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Two-electrode spherical system (Potential)

  1. Oct 21, 2011 #1
    1. The problem statement, all variables and given/known data

    See figure attached.

    2. Relevant equations



    3. The attempt at a solution

    See figure attached for the provided solution.

    I got everything up to what I put in a red box.

    Where does he get that negative from?

    Did he do that with the intentions of reversing the limits on his integral?
     

    Attached Files:

  2. jcsd
  3. Oct 21, 2011 #2

    SammyS

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    [itex]\displaystyle \Phi_r-\Phi_b=-\int_b^r\vec{E}\cdot\vec{dr}[/itex]
     
  4. Oct 21, 2011 #3
    Isn't that the potential at some radius r, with the reference set at radius b?

    In the question it says that the reference for the potential should be the potential on the inner conductor where the radius is a.

    I have the urge to write,

    [tex]\phi(r) = \int_{b}^{r} \vec{E} \cdot \vec{dr} + C[/tex]

    Where C is chosen such that,

    [tex]\phi(a) = 0[/tex]

    but we already know from the first part that it is already 0, so C=0.
     
    Last edited: Oct 21, 2011
  5. Oct 21, 2011 #4

    SammyS

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    From the first part, we know that Φb = Φa
     
  6. Oct 21, 2011 #5
    On the third line from the bottom, didn't he miss a negative when he evaluated his integral?

    It should be,

    [tex]\left(-\frac{1}{r} \right)^{r}_{c} = \frac{1}{c} - \frac{1}{r}[/tex]

    Correct?
     
  7. Oct 21, 2011 #6

    SammyS

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    After looking at it for some time, I agree with you.
     
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