1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Need guidance on finding charge.

  1. Aug 26, 2012 #1
    1. The problem statement, all variables and given/known data
    System with 3 charged particles red, yellow, and blue. The blue and red have a positive charge, while the yellow has a negative charge. The blue particle lies at (0,0). The red charge lies at (d1,0). The Yellow charge lies at (d2cos(θ), -d2sin(θ)). The Fon blue is (0,-F); F>0. q of yellow is 2q. What is the charge on the red particle in terms of q, d1, d2, and theta.


    2. Relevant equations
    Coulomb's law. F=(K*q1*q2)/d^2


    3. The attempt at a solution
    http://i249.photobucket.com/albums/gg221/cidr1337/394254_297418573698401_1519950841_n.jpg
    edited for kushan's question.
     
    Last edited: Aug 26, 2012
  2. jcsd
  3. Aug 26, 2012 #2
    Is the answer asking in terms of q , qr( charge of red) , theta , d1 , d2 .
    I have an intuition that q is charge of red particle . Can you check you question ?
     
  4. Aug 26, 2012 #3
    in terms of q, theta, d1, and d2.
     
  5. Aug 26, 2012 #4
    So what is the charge of red particle ?
     
  6. Aug 26, 2012 #5
    yes.
     
  7. Aug 26, 2012 #6
    You know the resultant of force is along -Fj . And you also know that force between blue and red would be along -i , and the force between blue and yellow makes and angle theta with positive x axis .
    Solve these vector equations .
     
  8. Aug 26, 2012 #7
    Not sure i understand... sorry.
    My new attempt i set 0 = Fredonblue+Fyellowonblue
    0=(K*qr*qb)/d1^2+ (k*qy*qb)/(d2cos(theta))^2
    -(K*qy*qb)/(d2cos(theta))^2=(K*qr*qb)/d1^2 qb and K cancels out and setting qy=2q yields
    |qr|= -(2q(d1)^2)/(d2cos(theta))^2
    does this make any sense or am i grabbing at straws?
     
    Last edited: Aug 26, 2012
  9. Aug 26, 2012 #8
    shameless self bump.
     
  10. Aug 26, 2012 #9
    I don't think that is true since it states that F>0.

    But I think that you have the right idea that F is just the sum of individual forces and then you just solve for qr in terms as stated.
     
  11. Aug 26, 2012 #10
    Can you help me figure out the force that yellow exerts on blue? I think I'm having a problem with the actual equation. In the x direction? I'm saying it is Fyonb= (K*|2q|*|qb|)/(d2)^2cos(theta) but it says the answer doesn't rely on K or qb
     
  12. Aug 26, 2012 #11
    [itex] \vec{F_{blue}} = \vec{F_{blue red}} + \vec{F_{blue yellow}} [/itex]

    I read the problem again and I think you're going to have to solve a system of equations to get it in the required terms.
    I'd start off by writing each of the forces acting on the other and then the net force for blue that you are given. Then see if you can set any of them equal to eliminate variables.

    I haven't started working out the problem yet so I can't make any promises but it seems like a reasonable approach.
     
  13. Aug 26, 2012 #12
    Thanks for trying to help me guys.
     
  14. Aug 26, 2012 #13

    TSny

    User Avatar
    Homework Helper
    Gold Member

    cidr1337,

    You are given that the net force on the blue charge has zero x-component. Do you see that the only way for that to happen is: The x-component of the force on the blue charge due to the yellow charge must cancel the x-component of the force on the blue charge due to the red charge?

    Can you construct expressions for these x-components of the forces?

    Can you use these expressions to set up an equation that expresses that the net x-component of the force on the blue charge is zero?
     
  15. Aug 26, 2012 #14
    *That is my problem. I can't set this up. I tried 0=Fyonb+Fronb but my y on be expression is wrong. I don't know how to express this. Apparently it doesn't depend on K or qb, I said F yonb = (K*|2q|*|qb|)/(d^2)*cos(theta).
    I'm probably going to feel stupid when I figure this out. :)
     
  16. Aug 26, 2012 #15

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Yes :smile:, that's the correct expression for the x-component of the force on the blue charge due to the yellow charge (where d = d2, right?)

    Now write out an expression for the x-component of the force on the blue charge due to the red charge (including the correct sign).
     
  17. Aug 26, 2012 #16
    My problem is I went into the hints and the question asked me, "Find the x component of the force that the yellow sphere exerts on the blue sphere.
    Express your answer in terms of q, d2and theta. I input Fyonb and it tells me the answer doesn't depend on K or qb. It's telling me it's not right :(.
     
  18. Aug 26, 2012 #17

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Your expression for the x component of Fyonb is correct. It does depend on k and qb, as well as on q, d2, and theta.

    You're ready to continue. So, go on and find an expression for the x component of Fronb.
     
  19. Aug 26, 2012 #18
    Ok so the x component is (K*(2q^2)*cos(theta))/(d2^2). idk how they got qb=q if i could have gotten that I would have got the problem.
     
  20. Aug 26, 2012 #19
    render.gif
    Can you guys help me figure out why qblue=q
     
  21. Aug 26, 2012 #20

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Where did you get this expression? Who is "they"? The expression is correct if qb = q, but your original statement of the problem did not specify a value for qb. Now, it actually doesn't matter what the value of qb is. You will see that if you set up the equation that states that the sum of the x-components of the two forces acting on the blue charge is 0, then you won't need to know the blue charge in order to find the red charge.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook