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Potential difference in a sphere relative to infinity

  1. Jun 1, 2012 #1
    1. The problem statement, all variables and given/known data
    A thin plastic spherical shell of radius a is rubbed all over with wool and gains a charge of -Q. What is the potential relative to infinity at location B, a distance a/3 from the centre of the sphere?


    2. Relevant equations

    [tex] \text{$\Delta $V}=\int _{\text{initial}}^{\text{final}}{E}.d{l} [/tex]


    3. The attempt at a solution
    since the potential difference between any 2 points in a spherical shell is 0, point B can be anywhere in the sphere (it doesnt have to be a/3 from the centre).

    so if we take point B as a point, an infinitesimal distance away from the shell (still inside the shell)

    then we can treat the shell as a point charge:
    [tex] \text{$\Delta $V}=\int _{\infty }^a\frac{1}{4\pi \epsilon }\frac{Q}{r^{2}}dr [/tex]

    which just gives:
    [tex] \frac{1}{4\pi \epsilon }\frac{Q}{a} [/tex]


    i dont think this answer is right. So, wheres the flaw in my logic, and how are you meant to do this question?

    (edit: the epsilons are meant to have a 0 subscript)
     
    Last edited: Jun 1, 2012
  2. jcsd
  3. Jun 1, 2012 #2
    Why do you think it is wrong??

    To be honest, I think that's the correct method. What you did was basically apply the Shell theorem and it does hold for any spherical shell with symmetric charge density...
     
  4. Jun 1, 2012 #3
    well, i thought my answer was too simple/easy to be right i guess.
     
  5. Jun 1, 2012 #4
    You have to take into account that the sphere has a charge of -Q not +Q, otherwise it's fine
     
  6. Jun 1, 2012 #5
    there should be a negative sign before this integral and the charge is negative. two negatives make a positive

    [tex] \text{$\Delta $V}=\int _{\infty }^a\frac{1}{4\pi \epsilon }\frac{Q}{r^{2}}dr [/tex]
     
  7. Jun 1, 2012 #6
    Yeah sorry, I just meant it for completeness I agree with the integral just if you do the integral you get an overall negative sign, which is helpful if you want to work out the work done moving this charge from infinity to a for example.
     
  8. Jun 1, 2012 #7

    TSny

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    Gold Member

    Looks good. But the integral of 1/r2 is -1/ r, so the answer will have opposite sign of what you wrote. The potential is negative near a negative charge.
     
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