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Net electric field

  1. Jan 18, 2009 #1
    1. The problem statement, all variables and given/known data
    if I have two charged particles A and B and are distanced 8 cm apart. At a point 4 cm from
    A, the net electric field is 0. The charge of A is 40 nC. What can we conclude about the charge of B?


    2. Relevant equations



    3. The attempt at a solution

    From my point of view B can be 40nC as well.. however I am not sure if there are some other answers as well. I am pretty sure there can be more than 1 answer
     
  2. jcsd
  3. Jan 18, 2009 #2

    Dick

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    There are two points that are 4cm from A where the fields can cancel. They can cancel at the midpoint of A and B or they can cancel 4cm away from A on the opposite side of A from B.
     
  4. Jan 18, 2009 #3
    hmm..not clear on what you meant here.. do you mind giving me a picture? how about B's charged magnitude? how large does it have to be?
     
  5. Jan 18, 2009 #4

    Dick

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    I'm not very good at pictures. Suppose the charges are on the x-axis. Put A at x=0cm and B at x=8cm. Then the third location could be either at x=4cm or x=(-4)cm. If x=4cm, I agree with you, the charges should be equal. How about at x=(-4)cm?
     
  6. Jan 18, 2009 #5
    well if it's at -4cm then the distance between A and B is only 4 cm away right? A is at 0 and B is at -4?
     
  7. Jan 18, 2009 #6

    Dick

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    No, A stays at 0cm and B stays at 8cm. Call the third point C=(-4)cm. C is 4cm from A and 12cm from B, right? What does the charge at B have to be to cancel the E field at C?
     
  8. Jan 18, 2009 #7
    I get the picture you're trying to say now. However...how do I calculate the magnitude of B based on that info??
     
  9. Jan 18, 2009 #8

    Dick

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    Use E=k*Q/r^2. You know the distance r for each charge and you know Q for A is 40nC. Write an equation that they sum to zero and solve for the charge at B.
     
  10. Jan 18, 2009 #9
    so EA + EB = 0
    k*40/4^2 + k*X/12^2 = 0

    and I need to solve for X?
     
  11. Jan 18, 2009 #10

    Dick

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    Yeah, sure. Solve for x!
     
  12. Jan 18, 2009 #11
    okay the answer I got is 360nC, is that right? and are those the only two points from A where the fields cancels?
     
  13. Jan 18, 2009 #12

    Dick

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    I get -360nC, don't you? The sign is important. At any other point besides those two there is an angle between the two E fields. Can two fields cancel if they are are at different angles?
     
  14. Jan 18, 2009 #13
    yes, I missed the negative signs.. as far as I know they can't cancel if it's at a different angles.. correct?
     
  15. Jan 18, 2009 #14

    Dick

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    Yes. Two vectors can only cancel if they are negatives of each other. So they have to point along the same line.
     
  16. Jan 18, 2009 #15
    okay.. so the conclusion is only these two, where B is 40nC and B is -360nC
     
  17. Jan 18, 2009 #16

    Dick

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    Seems so to me. Ok with you?
     
  18. Jan 19, 2009 #17
    Seems fine with me, I am just worried that I missed some answers.
     
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