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Electric potential and capacitance (semicircle problem)

  1. Feb 19, 2009 #1
    1. The problem statement, all variables and given/known data

    A uniformly charged insulating rod of length 14cm is bent into the shape of a semicircle (looks like a "C"). The rod has a total charge of (-7.50E-6C). find the electric potential at O, the center of the semicircle

    2. Relevant equations

    v= ke[tex]\int[/tex]dq/r

    3. The attempt at a solution
    after integrating I get: KeQ/r
    =Ke(-7.50E-6)/(0.14m)
    For the semicircle will the radius be 0.7m? I'm frustrated with this problem and maybe I'm just making a dumb calculation mistake.
    The answer is: -1.51MV
    Thank you in advance! :shy:
     
  2. jcsd
  3. Feb 19, 2009 #2

    Delphi51

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    Homework Helper

    No. The distance around the semicircle is (pi)*r = 14 cm
    It would be interesting to see how you set up your integral.
     
  4. Feb 19, 2009 #3
    Thanks for your reply.
    based on the formula v= ke [tex]\int[/tex]dq/r: I took (1/r) out of the integral and integrated only dq which is (1dq) =Q
     
  5. Feb 19, 2009 #4

    Delphi51

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    Oh - of course - you don't have much of a job integrating because every bit of the charge is the same distance from the center! So, have you now got the right answer? If not, let us know what radius you used and how it worked out.
     
  6. Feb 19, 2009 #5
    I used 14cm (0.14m) as the radius and got:
    {(8.99x10^9)(-7.50x10^-6)}/(.14m)= -481607.1429
    Thank you for your help :)
     
  7. Feb 19, 2009 #6
    Solved!!!

    {(8.99x10^9)(-7.50x10^-6)}/ (.14/pi)= -1513013.461= -1.51MV
     
  8. Feb 19, 2009 #7
    Thank you!!!!!!
     
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