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Surface Charge Density

  1. Feb 7, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-2-7_10-37-7.png

    2. Relevant equations
    E=σ/(2Eo)
    σ=2Eoma/e
    a=Δv/Δt

    3. The attempt at a solution
    So doing my best to read the velocity over time graph I came up with
    Δv/Δt=(-2E5m/s)/(10E-12s)=-2E17 m/s/s

    σ=2Eoma/e
    =2(8.99E-9C)(9.109E-31kg)(-2E17m/s/s)/(1.6022E-19)
    =-0.020444 C/m^2
    but by the first illustration you can see that the sheet has a positive charge so the charge density should be:
    0.020444 C/m^2 <----this final answer was incorrect
     
  2. jcsd
  3. Feb 7, 2017 #2

    haruspex

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    The question states 2.9E5m/s

    What is th sign of sn electron's charge?
     
  4. Feb 7, 2017 #3
    Okay let's try this again.
    A=(2*-2.9E5m/s)/(2E-11s)
    =-2.9E16
    So using that correct value along with the appropriate value for the charge of an electron:
    2(8.99E9C)(9.109E-31kg)(-2.9E16)/(-1.6022E-19)
    2.964E15 C/m^2
     
  5. Feb 7, 2017 #4

    haruspex

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    Looks ok to me now. Still marked as wrong?
     
  6. Feb 7, 2017 #5

    TSny

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    Is the green number correct?
     
  7. Feb 7, 2017 #6

    haruspex

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    I have a bad habit of assuming posters know what constants to use and what their values are.
     
  8. Feb 7, 2017 #7
    should this be the permittivity of free space?
     
  9. Feb 7, 2017 #8

    haruspex

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    I believe so.
     
  10. Feb 7, 2017 #9

    TSny

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    I would have overlooked it also, but his/her result for σ was humongous.
     
  11. Feb 7, 2017 #10
    Let's try this again
    The acceleration according to the graph is 2*(-2.9E5m/s)/2E-11= -2.9E16 m/s/s

    σ=2Eoma/e
    =2*(8.855E-12)(9.109E-31)(-2.9E16)/(-1.6022E-19)
    =3E-6 C/m^2
    Still not acceptable
     
  12. Feb 7, 2017 #11

    haruspex

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    You can either use Coulomb's constant, but as a divisor here, or the permittivity as a multiplier. But in the second you need a 4π too.
    Edit: no that's wrong... Let me check.

    Edit2: I should have written, you can either divide by Coulomb's constant and by 4π, or multiply by the permittivity, as you did. Anyway, TSny seems to have identified the remaining issue.
     
  13. Feb 7, 2017 #12

    TSny

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    Looks good. But the data was given to 3 significant figures.
     
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