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Electrical Potential

  1. Feb 21, 2007 #1
    1. The problem statement, all variables and given/known data
    Three equal point charges, each with a charge of 2.00 uC, are placed at the corners of an equilateral triangle whose sides have a length of 0.400 m. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)


    2. Relevant equations

    [tex] U_{tot} = \frac {Q^{2}} {4*pi*8.85*10^{-12}} * (\frac {A} {d1} + \frac {B} {d2} + \frac {C} {d3}) [/tex]


    3. The attempt at a solution

    [tex] U_{tot} = \frac {(2.00*10)^{-19}} {4*pi*8.85*10^{-12}} * (\frac {4} {0.400} + \frac {4} {0.400} + \frac {4} {0.400}) [/tex]

    The online program says i'm off by an additive constant once again and my work seems right?
     
  2. jcsd
  3. Feb 21, 2007 #2
    Think about assembling the charges one by one, bringing them each in from infinitely far away.
     
  4. Feb 21, 2007 #3
    When my professor tells us to think about things, usually it involves something going to 0? Are the charges 0 as you bring them closer?
     
  5. Feb 21, 2007 #4
    No. How much work does it take to bring in one charge from infinity if there are no other charges around?
     
  6. Feb 21, 2007 #5
    The work done would be U_a - U_b = -(U_b-U_a) = -ChangeU
     
  7. Feb 21, 2007 #6
    If there are no other charges around, there is no potential which you would have to do work against to bring in the first charge. Now what about the second charge?
     
  8. Feb 21, 2007 #7
    The work done should be the amount of work it takes to bring in the second charge?
     
  9. Feb 21, 2007 #8
    Calculate that work. Now you have two charges in a region. How much energy does it take to bring in the third charge? Then add.
     
  10. Feb 22, 2007 #9
    Is there an equation for work in this case? I thought my original equation was it?
     
  11. Feb 23, 2007 #10
    Should it not be:

    [tex] U_{tot} = 3*( (\frac {1} {4*pi*8.85*10^{12}}) * (\frac {2.00*10^-{19}} {0.400}) )[/tex]


    I know the amount of work is just the summation of amount of work to move one charge.
     
  12. Feb 23, 2007 #11

    Hootenanny

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    No. As Robb suggested, think about assembling the array of charges. So, lets start with just one charge in free space; there is no external electric field and no other charges, so the work done to place the charge is zero. No consider bringing a second charge from d = [itex]\infty[/itex] to d = 0.4 How much work must you do on this second charge?
     
  13. Feb 23, 2007 #12
    Amount of work would be:

    [tex] V = \frac {K*Q} {r} [/tex]
     
  14. Feb 23, 2007 #13

    Hootenanny

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    Almost, but not quite. Note that electrical potential is not the same as electric potential energy.
     
  15. Feb 23, 2007 #14
    [tex] U = \frac {1} {4*pi*E_{0}} * \frac {Q_{1}*Q_{2}} {r} [/tex]
     
  16. Feb 23, 2007 #15

    Hootenanny

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    Correct, and since Q1 = Q1 := Q, we can say that

    [tex]U = k\frac{Q^2}{r}[/tex]

    So, now we have two charges in place and have done U joules of work. So how much work must we do on the third charge (Q) to bring it from d = [itex]\infty[/itex] to d = 0.4?
     
  17. Feb 23, 2007 #16
    [tex]U = k\frac{Q^3}{r}[/tex]
     
  18. Feb 23, 2007 #17
    Careful, your units aren't even correct here.
    First, conceptually, will it take more work to bring in the third charge compared to the second, or less work?
     
  19. Feb 23, 2007 #18
    You just said this equation was right, now you say it's incorrect.
     
  20. Feb 23, 2007 #19
    Umm, Q*Q is not the same units as Q*Q*Q.
     
  21. Feb 23, 2007 #20
    Well I'm lost in what you're saying now.

    You said to move 1 charge the potential is:

    [tex]U = k\frac{Q^2}{r}[/tex]

    After that I have no idea what you're talking about anymore
     
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