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The electric field due to a point charge

  1. Sep 5, 2011 #1
    http://www.monmsci.net/~fasano/phys2/Chap22_10.pdf
    The question is on this PDF File. It's 22.8. I get the logic of it, the electric fields will be 0 only when they both have the same magnitude. The math is shown below the problem, but i don't understand it. Could someone solve this and show me how to do this step by step?

    (You can see I'm not cheating, because if I was, I would just copy the answer out of that page. I want to understand it and do well in this class.)
     
  2. jcsd
  3. Sep 5, 2011 #2

    Doc Al

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    The way things work here is that you give it a shot and we provide hints and help.

    Are you familiar with the field from a single point charge?
     
  4. Sep 5, 2011 #3
    Yeah, It's E=((k)(Q))/r^2
     
  5. Sep 5, 2011 #4

    tiny-tim

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    Welcome to PF!

    Hi Jay9313! Welcome to PF! :smile:

    Which bit do you not understand?

    Do you understand (ignoring the constants) 5q/x2 = 2q/(L-x)2 ?
     
  6. Sep 5, 2011 #5
    The way I was setting this problem up is
    (2)/((4 Pi Episilon zero)(x^2)) = (5)/((4 Pi Epsilon zero)(x+L)^2)
    Then I would solve, and I used the quadratic formula. I got
    (-10 +_ 6.3L)/6

    I have no clue how to solve for L, and I don't even know if my math is right
     
  7. Sep 5, 2011 #6

    Doc Al

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    Good.

    So, at point (x,0), how would you express the field from each charge? What's the distance to each in terms of the givens?
     
  8. Sep 5, 2011 #7
    I ignored the constants, that was easy. I got to the point where I got a quadratic. 3x^2 + 10xL + 5L^2

    Is the quadratic right? How do you solve for a quadratic with two variables? =(
     
  9. Sep 5, 2011 #8
    Well the fields would be equal to each other. The distance would be (x+L)^2 right?
     
  10. Sep 5, 2011 #9

    Doc Al

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    What two variables? L is a constant. You'll express your answer in terms of L.
     
  11. Sep 5, 2011 #10
    I got (-10L +_ 6.3L)/6 = 0
    How would you solve that? Can you even solve it?
     
  12. Sep 5, 2011 #11

    Doc Al

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    x is the distance from the point to the origin, where q1 is. So what's the distance to q1?

    L is the distance between the charges, so what's the distance to q2?
     
  13. Sep 5, 2011 #12
    Oh! Is is L and x + L?
     
  14. Sep 5, 2011 #13

    Doc Al

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    Not quite. The distance to q1 is x. Is the distance to q2 shorter or longer?
     
  15. Sep 5, 2011 #14
    Oh ok, it's x and x + L , but since it is a function that is squared, it would have to be x-L. So you get (5)/((4 Pi Epsilon zero)(X^2) and you get (2)/(( 4 Pi Epsilon zero)(X-L)^2) right? =D

    Then once you cancel terms, you get
    5x^2-10xL+5L^2= 2r^2
    You then solve the above equation using the quadratic formula, and you get
    (10L+2LSqrt(10))/6
    The above equation can be simplified by first reducing the 6..
    (5L+LSqrt(10))/3)
    You can then pull that 3 out to make it look nicer.
    (1/3)(5L+LSqrt(10))
    Then you can factor out an L
    (1/3)(L)(5+Sqrt(10))
    And it's not plus or minus Sqrt(10) because it's length, and you can't have a negative length. Am I right? =D
     
  16. Sep 5, 2011 #15
    Thank you so much =D I sat down for lunch, and I was waiting for the food, and I was just thinking about it non stop. Then it slowly started to fit together. =D
     
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