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Calculation the resistance of a spherical shape

  1. Aug 18, 2010 #1
    hello there!
    I was reading about ohm rule in a MIT physics course and they calculate the resistance of a nested spherical shells like this :

    attachment.php?attachmentid=27628&stc=1&d=1282143758.jpg

    and they but the microscopic form of ohm's law which is :
    [tex]J=\sigma_q E[/tex]
    and
    [tex]I=A J[/tex]
    so
    [tex]I=(4 \pi r^2 ) (\sigma_q E)[/tex]
    and
    [tex]E=-\frac{\partial V}{\partial r}[/tex]
    and he said that the potential must be like this form
    [tex]V=\frac{C}{r}+D[/tex]
    where : C and D are constant .
    so my question is what is this form and why he did that and i could calculate the potential easily :
    [tex]V_{ab}=\int E . dr[/tex].
    thank you
     

    Attached Files:

  2. jcsd
  3. Aug 18, 2010 #2
    For symmetry, the electric field is directed radially. Since there are no charges inside the conductor, the flux

    [tex]\Phi=4\pi r^2 E[/tex]

    must be constant. The difference of potential between a and b is

    [tex]\Delta V=\int_a^b\frac{\Phi dr}{4\pi r^2}=\frac{\Phi(b-a)}{4\pi ab}[/tex].

    But, as you said, [tex]I=\sigma\Phi[/tex], so

    [tex]\Delta V=\frac{I(b-a)}{4\pi ab\sigma}\equiv RI[/tex]

    and hence

    [tex]R=\frac{(b-a)}{4\pi ab\sigma}[/tex]
     
  4. Aug 18, 2010 #3
    thank you i think it is solved now .
     
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