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Simple electric field question

  1. Feb 4, 2004 #1
    Here is the question: A positive point charge Q1 = 3.00×10-5 C is fixed at the origin of coordinates, and a negative charge Q2 = -8.86×10-6 C is fixed to the x axis at x = +1.97 m. Find the location of the place along the x axis where the electric field due to these two charges is zero.

    I am puzzled.. if both Q1 and Q2 are on the X axis, and they are of opposing signs, doesnt there ALWAYS exist an electric field, since the field will flow from the positive to the negative??

    Can someone point out what I am missing? (b/c I am sure I'm missing something very obvious)

    thnx
     
  2. jcsd
  3. Feb 4, 2004 #2

    Doc Al

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    Staff: Mentor

    That's true between the two charges. So what does that tell you?
     
  4. Feb 4, 2004 #3
    it is on the outside of these two charges..



    X is to the left of Q2 since Q1 is much stronger than Q2 and they are of opposite charges.. thus the picture would look like:
    Q1---Q2----X

    I set up Etotal=E1 + E2 = 0
    thus kQ1/R1^2 = kQ2/R2^2, cancelling out K..
    if Q1=3x10^-5 C and is at X=0 thus R1 (relative to x) is X+1.97 and
    Q2=-8.86x10^-6 and is at X=1.97 thus R2 is X

    thus my equation is: (3x10^-5)/(X+1.97)^2 = (-8.86x10^-6)/(X)^2
    Solving for X, I get .18819, which would place X at 1.97+.18819=2.16m.

    But this is wrong.. can someone give me a hint to where I went wrong? This question is killing me..
     
  5. Feb 4, 2004 #4
    Your analysis is correct. You have an error in solving the equation.
     
  6. Feb 5, 2004 #5

    Doc Al

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    You mixed up Q1 and Q2 in your equations. According to your picture, Q2 must be the smaller charge.
     
  7. Feb 5, 2004 #6
    [?]

    Q2 is the smaller charge. I think he just got confused when he wrote "left", since he drew it with position "x" correctly on the right. The way he set up the equation is consistent with the drawing and does give the correct result. He simply made some arithmetical error in solving it.
     
  8. Feb 5, 2004 #7
    I figured it out. I probably spent 4 hours doing this problem (and every variation of solving it under the sun) before I figured out I was taking the Square instead of the square root!

    thanks for all of your help!
     
  9. Feb 5, 2004 #8

    Doc Al

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    You are absolutely right! (I wasn't paying attention to the exponents. ) Thanks for the correction.
     
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