1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Stuck on surface change density problem.

  1. Feb 16, 2004 #1
    Stuck on surface charge density problem.

    Calculate the surface charge density for a solid spherical conductor of radius .438m if the potential is .876m from the center of the sphere is 1.87kV. Answer in units of C/m^2

    Ok I did this:
    V= Kq/r
    q= V*r/k
    q= 1.87e-3*.876/8.99e9
    q= 1.822e-13

    E= kq/r^2
    E= 8.99e9*1.88e-13/.438^2
    E= 8.809e-3

    SCD= 2*epsilon*E
    SCD= 1.559e-13

    But it's wrong.

    The only thing I'm not sure about is the Electric field.
     
    Last edited: Feb 16, 2004
  2. jcsd
  3. Feb 16, 2004 #2
    I am learning this material right now, but here goes a stab:

    I think you calculated the total charge of the sphere correctly.
    Assuming that q=1.822*10^-13, I think the next step is to simply divide this by the area of the sphere:
    SCD=Q/A
    SCD=1.822*10^-13 (C)/ 4*pi*r^2
    SCD=1.822*10^-13 (C)/ 4*pi*(.438m)^2
    SCD=1.822*10^-13 (C)/ 4*pi*(.191844m^2)
    SCD=7.55*10^-14 (C/m^2)

    I hope that is the answer. If not, I would really like someone else to let us know where our errors are. I am learning this material too, but figured I would give it a try.

    Thanks.
     
  4. Feb 16, 2004 #3
    Your radial distance in your potential equation is incorrect. Also, why are you trying to find the E-field at the surface of the sphere? That is going to be ill-defined. Listen to paul he has the correct method. Also 1.87 kV = 1.87e+3 V not to e-3.
    Good luck.
     
  5. Feb 16, 2004 #4
    ...

    Forgot to tell that I also tried that. Thanks for trying though.

    When using V in calculations, it should be in V not KV so that's why it's 1.87e-3
     
  6. Feb 16, 2004 #5
    At the risk of sounding dumb, what should his radius be in the potential equation?
     
  7. Feb 16, 2004 #6
    .876m because it's telling the potential from that distance. You did the same calculations as I did.
     
  8. Feb 16, 2004 #7
    Since you seem to know E JUST outside the conductor, I found in my text where:

    E=SCD/epsilon

    So can we use this rearranged to solve for SCD?
     
  9. Feb 16, 2004 #8
    I did that, it's in the first post but it's wrong I don't know why

     
  10. Feb 16, 2004 #9
    Right, I should have realized before I posted that is where the 2*epsilon*E came from. I guess I thought I was onto something and got excited. Should have ran the numbers first.

    Does you book have an answer for this problem? Maybe we can back into it?
     
  11. Feb 16, 2004 #10
    This is an online howework question. Answers aren't posted until after the due date.
     
  12. Feb 16, 2004 #11
    Where does the 2 come from?
     
  13. Feb 16, 2004 #12
    I am calling it a night. I will pick up with this one in the morning. Hopefully I will be a little sharper.
     
  14. Feb 16, 2004 #13
    Never mind my friend did it for me, he got 7.558366e-8 and it's right
     
  15. Feb 16, 2004 #14
    Can you let me know how he got that? I am curious now.
     
  16. Feb 16, 2004 #15
    Q = ( bigR * V ) / K

    then Q / a
    a = 4pi small r ^2

    V= 1870

    that's where we went wrong
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Stuck on surface change density problem.
  1. Stuck on this problem (Replies: 1)

  2. Surface density. (Replies: 0)

  3. Stuck in a problem (Replies: 3)

Loading...