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Potential Difference problem with sphere and capacitor

  1. Feb 13, 2014 #1
    1. The problem statement, all variables and given/known data

    physics_problem.jpg

    A thin spherical shell made of plastic carries a uniformly distributed negative charge -6e-10 coulombs (indicated as -Q1 in the diagram). Two large thin disks made of glass carry uniformly distributed positive and negative charges 1.5e-05 coulombs and -1.5e-05 coulombs (indicated as +Q2 and -Q2 in the figure). The radius R1 of the plastic spherical shell is 6 mm, and the radius R2 of the glass disks is 4 meters. The distance d from the center of the spherical shell to the positive disk is 18 mm.

    (a) Find the potential difference V1 - V2. Point 1 is at the center of the plastic sphere, and point 2 is just outside the sphere.


    (b) Find the potential difference V2 - V3. Point 2 is just below the sphere, and point 3 is right beside the positive glass disk.


    Remember that the sign of the potential difference is important.


    2. Relevant equations
    Esphere = (1/4ε0)*(q/r2)
    Ecapacitor = Q/A/ε0
    ΔV = - (E*ΔL)


    3. The attempt at a solution

    a)
    Ecapacitor = Q/A/ε0
    = (1.5e-5)/(4^2∏)/(8.85e-12) = 3.4e4 N/C

    ΔV = - (E*ΔL)
    = - (3.4e4 * .006) = -202.32 V/m


    b)

    I don't understand how to do part b. I really could use some help. Part A is correct.
     
    Last edited: Feb 13, 2014
  2. jcsd
  3. Feb 13, 2014 #2
    Please provide the figure.
     
  4. Feb 13, 2014 #3
    I added it.
     
  5. Feb 13, 2014 #4

    BvU

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    Is it unreasonable to think you can separately calculate the potential differences and add them up ?
    (and take good care of the signs!)
     
  6. Feb 13, 2014 #5
    I've tried that, and it could be that I'm doing it wrong, maybe mixing up signs, but I'll take another look at it!
     
  7. Feb 14, 2014 #6

    BvU

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    Well, if we find something in the category "The attempt at a solution" in the next post, maybe we can help. That's what PFHW is for.
     
  8. Feb 14, 2014 #7
    Sorry, I meant to respond again, I figured out the problem. I was using the wrong equation for finding the potential difference due to the sphere.
     
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