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Finding energy in a 4 object charge system

  1. Mar 28, 2016 #1
    1. The problem statement, all variables and given/known data
    An object with charge +2.0×10−5 C is moved from position C to position D in the figure (Figure 1) . q1 = q3 = +10.0×10−5 C and q2 = −20.0×10−5 C. All four charged objects are the system.

    Here's a picture to the problem
    upload_2016-3-28_18-27-18.png
    2. Relevant equations
    $$ F = \frac {kq_1 q_2}{r^2} $$
    $$ U_e = \frac {kq_1 q_2}{r} $$
    $$ a^2+b^2=c^2 $$

    3. The attempt at a solution
    I first attempted finding all the energy for the system by finding the initial potential electric energy for each pair of charged objects and added them together. I managed to find the distance from ## q_2 ## to ## q_1## and ##q_3## using Pythagorean theorem.$$ c → q_1, c → q_2, c → q_3, q_2 → q_1, q_2 → q_3 $$
    Unfortunately after putting the added energies it came up as wrong.
     
  2. jcsd
  3. Mar 28, 2016 #2

    BvU

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    Hello Anthony, :welcome:

    I suspect something went wrong with your calculations, can you show them, please ?
    Note that 'the system' consists of four charges!
     
  4. Mar 28, 2016 #3

    ehild

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    This is the picture :

    upload_2016-3-29_0-26-6.png
    Have you left out the pair q1-q3?
     
  5. Mar 28, 2016 #4
    Ah yes! It appears I did. However, even without having it I was getting 126 J. It says that the answer is -105 J. So that tells me I messed up somewhere while calculating with all the ##U_e##'s.
    I dont have access to my work atm as I am not home. I will make sure to post it as soon I get the chance!
     
  6. Mar 28, 2016 #5
    These are the numbers I got along with the added ##q_1, q_3##
    $$ U_{c,q1} = \frac {(9*10^9)(+2.0*10^{-5}) (+10.0*10^{-5})} {2.0} = 9\ J$$
    $$ U_{c,q2} = \frac {(9*10^9)(+2.0*10^{-5}) (-20.0*10^{-5})} {2.0} = 18\ J$$
    $$ U_{c,q3} = \frac {(9*10^9)(+2.0*10^{-5}) (+10.0*10^{-5})} {2.0} = 9\ J$$
    $$ U_{q2,q1} = \frac {(9*10^9)(-20.0*10^{-5}) (+10.0*10^{-5})} {2\sqrt2} = 45\ J$$
    $$ U_{q2,q3} = \frac {(9*10^9)(-20.0*10^{-5}) (+10.0*10^{-5})} {2\sqrt2} = 45\ J$$
    $$ U_{q1,q3} = \frac {(9*10^9)(+10.0*10^{-5}) (+10.0*10^{-5})} {4.0} = 22.5\ J$$
     
  7. Mar 29, 2016 #6

    ehild

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    You ignored all the negative signs.
     
  8. Mar 29, 2016 #7
    I've tried it with the negative signs, but I end up getting -68. The correct answer shown was -105, so I don't know exactly where I went wrong.
     
  9. Mar 29, 2016 #8

    ehild

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    You must include the signs of the charges,
    And even the numerical values of U(q2,q1) and U(q2,q3) are wrong.
     
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