Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Electric Potential Energy of a System of Charges

  1. Feb 19, 2010 #1
    1. The problem statement, all variables and given/known data

    A square of side a has a charge +Q at each corner. What is the electric potential energy of this system of charges?

    Express your answer in terms of the variables a, Q and appropriate constants.

    2. Relevant equations

    U=kq1q2/r

    3. The attempt at a solution

    I figured I could find the electric potential energy of one charge and multiply this quantity by 4 to obtain the total electric potential energy of the system. Here's what I did:

    U = 2(kQ2/a) + kQ2/sqrt(2)*a (1)

    How this makes sense to me:
    2(kQ2/a) -----> this part is for two of the +Q charges, one parallel to the charge we're looking at and the other is perpendicular. The distance between the charges is a.

    kQ2/sqrt(2)*a ------> this part is for the +Q charge that is diagonal from the charge we're looking at. The distance separating them is sqrt (a2+a2) which simplifies to sqrt(2)*a

    Then I took (1) and multplied it by 4. I simplified the expression and ended up with
    U = (8+sqrt(8)kQ2)/a
    When I inputted this answer, I got a message saying "Your answer either contains an incorrect numerical multiplier or is missing one."

    Does anyone know what I'm doing wrong?
     
  2. jcsd
  3. Feb 19, 2010 #2
    Woops sorry, the line that says U = (8+sqrt(8)kQ2)/a is supposed to say:

    U = ((8+sqrt(8))kQ2)/a
     
  4. Feb 19, 2010 #3
    That is incorrect. Note that you're counting every interaction twice!

    Try it like this:
    When you bring in the first charge from infinity, you do 0 work, since there's no force to oppose you.

    When you bring the second charge in from infinity, you have to do work against the E field of the first charge.

    For the third, you have two charges opposing the motion, and for the last, you've got 3.

    See where that gets you, you'll be surprised at how close you were.
     
  5. Feb 19, 2010 #4
    Oh I see...that makes sense. You're right, I was counting each interaction twice. I worked on the problem again and got (4+sqrt(2))kQ2/a...can anyone confirm this?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook