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Homework Help: Please help me with this question!

  1. Nov 4, 2004 #1
    There is a spherical resistor witn an inner radius a and outer radius of b. The outer surface and the inner surface are covered with conducting sheets. Find the resistance betwen the two surfaces assuming uniform resistivity [tex] \rho [/tex]


    I know [tex] R = \rho \frac{L}{A} [/tex]
    If i divided the sphere into many cylinders then each cylinder would have a length of (b-a) and then the area would be ab times the radius (b-a)?

    so then i end up with [tex] R = \rho \frac{1}{(b-a)ab} [/tex]

    but that isnt right because i know the answer has someting to do with 4pi? plase help!
    Last edited: Nov 4, 2004
  2. jcsd
  3. Nov 4, 2004 #2
    This is an example of a continuous distribution of resistive material. That should give you a hint of using calculus. So instead of R = rho * L/A you will get

    [tex]dR = \rho\frac{dL}{A}[/tex]

    and proceed.
  4. Nov 4, 2004 #3
    but what is the expression for A? area of this box would be (b-a) times ....??
  5. Nov 5, 2004 #4


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    From first principles the current density j (current per unit area) is

    [tex]j = \sigma E = -\frac {1}{\rho} \frac {\delta V}{\delta r}[/tex]

    where [itex]\sigma = 1/\rho[/itex] is the conductivity. Now

    [tex]j = \frac {I}{4\pi r^2}[/tex]

    for your problem and it follows that

    [tex]\Delta V = -\int_{a}^{b} \frac {\rho I}{4 \pi r^2} \delta r[/tex]


    [tex]\Delta V = \frac {\rho I}{4 \pi} \left( \frac {1}{a} - \frac {1}{b}\right)[/tex]

    The resistance is just the right hand side divided by I. Notice that in your original expression the units are not correct which is a hint that something was amiss.
  6. Nov 5, 2004 #5
    thank you very much!
    that wasn't an approach i was thinking about but it works , and how!

    thank you very much again!
  7. Nov 5, 2004 #6
    The dR approach is equivalent to Tide's approach. When I gave you the hint to use calculus, you should've realized that I am referring to the area of a spherical surface which would be 4pi (r squared).
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