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Homework Help: Finding charge based on net force and other charges

  1. Sep 22, 2008 #1
    1. The problem statement, all variables and given/known data

    Blue charge is at origin with positive q charge
    Red charge at point (d1,0) with unknown positive charge q_red
    Yellow charge at point (d2cos(theta),-d2sin(theta)) with negative 2q charge

    The net electric force on the blue sphere has a magnitude F and is directed in the - y direction.

    Suppose that the magnitude of the charge on the yellow sphere is determined to be 2q. Calculate the charge q_red on the red sphere. Express your answer in terms of q, d1, d2, and theta.

    2. Relevant equations

    electric force F = kq_1q_2/r^2 where k = 9*10^9, q_1 and q_2 represent point charges, and r is distance between point charges

    3. The attempt at a solution

    F = [(k*q_yellow*q_blue)/d2^2 ] + [(k*q_red*q_blue)/d1^2]
    F = [(k*(-2q)*q)/d2^2] + [(k*q_red*q)/d1^2]
    d1^2[F - ((k*(-2q)*q)/d2^2)] / kq = q_red

    i think i'm on the right track but i did not use theta which i need to, where did i go wrong
     

    Attached Files:

  2. jcsd
  3. Sep 23, 2008 #2
    does this involve the coordinate location of the yellow charge because it contains theta which i think i need in my final answer, but the distance is stated as d2 so why should i need it?
     
  4. Sep 23, 2008 #3

    LowlyPion

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    I can't see your picture as yet, but I would suggest that you separate the forces into their x,y components and then add them. They tell you the result vector is acting in the -Y direction only, so x components must add to 0. Force is a vector and adding the magnitudes if they are not acting along the same line is not the way to do it.
     
  5. Sep 24, 2008 #4
    how about this, i put the net forces into components:

    fnet_x = [k(2q)(q)/(d2cos(theta))^2] + [k(q_red)(q)/(d1^2)]
    fnet_y = [k(2q)(q)/(d2sin(theta))^2] + 0

    F = sqrt((fnet_x)^2 + (fnet_y)^2)
    [sqrt[F^2 - (fnet_x)^2](d2sin(theta))]/(2q)(k) = q_red

    is that correct now?

    how do i factor in the -y net force direction?
     
    Last edited: Sep 24, 2008
  6. Sep 24, 2008 #5

    LowlyPion

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    Not quite.

    First simply identify the force between the Blue/Red and the Blue/Yellow.
    These are the forces that you must treat as vectors.

    Hence
    F(b/r) = kqb*qr/(d1)2*x-hat + 0*y-hat
    F(b/y) = kqb*qy/(d2)2*Cosθ*x-hat + kqb*qy/(d2)2*Sinθ*y-hat

    Now when you add F(b/r) and F(b/y) you add the components.
    But you also know that the x-components (x-hat terms) must add to 0
    And you also know that the charge on Yellow is -2*q and the charge on Blue is +1*q. The qr is the one that is unknown. Figure it must be a positive charge since Red is positive Yellow negative and otherwise they could never add to 0.
     
  7. Sep 24, 2008 #6
    so this is what i got, since we know the x components must sum to zero soo...

    [(q_red)(+q)k]/d1^2 + [(+q)(-2q)k]/d2cos(theta)^2 = 0
    so q_red = [-k(+q)(-2q)(d1^2)]/[(d2cos(theta)^2)(+q)(k)]

    is this right?
     
  8. Sep 24, 2008 #7

    LowlyPion

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    No. I don't think you read the equations I gave you carefully.

    [tex] \vec{F_{BR}} = \frac{k*Q_B*Q_R}{d_1^2} *\hat{x} + 0*\hat{y}[/tex]

    [tex] \vec{F_{BY}} = \frac{k*Q_B*Q_Y}{d_2^2}*Cos\theta* \hat{x} + \frac{k*Q_B*Q_Y}{d_2^2}*Sin\theta *\hat{y} [/tex]
     
    Last edited: Sep 24, 2008
  9. Sep 24, 2008 #8
    okay i see now, aside from the issue with the sine and cosine, was my approach for solving for q_red correct?

    here it is with corrections

    [(q_red)(+q)k]/d1^2 + [(+q)(-2q)(k)(cos(theta))]/d2^2 = 0

    q_red = -[k(+q)(-2q)(d1^2)(cos(theta))]/[(d2^2)(+q)(k)]

    q_red = -[(-2q)(d1^2)(cos(theta))]/[(d2^2)] = (2q)(d1^2)(cos(theta))]/[(d2^2)

    how about now?
     
  10. Sep 24, 2008 #9

    LowlyPion

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    That looks more like it.
     
  11. Sep 24, 2008 #10
    thanks so much, it was right!!!
     
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