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X Coordinate in Electric Field System

  1. Jan 21, 2010 #1
    1. The problem statement, all variables and given/known data

    Locate the x coordinate such that E=0?
    Coulomb constant is 8.98755x10^9 N m^2/C^2. The 1.68x10^-6 charge is at the origin and a -8.54x10^-6 charge is 10 cm to the right, as shown in the figure.

    http://i325.photobucket.com/albums/k398/bdh1613/018.jpg

    Locate the x coordinate such that E = 0. Note: q1 is at the origin O. Answer in units of cm.



    2. Relevant equations

    E=Fe/q
    E=Kc(q/d^2)
    F=Kc(q1q2/d^2)
    F=1/(4pi(emf))*(q1q2/d^2)


    3. The attempt at a solution
    I've established that q1/(d)^2 should equal q2/(d+10)^2, but it seems I'm missing a component somewhere. I repeatedly get extremely small values, none of which satisfy the answer. I know that the placement of the charge will be left of the positive charge. Thanks in advanced if anyone knows what to do.
     

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  3. Jan 21, 2010 #2

    kuruman

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    In which of the three regions I, II or III do you think is the point that you are looking for? Also, you need to show a bit more of what you did so that we can find where you went wrong.
     
  4. Jan 21, 2010 #3
    I'm sorry for the confusion. I believe that the point that it is asking for is located in region I. As for what I've done so far, I have set the equation for each electric field upon each charge equal to each other. (E=Kcq/d^2). I set the distance from q1 to the point = to d and the distance from q2 to the point equal to d+10cm. So my equation yields to (1.68e-6 / d^2)=(8.54e-6/ (d+10)^2). (I dropped the negative sign on the q2 charge because it resulted in a non-real answer and only signifies direction or charge type.) When solved through I found that d equaled -3.072550718cm and 7.970509901cm. I think that I possibly selected the wrong region.
     
    Last edited: Jan 21, 2010
  5. Jan 22, 2010 #4

    kuruman

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    You did the problem correctly. The two answers, -3.07 cm and +7.97 cm is what you should get when you solve the quadratic. Which of the two is appropriate for this problem? Answer: 7.97 cm because the distance from the 8.54 μC charge is 10+7.97 = 17.97 cm, i.e. in region I, to the left of both charges. If you choose -3.07 cm this puts the point between the two charges at 10+(-3.07) = 6.93 cm from the 8.54 μC charge which you know can't be.

    By the way, that second point between the charges would be the point of zero electric field when the two charges have the same sign, either both positive or both negative. That's because in this case you would set up the equation exactly the same way and the quadratic would give you the same two solutions.
     
  6. Jan 22, 2010 #5
    I realize where I went wrong. I was entering the 7.97 meters as a positive value. This value was positive because it was asking for distances in terms of their values between the charges. The problem asks for a coordinate point. This makes the answer negative. So the coordinate is x= -7.97cm. Thank you for verifying my steps :)
     
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