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The energy needed to keep charges in place

  1. May 31, 2014 #1
    How much energy is needed to place four positive charges, each of magnitude + 5.0mC, at the vertices of a square of side 2.5cm ?

    I am not entirely sure where to start here. Am I looking for W=? like the work it takes to place it ??
     
  2. jcsd
  3. May 31, 2014 #2

    Simon Bridge

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    Hint: work-energy relation.

    Note: you don't need energy to keep charges in place, just to assemble them in the first place.
     
  4. May 31, 2014 #3
    Have you learned the energy associated with a pair of charged particles? If you haven't than first you should make sure you understand how to calculate the energy associated with a pair of particles before trying to solve a problem with multiple particles.
     
  5. Jun 3, 2014 #4
    I am doing the work that was assigned to me, after the unit was done!
     
  6. Jun 3, 2014 #5
    I am understanding to use the W=q( Vb-Va) = q( kQ/rb - kQ/ra) , and i am thinking you have to calculate like putting one at a time. Am I on the right track?
     
  7. Jun 4, 2014 #6

    Simon Bridge

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    That is correct.
    The work needed to put the first one there is zero.
    The second charge has to do work, or have work done on it, associated with the potential from the first charge.
    The third charge has to deal with both the other charges etc.
     
  8. Jun 5, 2014 #7
    Thanks, sounds great :)
     
  9. Jun 5, 2014 #8
    Q1=Q2=Q3=Q4= + 5.0 mC=0.005C k=9X10^9 r=2.5cm=0.025m
    There is no repelling force on the first charge, therefore placing Q1 is zero energy
    W1=0
    The second charge would have the first charge is a repelling force on it.
    W2=kq1q2/r= 9000000J
    The third charge would have a repelling force of both charge 1 and charge 2.
    W3= { kq1q3/r} + (kq2q3/r}= 9000000J+9000000J= 18000000J
    The last charge will have all other three charges causing repelling forces on it.
    W4={kq1q4/r} + {kq2q4/r} + {kq3q4/r} =27000000J

    Total W= W1+W2+W3+W4= 0+ 9000000J+ 18000000J+ 27000000J=54000000J

    This look right??
     
  10. Jun 5, 2014 #9

    Simon Bridge

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    All charges magnitude q, and we want to assemble them into a square with side length a.

    Bring the 1st charge to the origin - no work.
    Bring the next charge to point (a,0) work = kq^2/a - you got that too, good.

    Bring the third charge to point (0,a) ... this charge is distance "a" from (0,0) but is a different distance from (a,0). Your calculation gave them all the same distance.

    Bring the fourth charge to point (a,a) to complete the square.
    Two of the three distances are the same but one is different.

    It helps to draw a picture.
     
  11. Jul 22, 2014 #10
    Right-Thanks- I missed that- Different r when its accros-
     
  12. Jul 22, 2014 #11

    Simon Bridge

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    Well done.
    You'll find that a lot of physics is done with diagrams.
     
  13. Jul 22, 2014 #12
    So, correct me if i'm wrong please, I got
    Adj^2 + Opp^2= Hyp^2 --> r(b)=SqRt{ 0.025m^2+ 0.025^2}
    r (b)= 0.035m( across the square , from one corner to another)
    The third charge would have a repelling force of both charge 1 and charge 2.
    W3= { kq1q3/r(a)} + (kq2q3/r(b)}= 9000000J+6428571J= 15 428 571 J
    The last charge will have all other three charges causing repelling forces on it.
    W4={kq1q4/r(a)} + {kq2q4/r(a)} + {kq3q4/r(b)} = 24 428 571 J

    Total W= W1+W2+W3+W4= 0+ 9000000J+ 15428 571J+ 24 428 571J=48 847 142J
     
  14. Jul 22, 2014 #13

    Simon Bridge

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    That looks good. It helps to do the whole thing in variables first though.
    For 4 charges q in a square sides x

    All charges same sign so all forces are repelling - need to do positive work to bring the charges to the square.
    All charges same magnitude, so the sum will look like:U=kq^2(some factors)
    The factors will depend on distances.

    repulsion along a side adds a factor of 1/x^2
    repulsion along a diagonal adds a factor of 1/2x^2 (pythagoras)
    ... thus two diagonals contributes the same as one side.

    The first charge has no contribution
    the next one adds a side (repulsion is only along one side of the square)
    the next adds a side and a diagonal
    the next adds two sides and a diagonal
    that's a total of 4 sides and 2 diagonals or 5 sides all together.

    U=5kq^2/x^2

    x=2.5cm
    q=5mC

    Interesting that the question is given in cm and mC ... why pick those units?
    k=8.988x109Nm^2/C^2 (10000 cm^2/m^2) (0.0001C^2/mC) = 8.988x109N (cm^2)/(mC^2)
    ... in other words - you don't have to convert to whole Coulombs and meters.
     
  15. Jul 23, 2014 #14

    ehild

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    1 mC= 0.001 C.

    ehild
     
  16. Jul 23, 2014 #15

    Simon Bridge

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    ... so there is...
    ... and so there are 0.001C/mC or 10^-6 C^2/mC^2 oh I missed a zero?

    k=8.988x10^8 N cm^2/mC^2.

    I think I need a vacation :(
     
  17. Jul 23, 2014 #16

    ehild

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    Two zeroes :devil:
    Happy vacation!

    ehild
     
  18. Jul 23, 2014 #17
    Thank you for all your help :D
     
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