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8 Charges on Square

  1. Jan 20, 2015 #1

    rlc

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    1. The problem statement, all variables and given/known data
    In the figure, four charges, given in multiples of 5.00×10-6 C form the corners of a square and four more charges lie at the midpoints of the sides of the square. The distance between adjacent charges on the perimeter of the square is d = 1.30×10-2 m. What are the magnitude and direction of the electric field at the center of the square?
    upload_2015-1-20_16-17-27.png
    The magnitude of E?
    Ex?
    Ey?

    2. Relevant equations
    Magnitude of Electric Field at point r: E=F/q0=(kQ)/r^2
    F=k(Qq0)/r^2

    3. The attempt at a solution
    The equations are ones I found in the textbook, but I'm not entirely sure how to use them or even if they would work for this particular problem. To find the charge on each charge, would I take the number of q's written next to it and multiply that by 5.00×10-6 C ? Would that become Q?
     
  2. jcsd
  3. Jan 20, 2015 #2

    CWatters

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    Homework Helper

    It's been a long time since I did this but my reading of the problem is that q=5.00×10-6 C, in other words the charges are specified in multiples of 5.00×10-6C. Best work out the problem in terms of q and substitute the value at the very end.

    I think the equation you have is correct. Replace Q with -5q, -2q, q, +3q etc

    Remember that the electric field is a vector that points away from +ve charge, towards -ve. So I would guess it's a matter of adding together 8 vectors.
     
  4. Jan 20, 2015 #3

    gneill

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    Staff: Mentor

    Take advantage of symmetry to simplify things. For example, the fields from the two +3q charges on the diagonal will cancel each other out at the center of the square. Several such "tricks" are available here.
     
  5. Jan 20, 2015 #4

    rlc

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    Ah, I didn't know that. So you can cancel out the +3q and the +5q.
    I got:
    (there are 3 +q): 3(5.00E-5)=1.5E-5
    -2(5.00E-5)=-1E-5

    (1.5E-5)+(-1E-5)=5E-6
     
  6. Jan 20, 2015 #5

    rlc

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    (8.99E9)(5E-6)q/(1.30E-2)^2

    Is this what I would do next? What would be q?
     
  7. Jan 20, 2015 #6

    gneill

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    Staff: Mentor

    ??? How would that work? They're not across from each other with the center of the square between them. And the two +3q charges are already "gone" from the picture, having cancelled each other out across the diagonal. Further, +3 does not equal +5, so they don't cancel numerically.

    For the fields from charges to cancel at a given location they must:
    1. Be the same distance away from the location
    2. Be the same sign
    3. Be the same magnitude
    4. Be located in opposite directions from the location (the two charges and the given location must be collinear).
     
  8. Jan 20, 2015 #7

    rlc

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    Sorry, I wrote what I meant wrong. The 2 +3q's cancel out, then I erroneously said that the 2 +5q's cancel out, but now I see that one is -5q, so I need to fix my math.
    So with those rules, the 2 +q's on the y axis can cancel out.
    That leaves:
    5(5E-6)=2.5E-5
    -2(5E-6)=-1E-5
    -5(5E-6)=-2.5E-5
    1(5E-6)=5E-6
    The sum of it is -5E-6

    (8.99E9)(-5E-6)q/(1.30E-2)^2
    Am I going about this right?
     
  9. Jan 20, 2015 #8

    gneill

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    Staff: Mentor

    I'd advise against plugging in any numbers yet, it just confuses things. There's more that can be done dealing with the q's.

    Here's another hint: The field produced by some charge Q at a given location is the same as that of a charge -Q located at the same distance in the other direction (again, collinear items). So this will allow you to sneakily move charges across the square (through the center!) and diagonally by changing their signs and adding them to the charges already there. You should be able to drastically reduce the number of actual charges that you have to deal with for the final field calculation.
     
  10. Jan 20, 2015 #9

    rlc

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    Are you referring to the +5q and the -5q?
     
  11. Jan 20, 2015 #10

    gneill

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    Staff: Mentor

    Sure. And any other charges that are symmetrically located across the center of the square. You should be able to reduce the original 8 charges down to just 2. Much fewer numbers to deal with! Saves the fingers and calculator buttons :)
     
  12. Jan 20, 2015 #11

    rlc

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    Stop me if I'm wrong, but would you add them?
    Move the -5q diagonally, and add it to +5q, making....+10q?
    Move the +q across the square to the -2q, making -3q?
     
  13. Jan 20, 2015 #12

    gneill

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    Yes!
     
  14. Jan 20, 2015 #13

    rlc

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    10(5.0E-6)=5E-5
    -3(5.0E-6)=-1.5E-5
    So the sum of that is 3.5E-5

    Does this look right? And now I use this formula?
    (8.99E9)(3.5E-5)q/(1.30E-2)^2
     
  15. Jan 20, 2015 #14

    gneill

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    No, not right. You've paired away the charges down to two remaining. Now you need to find and add their actual electric fields at the location in question (the center of the square).

    Fields are vector quantities, with both magnitude and direction. So you'll be finding the magnitudes using the kQ/r2 equation, and the directions are given by the location (and sign) of the charges with respect to the location in question (the center of the square in this case). Look up a picture of the field surrounding a charge to see the direction that the field points for a given charge sign.

    Add the vectors to find the net resultant of the field at that location.
     
  16. Jan 20, 2015 #15

    rlc

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    A positive charge's field lines radiate outward while a negative charge's field lines radiate inward.

    kQ/r2 ---replace Q with 10(5.0E-6)=5E-5 and -3(5.0E-6)=-1.5E-5 ?
     
  17. Jan 20, 2015 #16

    gneill

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    Staff: Mentor

    That'll work. Remember to attach units to your values.

    Although, come to think of it, I might've chosen to move the +5q from the lower left to the upper right to make -10q as the total there. Then both of the charges remaining would be in the first quadrant and have the same sign. Keeps all the trigonometry in the first quadrant where it's most familiar :)
     
  18. Jan 20, 2015 #17

    rlc

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    -10(5.0E-6)=-5E-5
    (9E9)(-5E-5)/(1.30E-2m)^2=-2662721893

    -3(5.0E-6)=-1.5E-5
    (9E9)(-1.5E-5)/(1.30E-2m)^2=-798816568

    Is this the right direction? Would I add them together now? (I changed the 10q into -10q like you suggested)
     
  19. Jan 20, 2015 #18

    gneill

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    Staff: Mentor

    Look at the diagram. Do the distances from the charges to the center look the same? What are the actual distances from the charges to the center of the square? You need to use the right distances in the field formula.

    Remember: you are adding vectors here. You need to separate the vectors into their components and add the like components to find the components of the resultant vector.
     
  20. Jan 20, 2015 #19

    rlc

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    Oh, right!
    the one with the -3 charge will stay the same, but the -10 charge is 1.84E-2 m away at that diagonal (pythagorean)
    -10(5.0E-6)=-5E-5
    (9E9)(-5E-5)/(1.30E-2m)^2=-2662721893

    -3(5.0E-6)=-1.5E-5
    (9E9)(-1.5E-5)/(1.838E-2m)^2=-7343031.963

    Right?
     
  21. Jan 20, 2015 #20

    gneill

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    Staff: Mentor

    Sure. I think you can drop the signs on the charges at this point for the math operations to follow. You know the directions of the field vectors based upon the signs and the placement of the charges with respect to the center of the square, so you really only need to deal with the magnitudes of the vectors. Sketch the vectors and assign their calculated magnitudes.

    Now you need to add the vectors find their resultant.
     
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