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Solving for Electric Potential Energy , given 4 point charges of equal value

  1. Jul 9, 2008 #1
    1. The problem statement, all variables and given/known data

    Four point charges are individually brought from infinity and placed at the corners of a square whose sides are 0.30 m each. Each charge has the identical value + 4.0 mC. What is the electric potential energy of these four charges?

    2. Relevant equations

    V=kq/r (or at least this is the equation I have been attempting to use to solve this)
    k= 9 x 10^9 Nm^2/C^2
    q= 4 x 10^-6 C
    r= 3.0 m

    3. The attempt at a solution

    [(9x10^9)(4x10^-6)]/.3=1.2X10^5

    Since all charges are the same I take that answer multiply by four to get the sum, which gives me 4.8x10^5 V.

    However, the answer is in V and I need an answer in Joules (J). So I am assuming I am not even using the correct equation to solve for EPE. I'm sure if I could figure out what equation to use to solve for EPE in Joules, I could do the math. My problem is not knowing what equation to use to solve the problem.
     
  2. jcsd
  3. Jul 10, 2008 #2
    Do you know the equation [tex]U_E = qV[/tex]?
     
  4. Jul 10, 2008 #3
    Yeah, I actually came across that a few minutes ago and tried to plug in what I knew.
    which would give me UE=qV, or (4x10^-6)(4.8x10^5)=1.92 Joules.

    I plugged in the answer for V that I got before, and used the given charge.

    Does this sound right?
     
  5. Jul 10, 2008 #4
    hmm I don't think so. Consider this: if a charge is brought from infinity to one of the corners of the square, that charge gains potential energy due to the presence of all the other charges. That means that the total potential energy will be the sum of energies between q1 and q2, q1 and q3, q2 and q3, q1 and q4, q2 and q4, and q3 and q4 (where q1-4 are the charges). Note that [tex]U_E = qV = \frac{k q_1q_2}{r}[/tex]
     
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