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Where is the neutral point? Electrostatics

  1. Feb 28, 2008 #1
    1. The problem statement, all variables and given/known data
    At the neutral point the electric field intensity is zero. Two point charges q1=+2nC and q2=-3nC are separated by 2.0m in air. Where is the neutral point?


    2. Relevant equations
    E=f/q
    coloumbs law
    E=E[q1] + E[q2]
    3. The attempt at a solution

    i dont have a solution for now since i cant figure out how to solve for d1 and d2
     
  2. jcsd
  3. Feb 28, 2008 #2
    What are [tex]d_1[/tex] and [tex]d_2[/tex] ??
     
  4. Feb 28, 2008 #3
    d1 is the distance of q1 from the neutral point , d2 is the distance of q2 from the neutral point
     
  5. Feb 28, 2008 #4
    Just try to look at it like a 1D problem. Along a line joining the two charges there will be a point where the sum of the fields is 0. Also using rules of repulsion/attraction for charges, you can figure out that the point will NOT be between the two charges. So the distance from one charge is d1, and the distance from the second charge is 2+d1.

    Drawing a picture always helps. :)
     
  6. Feb 28, 2008 #5
    Just take it as 'd' from q1. Once you calculate distance from q1, you can use your math to get the distance from q2.

    Actually, in this case there will be two neutral points. So you'll get a quadratic equation.

    Code (Text):
               *                 q1-------------------------------q2          *
    Code (Text):
                 |<----------->|
                       d    
    * is neutral point.






    [tex] E_{q1}=E_{q2}[/tex]



    [tex]\frac{k*2}{d^2}=\frac{k*3}{d+2} [/tex] <-----------------solve this.
     
  7. Feb 29, 2008 #6
    how can it happen that it wont be between the charges?

    [​IMG]

    why is it d+2.. ? where did the 2 come from?
     
    Last edited: Feb 29, 2008
  8. Mar 1, 2008 #7
    still need answer to this :( thanks to those who can help
     
  9. Mar 1, 2008 #8
    Just use your rules of repulsion/attraction between two charges. If the test change was placed between the two charges in the question, The negative would attract it left and the positive would push it away, towards the left. No matter how you add them, it would result in a vector pointing left.

    The 2 is the distance between the 2 charges. As I said earlier in my post, a diagram always helps, the red dot is the neutral point.
    [​IMG]
     
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