1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Work required to assemble charged particles

  1. Oct 13, 2008 #1
    1. The problem statement, all variables and given/known data
    How much work is required to assemble eight identical charged particles, each of magnitude q, at the corners of a cube of side s?


    2. Relevant equations
    W=deltaU

    delta U = kQq/r


    3. The attempt at a solution

    I've come up with those equations, and was trying to plug the information into delta U. However, all the charges are identical, so there are not two different values for q. But would it be delta U = kq^2/s ??
     
  2. jcsd
  3. Oct 13, 2008 #2
    I take it that the charges come from infinity.

    The work required to bring them in from infinity is the sum of the electrostatic energies of the pairs of charges. With 8 charges there are 8*7/2 = 28 such pairs. These can be viewed as the 12 sides of the cube, its 12 surface diagonals and its 4 body diagonals. Be sure to remember that the energy is inversely proportional to the distance, not (like the force) to the distance squared.
     
  4. Oct 13, 2008 #3
    Yeah it doesn't say where the charges are coming from.

    8*7/2 ... where does the 7/2 come from? How are you getting 28 pairs?

    12 sides of a cube? Aren't there only 6?
     
  5. Oct 13, 2008 #4
    Anyone?
     
  6. Oct 13, 2008 #5
    Does anyone know how they got the 28 pairs?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?