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EPE at a point due to two point charges

  1. Apr 13, 2012 #1
    CodeCogsEqn.gif

    The above equation gives the EPE of two point charges separated by a distance r.

    Firstly, I do not understand how this formula gives the TOTAL EPE of the system.

    Secondly, lets say I have three deuterium nuclei moving towards one another with initial speed V.

    They all stop instantaneously at the same point such that they are all separated from one another by distance r.

    How would the above formula change to give the total EPE of the system ? Would you consider each pair of nuclei ?

    So total EPE of system = (EPE of 1st+2nd) + (1st+3rd) + (2nd+3rd) ?

    Hence, the formula becomes CodeCogsEqn-1.gif
     
    Last edited: Apr 13, 2012
  2. jcsd
  3. Apr 13, 2012 #2

    tiny-tim

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    hi elemis! :smile:
    no work is done bringing the first charge from infinity

    so you only need to consider the work done by bringing the second charge
    potential energy is defined as minus the work done (per charge) …

    if there's more than one force, how does that affect the total work done? :wink:
     
  4. Apr 13, 2012 #3
    Hi TinyTim !! We meet again !

    attachment.php?attachmentid=141266&d=1334228946.png

    Please have a look at the above image i.e. part (a)

    You are telling me that no EPE is gained by one of the nuclei ?

    So we consider the EPE of only one?

    EDIT : How is that possible when both are moving towards one another and both are trying to overcome the repulsive force of the other ?
     
  5. Apr 13, 2012 #4

    tiny-tim

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    work done = force "dot" displacement

    if they're both moving, wouldn't you expect the displacement of one to be halved, and the total distance to remain the same? :wink:
     
  6. Apr 13, 2012 #5
    Could you elaborate ? Are you saying r in the very first formula at the top of the page is in fact r/2
     
  7. Apr 13, 2012 #6
    So is it the sum of the forces of all the nuclei multiplied by their displacement ?
     
  8. Apr 13, 2012 #7

    tiny-tim

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    no, the result will be the same

    but you can obtain it either by bringing one charge from infinity (zero work), then bringing the other charge (∫ F(r).dr)

    or by bringing both charges from infinity together (∫ F(r).d(r/2) + ∫F(r).d(r/2)) :wink:
    yes, but you'd be crazy to try to do it that way :yuck: …

    bring each charge from infinity one at a time :smile:
     
  9. Apr 13, 2012 #8
    Before we all get confused, lets deal with the question in post no. 3 first.

    Its alright to do that i.e. pretend one nuclei moves to the collision point and calculate the change in EPE (zero for the first one because both are at infinity at feel no repulsion) and calculate the change in EPE for the other ?
     
  10. Apr 13, 2012 #9

    tiny-tim

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    yes :smile:

    (and there's no problem with the "fixed" one having speed v … we can bring the other one in as fast or as slow as we like, and the speed makes no difference to the force (if we're ignoring relativity))
     
  11. Apr 13, 2012 #10
    With regards to this setup of two point charges, I was thinking the other day (assuming both charges are the same magnitude and opposite sign...) What is the equation of motion? If we write Coulomb's Force Law then the equation seems hard to solve. Can we use energy/ work done equations to solve the equation of motion more simply? Or use Lagrangian mechanics methods??
     
  12. Apr 13, 2012 #11
    Thank you very much TinyTim ! As always, I am much obliged.

    You have literally answered a query that my teacher couldn't explain and has been 'thinking' about for the last six months.
     
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