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Electric force distance problem

  1. Apr 26, 2006 #1
    Two point charges lie on the y-axis. A third charge q is placed somewhere in space such that the resultant force on it is zero. What distance from the origin must this third charge be placed such that the resultant force on it is zero? Answer in units of m.
    The first charge is 8[tex] \mu C[/tex] and is located at 0.5 m.
    The second charge is -37 [tex] \mu C[/tex] and is located at -1 m
    The third charge is 8 [tex] \mu C [/tex].
    I know that this third charge needs to be on the y-axis and for the force to be zero, the forces have to cancel each other out.
    so I found the F on 1 & 2 by k(8e-6 *37e-6)/ (1.5^2)
    and got this to be 1.184
    Then I found F from 1 & 3 by k(8e-6*8e-6)/r^2
    solving for r gave me .697, which I added to .5, b/c the third charge is above the 1st one. This gave me 1.2 m,which isn't right.. can someone please help?
     
  2. jcsd
  3. Apr 27, 2006 #2
    when you place the third charge at position r with respect to the origin of the XY-plane, you will have THREE forces in total that have to cancel out.

    1) F on charge 1 and 2 (the one you calculated first)
    2) F on charge 1 and 3 : what is the equation ?
    3) F on charge 2 and 3 : what is the equation ?

    Then 1) + 2) + 3) = 0 and solve for r. You will get two r-values of which one shall be correct.

    regards
    marlon
     
  4. Apr 27, 2006 #3
    Ok so for F_13 I used k(8e-6*8e-6) /R^2 = .576/r^2
    and for F_ 23 I used k(37e-6 * 8e-6) / R^2 = 2.664/r^2
    then I added 1.184, which was my previous answer for F_12, and .576/R^2 and 2.664/r^2 = 0. Solving for R gave me 1.65... am I doing this right?
     
  5. Apr 27, 2006 #4

    Doc Al

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    You are trying to make the net force on the third charge equal zero, so there are only two forces to consider: F_13 & F_23. (F_12 is irrelevant.)

    First question to answer is in what region must the third charge be placed to possibly get zero net force on it? Only three choices: y > 0.5; 0.5 > y > -1.0; y < -1.0. Hint: What must be the relative direction of the two forces in order for them to cancel?

    Once you've figured out the right region, then you can set up your force equation to solve for the exact position. (Hint: The coulomb constant and the micros will all cancel out, so don't do any arithmetic until the last moment.)
     
  6. Apr 27, 2006 #5
    I think that it has to be in the positive y-axis, so >0.5.
    then would I just do:k*q1*q3 /r^2 = -k*q2*q3/r^2 ?
     
  7. Apr 27, 2006 #6

    Doc Al

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    Right.

    No, the distances are not equal. (That equation simplifies to q1 = -q2; does that make sense?)

    Instead, call the position of the third charge y. Now find the distance between y and the other charges. Then set the magnitude of the two forces equal and solve for y.
     
  8. Apr 27, 2006 #7
    So would it be q1*q3/ (0.5 + Y)= q2*q3/ (1.5 + Y)?
     
  9. Apr 28, 2006 #8

    Doc Al

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    That will work. You've defined "Y" as the distance above the first charge--nothing wrong with that. But be sure to express your final answer in terms of distance from the origin as requested in the problem statement.

    Edit: You forgot to square your distances; Your equation should read:
    q1*q3/ (0.5 + Y)^2 = q2*q3/ (1.5 + Y)^2
     
    Last edited: Apr 28, 2006
  10. Apr 28, 2006 #9
    Solving for y gave me .224, I add this to 0.5 right?
     
  11. Apr 28, 2006 #10

    Doc Al

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    If that's the answer for Y, then you add it to 0.5. But I would check that answer.

    Note: I failed to point out that you left out the squares in your last equation. (I assume you were aware of that.)
     
  12. Apr 29, 2006 #11
    I forgot the squares earlier...
    so (8e-6)(8e-6)/(.5+ y)^2= (8e-6)(37e-6)/ (1.5+ y)^2
    Solving for y, gave me .369, which I added to .5 to get .869 which still isn't right... what am I doing wrong now?
     
  13. Apr 29, 2006 #12

    Doc Al

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    I get a slightly different answer. Write out your quadratic equation so we can take a look.
     
  14. Apr 29, 2006 #13
    I cross multiplied and got,
    (6.4e-11)(2.25+3y+y^2) = 2.96e-10(.25+y+y^2)
    1.44e-10+1.92e-10y+6.4e-11y^2= 7.4e-11+2.96e-10y +2.96e-10y^2
    Then my quadratic came out to be -2.32e-10y^2 -1.04e-10y + 7e-11= 0Then by the quadratic formula I got .369
     
  15. Apr 29, 2006 #14

    Doc Al

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    Your equation looks OK to me, but your solution does not. Maybe you messed up in plugging into the formula.
    You can save yourself a lot of exponents (and opportunity for arithmetic error) by canceling things out before going any further. I'd write this step as:
    8(2.25+3y+y^2) = 37(.25+y+y^2)

    Do you see what I did?
     
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