be appreciated.
     
    Last edited: Oct 20, 2009
  2. jcsd
  3. Oct 20, 2009 #2

    turin

    User Avatar
    Homework Helper

    I think you accidently used the cylindrical version of the divergence.

    Why are you integrating?
     
  4. Oct 20, 2009 #3
    I have used the cylindrical version, though this problem only depends on r so maybe use spherical? I integrated the partial differential with respect to r.

    if I assume that Er is not a function of r but a constant then,

    [tex] \frac{\partial (r E_{r})}{\partial r} = E_{r} [/tex]

    or if i use a radial divergence from the beginning:

    [tex] \nabla . E = \frac{\partial (E)}{\partial r} [/tex]

    but by using that I get:

    [tex] E = \frac{\rho r}{\epsilon_0} [/tex]

    i.e missing a factor of 3
     
    Last edited: Oct 20, 2009
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