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Work required?

  1. Mar 9, 2016 #1
    1. The problem statement, all variables and given/known data
    two particles with charges 4e and -4e are fixed at the vertices of an equilateral triangle with sides of length a. If k=1/4 pi Ԑ what quantity of work is required to move a particle with a charge q from the other vertex to the center of the line joining the fixed charges?

    2. Relevant equations
    W = -delta U

    3. The attempt at a solution
    -(4kQq / a - 2kQq / a) = -2kQq / a => 2kQq / a
    I know this is for when the charges are equal (both are Q) but Im not sure how to translate that into my problem...
     
  2. jcsd
  3. Mar 9, 2016 #2
    would it be (( 4*1/4 pi Ԑ *4) - (2*1/4 pi Ԑ *-4))/a?
     
  4. Mar 9, 2016 #3

    haruspex

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    No, it's for one charge of q and one of Q.
    No. Please post your reasoning and working.
     
  5. Mar 10, 2016 #4
    I think I am just getting more confused and that I should start from the beginning... What is the best way to go about this problem?
     
  6. Mar 10, 2016 #5

    haruspex

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    What is the potential where the charge starts?
     
  7. Mar 10, 2016 #6
    kQq/r
     
  8. Mar 10, 2016 #7
    And the potential energy, at the time it starts?

    You should also be using the brackets appropriately - 1/4πε0 is read as (1/4)*πε0.
     
  9. Mar 10, 2016 #8
    Also, which work is this?
     
  10. Mar 10, 2016 #9
    Wouldn't that be the work that I'm looking for?
     
  11. Mar 10, 2016 #10
    So would it be ((4*1/4πε)-(-4*1/4πε))/a?
     
  12. Mar 10, 2016 #11
    Is that the work done by the external force, or the electric force?
    As hurspex pointed out, that is wrong. I was merely requesting you to write in a more, correct fashion. You will need to show your working.
     
  13. Mar 10, 2016 #12
    at first i thought i should be using (1/(4*πε0))*((q1q12)/r) is this correct? and should my answer have an 'a' is it?
     
  14. Mar 10, 2016 #13
    or do i need to look at it as a dipole moment?
     
  15. Mar 10, 2016 #14

    haruspex

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    I meant the total potential, due to the two fixed charges.
     
  16. Mar 10, 2016 #15

    haruspex

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    You do not care about forces. This is just about potentials.
     
  17. Mar 10, 2016 #16
    Can you consider a +4e and -4e as a dipole for any distance from the dipole? What is their constraint?
     
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