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anonymous

  • 4 years ago

how can i calculate the charge on a point q, given the net electric force and the values of 2 other charges?

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  1. anonymous
    • 4 years ago
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    you've got to show the whole question mam

  2. NotTim
    • 4 years ago
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    You should add the given info into a diagram and/or redraw the diagram.

  3. anonymous
    • 4 years ago
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    maybe this will help ;)

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  4. NotTim
    • 4 years ago
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    Yup. I can't really much though. I can't remember much from this,=.

  5. NotTim
    • 4 years ago
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    Eashmore, can you do anything here?

  6. anonymous
    • 4 years ago
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    I can do it.

  7. anonymous
    • 4 years ago
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    We have to balance forces. We know the expression for electromagnetic force is\[F_E = k_e {q_1 q_2 \over r^2}\] First, let's find the component of the net force that acts in the x and y-directions. \[F_{x,3} = F_3 \cos(\theta)\]\[F_{y,3} = F_3 \sin(\theta)\] We note that both forces act positive in their respective directions. Since \(q_1\) is positive and located to the left of \(q_3\), \(q_3\) must be positive if the force from \(q_1\) on \(q_3\) is to be in the positive x-direction. The same thought process verifies this when relating \(q_2\) and \(q_3\). Additionally, let's note that the force from \(q_1\) acts on \(q_3\) solely in the x-direction, and \(q_2\) acts on \(q_3\) solely in the y-direction. Let's balance the forces in x and y-directions now. \[F_{x,3} = k_e {q_1 q_3 \over r^2}\]and\[F_{y,3} = k_e {q_2 q_3 \over r^2}\]

  8. anonymous
    • 4 years ago
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    ok this makes sense so far

  9. anonymous
    • 4 years ago
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    You should be able to rearrange the last two equations for \(q_3\). You should get the same value.

  10. anonymous
    • 4 years ago
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    what do you mean? set them equal? or factor q3 out of the equation?

  11. anonymous
    • 4 years ago
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    Pick one and solve it for \(q_3\). If we solved both for \(q_3\), both equations should produce the same value for \(q_3\).

  12. anonymous
    • 4 years ago
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    Fx^2 + Fy^2 = 30.187 ^2 ? and sub both equations in for Fx and Fy

  13. anonymous
    • 4 years ago
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    I have two unknowns in the above equations yes? so how do I solve for Q3?

  14. anonymous
    • 4 years ago
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  15. anonymous
    • 4 years ago
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    factoring?

  16. JamesJ
    • 4 years ago
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    You only have one unknown: \( q_3 \). You know q1 and q2; you know r for each equation; and you know the constant \( k_e \). So you only need to solve for q3.

  17. JamesJ
    • 4 years ago
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    (Notice the r=1 m for one of the forces, and r=2m for another force)

  18. JamesJ
    • 4 years ago
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    In other words, for both of the equations .... the only variable you don't know is q3. So you can use either equation to solve for q3.

  19. JamesJ
    • 4 years ago
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    talk to me ... what's not making sense?

  20. JamesJ
    • 4 years ago
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    ok ...I'm out here of here then.

  21. anonymous
    • 4 years ago
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    i do not know Fx or Fy I only know the sum of both..sorry having internet problems

  22. anonymous
    • 4 years ago
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    You know \(F_x\) and \(F_y\). Take the total force as being \(F_3\), then use the trigonometric relations I gave you for \(F_x\) and \(F_y\). \[F_x = F_3 \cos(\theta) ~ {\rm and} ~ F_y = F_3 \sin(\theta)\]

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