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Zero Electric Charge

  1. Apr 7, 2009 #1
    1. The problem statement, all variables and given/known data
    A charge Q1 = 0.0306 μC is located a distance d = 4.17 cm to the left of another charge Q2 = - 0.0206 μC.

    a. At what point along the line joining the charges is the electric field zero? Give your answer as the distance x from the charge Q2, positive if the point lies to the right of Q2 and negative if to the left.
    HELP: First figure out the approximate region in which the third charge can be placed so that the electric forces from the first two charges will balance.
    HELP: Set the electric fields from Q1 and Q2 equal at position x--you will have to solve a quadratic equation to find x.



    2. Relevant equations
    E=kQ/r2
    Quadratic equation


    3. The attempt at a solution
    First I solved the for the electric field of both charges, but I'm not sure if that is relevant to my problem or not.
    Then, I followed the Help's advice, and set my two equations equal to each other:
    -((8.99x109)(.0206x10-6))/(.0417-x)2=((8.99x109)(.0306x10-6))/x2
    I ended up with x = 51.641 or -3.643. Neither is correct as confirmed both by my plugging in the numbers and checking, and submitting the answers in my answer box. :/
    I'm confused about whether x should be squared or not, I'm pretty sure it should be. If the E-field of my negative charge should be negative or not... And if i should be doing .0417-x as the distance in said E-field...
     
  2. jcsd
  3. Apr 7, 2009 #2

    LowlyPion

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    Unfortunately you didn't follow the first hint. Since the field between the charges is not 0 anywhere in between, since 1 is positive and the other is negative, then constructing your equation as x and .0417 - x won't work.

    If there is a null point anywhere, then it must be to the other side of wherever the other charge. Hence if the distance to Q2 is x, then the distance to Q1 would be .0417 + x.
     
  4. Apr 7, 2009 #3
    alright, I just tried that out, too, and it didn't work either. I'm still getting it wrong, and I'm not quite sure where things are going wrong. I'm fairly positive it must be with my algebra, because things just aren't working out algebraically. :/
     
  5. Apr 7, 2009 #4

    LowlyPion

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    You should end with the simple equation that

    Q1/(.0417 + X)2 = Q2/X2
     
  6. Apr 7, 2009 #5
    Okay, I just tried that, and I'm still getting it wrong. It's closer than it was before, because the point isn't meters away from the charges.
    I ended up with x being equal to either 5.246x10-7 or -5.318x10-6.
     
  7. Apr 7, 2009 #6

    LowlyPion

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  8. Apr 7, 2009 #7
    Except that the charges aren't 306 and 206. They'd be .0306x10^-6... :/
     
  9. Apr 7, 2009 #8

    LowlyPion

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    Yeah ... but.

    The ratio is all you care about.
     
  10. Apr 7, 2009 #9
    ... You've lost me.
     
  11. Apr 7, 2009 #10

    LowlyPion

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    Q1/(.0417 + X)2 = Q2/X2

    (Q1/Q2)*x2 = (.0417 + X)2

    Regardless of what the charges are, you are only concerned with their ratio. The rest of it just cancels out.
     
  12. Apr 7, 2009 #11
    I don't think I understand why I'm only concerned with the ratio. If I need to find a specific point that would only happen with these two charges, wouldn't I need to use the exact charges?

    EDIT: Either way, I tried out the numbers you came up with above in the quadratic equation, and those are the exact same numbers I came up with originally when I tried this. Unfortunately, I'm fairly positive the distance is not 500 and some odd cm.
     
    Last edited: Apr 7, 2009
  13. Apr 7, 2009 #12

    LowlyPion

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    OK.

    Which result is different?

    306/206 =

    or

    .0306*10-6/.0206*10-6 =
     
  14. Apr 8, 2009 #13

    LowlyPion

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    0 = .001739 +.0834X - .485X2

    No you did not try these coefficients in the quadratic equation.

    You get no where near 500cm.
     
  15. Apr 8, 2009 #14
    I copy and paste the coefficients into the quadratic calculator.
    X1=5.242314636470677
    X2= -53.20091153123778

    The answer should be in meters, therefore it is 500 some cm.

    Unless for some reason it isn't in meters, which I don't really think is the case, since the original equation calls for meters.

    I do see what you mean about the ratios, though. Those are exactly the same.
     
  16. Apr 8, 2009 #15

    LowlyPion

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  17. Apr 8, 2009 #16
    You're completely right. I need to make a note of what is a and c for the quadratic equation
    Fantastic- thank you! :)
     
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