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Electrostatics: Two charged balls with attraction

  1. Jun 10, 2009 #1
    Hi, I have an exam coming up soon. But I thought I would clear my doubts ahead of time.

    1. A fixed conducting ball has charge q1= 3X10^-6C. An identical ball with charge q2 is held at a distance x away from q1. The two balls attract each other with a force of 13.5N. The balls are then connected by a conducting wire. After the wire is removed, the balls repel each other with a force of 0.9N.

    a) what was the charge of q2 of the second ball?
    b) What is the separation x between the balls?


    2. At what position between particles 1 and 2 will particle 3 experience no net force?
    q1= 2x10-9 q2=3x10-9 q3= -2x10-9. The distance between particle 1 and 2 is 0.04.


    Thanks
     
  2. jcsd
  3. Jun 10, 2009 #2

    LowlyPion

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    Re: Electrostatics

    Welcome to PF.

    Where did you get stuck on the first problem?
     
  4. Jun 10, 2009 #3
    Re: Electrostatics

    well I know that in the beginning the q2 should have a - charge since it is attracted to the q1 and then after the wire it should become -. I'm not really sure how to isolate for
    q2 from the equation.
     
  5. Jun 10, 2009 #4

    LowlyPion

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    Re: Electrostatics

    If the charge from both is the same it will be repulsive.

    That means in the first you a force made up of the given charge times the unknown. The second force is derived from the average of the 2 charges squared.

    F = kq1*q2/r2

    After

    F = k(q1 + q2)2/(4*r2)

    Where (q1 + q2) / 2 is the average that gets squared.
     
  6. Jun 10, 2009 #5
    Re: Electrostatics

    how did you get F=Kq(q1+q2)/(4*r^2) ? where does the extra q come from?
     
  7. Jun 10, 2009 #6
    Re: Electrostatics

    btw the answer to this problem is apparently
    a) -5x10-6 C
    b) 0.10m

    But I really just want to understand what i'm doing with the steps.
     
  8. Jun 10, 2009 #7

    LowlyPion

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    Re: Electrostatics

    Sorry, there is no extra q. I noticed the typo and edited it apparently after you picked up the post.
    Note that the correct formula is

    F = k(q1 + q2)2/(4*r2)
     
  9. Jun 10, 2009 #8
    Re: Electrostatics

    So I'm going to let r^2 equal each other

    kq1q2/F = k(q1 +q2)^2/F(4)

    So I isolate for q2 right?

    or do I say that kq1q2/F = k(q1 +q2 divide by 2)^2/F(4)
     
  10. Jun 10, 2009 #9

    ideasrule

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    Re: Electrostatics

    The factor of 4 came from q=q1+q2 divide 2, so your first equation is correct, but not your second. Note that the two "F"'s are different: in kq1q2/F, F=-13.5 N, but in k(q1+q2)^2/4F, F=0.9 N.

    Otherwise, you're on the right track. Cancel out the k, expand the right side, and you'll get a quadratic equation. It should be easy from there. Remember that q1, F, and the other F are all given.
     
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