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Finding potential between identical spheres

  1. Feb 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Two oppositely charged identical insulating spheres, each 56.0cm in diameter and carrying a uniform charge of magnitude 165μC , are placed 1.20m apart center to center.
    If a voltmeter is connected between the nearest points (a and b) on their surfaces, what will it read?
    a1r0h1.jpg


    2. Relevant equations
    V=k((q1/r1)-q2/r2)


    3. The attempt at a solution
    At first I thought that the potential between the two would be zero because they are equal but opposite. However, that was incorrect. I have let r=0.6 and r=0.88 however, using both was still incorrect. I do not know where to go from here.
     
  2. jcsd
  3. Feb 2, 2013 #2

    gneill

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    Staff: Mentor

    Start by finding the individual potentials at points a and b near the surfaces of the spheres.
    Show your work.
     
  4. Feb 2, 2013 #3
    Va=k(q/r)=(8.99*109)((165*10-6)/.28)=5.298*106

    The potential for Vb would be the same but opposite. Is this correct?
     
  5. Feb 2, 2013 #4

    gneill

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    What about the contribution of the field of the other sphere at a? Same goes for point b where the fields from both spheres contribute to the total.
     
  6. Feb 2, 2013 #5
    This is where I am confused. Would you use the same equation but make r=1.20m?
     
  7. Feb 2, 2013 #6

    gneill

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    Staff: Mentor

    The distance is always from the center of the charge to the location in question.
     
  8. Feb 2, 2013 #7
    So
    Va=k(q/r)=(8.99*109)((165*10-6)/1.48)=1.002*106?
    and the same for Vb?
     
  9. Feb 2, 2013 #8

    gneill

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    How did you arrive at the distance of 1.48m ? Also, there should be contributions from both charges.
     
  10. Feb 2, 2013 #9
    The distance between the two is r=1.2m plus from the center of the sphere is 0.28m. Therefore the total distance would be 1.48m. If there are contributions from both charges am I not using the right equation?
     
  11. Feb 2, 2013 #10

    gneill

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    The distance between the spheres is specified center-to-center.

    For a given location in space you need to find the contribution of both charges. That means applying your equation twice for each location, once for each charge that contributes.
     
  12. Feb 2, 2013 #11
    So would it be Va=(8.99*109)((165*10-6)/1.76) plus the contribution from Vb therefore the answer would be V=1.7*106V?
     
  13. Feb 2, 2013 #12

    gneill

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    I think you meant 0.28 rather than 0.88 in the above, as your result value looks okay.

    That takes care of the potential at point a. As you said previously, the potential at b will be the same but with opposite sign. What then is the potential difference between the two points?

    EDIT: I take it back, your result looks a bit low. Re-check your calculations.
     
    Last edited: Feb 2, 2013
  14. Feb 2, 2013 #13
    The difference would then be zero?
     
  15. Feb 2, 2013 #14

    gneill

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    I just realized that your result for the potential at point a still looks a bit off (a bit low). Recheck your calculations. What are the individual contributions from the two charges at point a?

    Regarding the difference value, if the potentials have opposite signs, how can the difference be zero?
     
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