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Electric Potential (charged conducting sphere)

  1. Apr 5, 2009 #1
    1. The problem statement, all variables and given/known data
    A 33 -cm-diameter conducting sphere is charged to 457 V (relative to a point an infinite distance from the sphere where the potential is zero).

    a. What is the surface charge density σ?



    2. Relevant equations
    Surface Area= 4(pi)r^2
    Surface Charge Density= Q/A
    V=kQ/r



    3. The attempt at a solution
    So initially I found the surface area (it's 3421.1944 cm^2) and then divided the electric potential by that. This is obviously wrong.
    Then, I used the formula for electric potential and solved for Q to get the charge of the sphere. I got 8.388x10^-9 C. I think I may have done something wrong because my units didn't cancel out quite right, but I didn't think too much about it. I then plugged this into my formula for surface charge density, and got 2.452x10^-10 C/m^2. My answer is supposed to be in uC/m^2 so I multiplied the answer by 10^6 and got 2.452x10^-4 uC/m^2.
    I put this into the answer box, and it is wrong.
    So thenI thought maybe I plugged in the wrong distance before (I used the radius of the sphere and not the diameter though i'm sure the radius is correct) and did the same as above, and ended up with 4.412x10^-3 uC/m^2.
    So right now, I'm simply at a loss as for what I'm doing wrong.

    I'd like to note that I'm in an algebra-based physics class for a reason- calculus is lost on me, I'm sorry.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 5, 2009 #2

    LowlyPion

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    First of all maybe express your total charge in μC.

    V = kQ/r

    Q = V*r/k = 457*.165/9*109 = 8.37*10-9C = .008378 μC
     
  4. Apr 5, 2009 #3
    that's what I meant by 'uC'. I'm not sure how to make the symbol on my keyboard.

    Did I change units wrong? to go from Coloumbs to μC you would multiply by 106, right?

    EDIT: I just tried that and I come up with the same answer.
     
    Last edited: Apr 5, 2009
  5. Apr 5, 2009 #4

    LowlyPion

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    Only if you are simultaneously changing to μC which are 10-6C

    So you can convert 10-6 to μ, yes.
     
  6. Apr 5, 2009 #5

    LowlyPion

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    I get a different answer when I divide .008378 by the surface area .3421 m2
     
  7. Apr 5, 2009 #6
    Wait- .3421 m2? Wouldn't 3421.1944 cm2 be 34.211944 m2? This may be where my arithmetic is messing up.
     
  8. Apr 5, 2009 #7
    Ah- I got it now, I was converting my cm2 to m2 wrong. I've got the answer now, thank you!
     
  9. Apr 5, 2009 #8

    LowlyPion

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    Maybe it's better to work in m from the beginning?

    Good luck.
     
  10. Apr 5, 2009 #9
    This is a good idea. It's easy to forget for me, but I'll have to keep it in mind.
    Thanks again.
     
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