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Gaussian Surface

  1. Sep 17, 2016 #1
    1. The problem statement, all variables and given/known data
    A thin spherical shell with radius R1 = 2.00 cm is concentric with a larger thin spherical shell with radius 6.00 cm . Both shells are made of insulating material. The smaller shell has charge q1=+6.00nC distributed uniformly over its surface, and the larger shell has charge q2=−9.00nC distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells.

    Part A: What is the electric potential due to the two shells at the following distance from their common center: r = 0?
    Part B: What is the electric potential due to the two shells at the following distance from their common center: r = 3.00 cm ?
    Part C: What is the electric potential due to the two shells at the following distance from their common center: r = 7.00 cm ?


    2. Relevant equations
    V = ∫E. dl = (k q) / r
    Φ = q(enc) / ε0 = ∫E.dA



    3. The attempt at a solution
    I got Part A:
    V for inner shell: V= (k q) / r = (8.99E9 * 6.00E-9) / 0.02 = 2700
    V for outer shell: V = (k q) / r = (8.99E9 * -9.00E-9) / 0.06 = -1350
    ΣV = 2700 + (-1350) = 1350V

    I eventually got Part B:
    V = -k q1 (1/0.03 - 1/0.02) = 899V
    V = -k q2 (1/0.06 - 1/ 0.03) = -1350
    ΣV = 899 - 1350 = -450V (Although the answer had to be changed to +450, not really sure why...Maybe V is magnitude)

    I tried doing the same thing Idid for Part B for Part C and got stuck:
    V = -k q1 (1/0.07 - 1/0.02) = 1930V
    V = -k q2 (1/0.07 - 1/ 0.06) = -192V
    ΣV = 1930 - 192 = 1740V, but that is apparently wrong.....

    I'm thoroughly confused :/

     
  2. jcsd
  3. Sep 17, 2016 #2

    gneill

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    Staff: Mentor

    Hi rsatchel, Welcome to Physics Forums.

    For part B, the location of interest lies between the shells. So it has the potential of the outer shell's surface plus whatever potential it experiences from the inner shell at its distance from that shell.

    For part C, the location is outside of both shells so the easiest approach is to draw a Gaussian surface at the same radius and apply Gauss' Law. What's the total charge contained within the Gaussian surface?
     
  4. Sep 17, 2016 #3
    Ah, OK.
    So I did q(net) = -3nC
    V = (k q)( 1/inf - 1/0.07)
    V = -(8.99E9 * -3E-9)(-1/0.007) = -385V
     
  5. Sep 18, 2016 #4
    Ok, so now there's another part

    What is the magnitude of the potential difference between the surfaces of the two shells?

    I thought I would need to use:
    V = ∫ E. dl = ∫ -(k q)/r dr (from A to B)
    V = (-k q) / r again,
    V = -k q( 1/B - 1/A)
    but what would the limits be?

    I thought maybe I should draw a Gaussian Surface at r=7cm and use the net charge to calculate the outer shell's potential and then do the same for the inner shell - didnt work.
    I also tried doing the limits from the inner shell to the outer shell - didnt work.

    Not sure what else to try :/
     
  6. Sep 18, 2016 #5

    gneill

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    Staff: Mentor

    The potential at the surface of the outer shell will be due the total charge enclosed in a Gaussian surface immediately surrounding that shell.

    The potential at the surface of the inner shell will be the potential due to that shell alone at its surface plus the potential due to the outer shell alone at that shell's surface (in passing into the interior space within the outer shell, the potential due to that outer shell is "remembered").

    Have a look here for more details.
     
  7. Sep 18, 2016 #6
    Thank you, that helped a lot!
     
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