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Energy required to remove point charges

  1. Aug 7, 2013 #1
    1. The problem statement, all variables and given/known data

    Four charges are placed at the corners of a square of side length a, as follows : -Q on the top left corner, +Q on the top right corner, -Q on the bottom left corner and +Q on the bottom right corner. Determine the electric field at the centre of the square and determine the energy required to remove one of the negative charges from this configuration for Q=10 uC and a=1 m.

    2. Relevant equations

    E=(k|q|)/r^2
    V=kq/r
    W=qV


    3. The attempt at a solution

    First, for the electric field, I worked out r at the center of the square using Pythagoras' theorem and found r to be (a/sqrt(2)). I then substituted that into the equation for Electric field and got (2kq/(a^2)). This is for one charge only. (I took the top left). From this configuration, I thought that the y components of the electric field would cancel due to the four charges leaving me with only the x components to deal with. The x components of the field for this particular charge is -(2kq/(a^2))cos 45 which gives -sqrt(2)kq/(a^2). So, the net electric field due to the four charges would be 4 times this. Could you please check this?

    For the second part of the question, I decided to work out the potential at the negative charge location (bottom left) due to the other charges using V=kq/r. I got the three potentials as being -89 880 V, 89 880 V and 63554.76V (?). I then added the three potentials together to get 63554.76 V and so the energy required to remove one of the negative charges (bottom left in my case) equals qV. And so, I got the final answer as being -6.355x10^-11 which seems like a strange answer. I would need some help, please

    Thanks
     
  2. jcsd
  3. Aug 7, 2013 #2

    mfb

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    Right.

    Looks good.
    It is a strange answer as it does not have units. And I don't see how you get this numerical value and its sign.
     
  4. Aug 7, 2013 #3
    Sorry, I forgot to put the unit. it is the Joule (J) and I got the numerical value by using Q=10 uC and V=V1 +V2 +V3 (the three potentials added together). So, the net potential is 63554.76 V and so qV= -10*10^-6*63554.76 which gives this answer. The sign is negative because q is negative. Is this wrong?
     
  5. Aug 7, 2013 #4

    haruspex

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    You really want qΔV. What is the change in potential for the charge moved to infinity?
     
  6. Aug 7, 2013 #5
    The potential at infinity is defined to be 0, right? And the potential at the current location is 63554.76 V is it not? So, is the change in potential -63554.76 V??
     
  7. Aug 7, 2013 #6

    haruspex

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    Yes.
     
  8. Aug 8, 2013 #7
    I think I am almost there. The final answer is q*delta V. So, it is -10*10^-6*-63554.76. So, this equals 6.355*10^-11 J. Is this correct?
     
  9. Aug 8, 2013 #8

    mfb

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    That value is wrong. How did you calculate it?
    Please use brackets:
    -10*10^(-6)*(-63554.76)
     
  10. Aug 8, 2013 #9
    Ok, The answer should be 0.64 J. I got the brackets wrong. Is this correct now?
     
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