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Electric field problem

  1. Aug 30, 2008 #1
    1. The problem statement, all variables and given/known data

    Two point charges +q and +2q are held along the x-axis, +q at x=0, +2q at x=3d.
    i) which region(s) on the x-axis where the electric field due to the two charges is zero?
    ii) Find the exact location in terms of d

    +q_____P_______+2q
    x=0 x=d x=3d

    2. Relevant equations

    E = k(q/r) Hint: solve the quadratic equation: K(q/x2)=K(2q/(3d-x)2

    3. The attempt at a solution
    For part i, the choices were: a) to the right of +2. b) between +2q and point P. c) between +q and point P. d) to the left of +q. e) both a and c. f) both b and d.

    Since both charges are positive they would be repulsing each other so I believe that there would be a charge both to the left of +q and to the right of +2q so this cancels out a and d and consequently e and f. I chose c because I figured that +2q is a stronger charge so zero would be closer to +q.

    for part ii, I attempted to solve the quadratic equation cancelling out K on both sides leaving: q/x2= 2q/(3d-x)2. I took the square of both side of the equation and ended up with: square root q/x=square root 2q/3d-x and now I'm stuck.

    I'm really bad at working problems without numbers and am not very good at physics, so any help would be greatly appreciated!

    Thanks in advance :smile:
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 30, 2008 #2

    Redbelly98

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    Staff Emeritus
    Science Advisor
    Homework Helper

    You're on the right track.

    So you have

    [tex]
    \frac{\sqrt{q}}{x} = \frac{\sqrt{2q}}{3d-x}
    [/tex]

    and you want to solve that for x. That's difficult when the "x" terms are in the denominator. Try to think of how to get the x's into the numerator, or how to get rid of the fractions altogether.
     
  4. Sep 1, 2008 #3
    That worked! Thanks a bunch! :)
     
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