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Electrostatic potential problem (got 1/3) correct

  1. Oct 8, 2009 #1
    Point charges q1, q2, and q3 are fixed at the vertices of an equilateral triangle whose sides are 2.50 m long. Find the electrostatic potential energy of this system of charges for the following charge values.
    a.)q1=q2=q3=3.90uC
    I got the right answer for this one 164mJ

    b.)q1=q2=3.90uC,q3=-3.90uC
    I used [tex]\Delta[/tex]V=k[(3.90uC^2/2.5m)-3.90uC/2.5]=14080 (or something like that)
    then V=q(sum of the charges) which is just 3.90uC * [tex]\Delta[/tex]V
    ans I got was 54.7mJ

    c.)q1=q2=-3.90uC, q3=3.90uC
    Same method as above.. same answer too.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Oct 8, 2009
  2. jcsd
  3. Oct 8, 2009 #2

    rl.bhat

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    Hi hitman0097, welcome to PF.
    Your answers appear to be correct.
     
  4. Oct 9, 2009 #3
    hmmm, I got the correct magnitude but the wrong sign? This is a online hw thing. But I don't know how you can minus energy as a answer. Like I know it's the change in energy and you can lose and gain it. But it can't ever be less then zero right?
     
  5. Oct 9, 2009 #4

    rl.bhat

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    In (c) there are two negative charges and one positive charge, So the net potential energy is negative. If you want to keep two positive charges at a certain distance, you have to push them towards each other.
    If you want to keep one positive charges and one negative charge at a certain distance, you have to pull them apart from each other. First one you call it as positive PE and second one you call it as negative PE.
     
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