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Position of a charged particle

  1. Sep 11, 2016 #1
    1. The problem statement, all variables and given/known data
    Two particles 1 and 2, each carrying 6.0 nC of charge, are located along an x axis, one particle atx=−30 mm and the other at x = 30 mm. Where along the y axis is a particle 3 carrying a charge of +2.0 nC if it experiences an electric force of 6.9 × 10^−5 N ȷ^?

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

    Pythagorean Theorm: c = SQRT((a^2) + b^2))

    3. The attempt at a solution
    So I know the answer is 10mm and 40mm. However, throughout my work, I am not even close to said values. Here is the work I have done:

    Some notes on my progress:

    1.) I first began by finding Fx (since Fy was defined in the problem as 6.9x10^-5, and was split since I only need the force of a single 6.0 nC charge).
    2.) After calculating this value, I took both force vectors and added them together to obtain the final total Force between particle 1 or 2 and particle 3.
    3.) Taking this force value, I plugged it back into Coulomb's law in an attempt to obtain the value of d (In this problem, x = 30mm, y is what I am trying to find. So by finding d, I can use Pythagorean to calculate it). However... this is where things break apart. When I find d, I realize that this value is SMALLER than my x value... which makes 0 sense. From here I became absolutely lost.
    https://scontent.fewr1-2.fna.fbcdn.net/v/t34.0-12/14237562_1253883344636041_2264484480519635381_n.jpg?oh=5f8d0bafacc2fd1b915de0e148810ab5&oe=57D91A9E
    https://scontent.fewr1-2.fna.fbcdn.net/v/t34.0-12/14330114_1253883357969373_2960711846063879329_n.jpg?oh=a12feff8c85bc3b70ac88aebe43c94cb&oe=57D84E03

    If someone could guide me from this point, I would greatly appreciate it! Thank you!
     
  2. jcsd
  3. Sep 11, 2016 #2

    kuruman

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    Hi boxybrownmd and welcome to PF

    :welcome:

    I think you need to start over and draw a diagram of the three charges showing the forces on the third charge due to the other two charges. Then add these forces as vectors. In what direction do you think is the resultant?
     
  4. Sep 11, 2016 #3
    Well the issue is that I cannot draw it exactly since I do not know where the third particle technically would be?
     
  5. Sep 11, 2016 #4

    kuruman

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    You know it's on the y-axis so draw it at some distance on the axis and label it y. Then draw your vectors.
     
  6. Sep 11, 2016 #5
    https://scontent.fewr1-2.fna.fbcdn.net/v/t34.0-12/14344969_1253913204633055_3281124858787924149_n.jpg?oh=6d016711bbcdadc3abae4e72aef365df&oe=57D9162B
    Edit: Forgot the picture :)

    I have attempted this but I am still lost on where to progress... I feel still at a loss on how to progress at this point. I mean, from a mathematical standpoint, its impossible for the overall Force value to have a smaller distance than x or y, but based on my math above, points towards that.
     
    Last edited: Sep 11, 2016
  7. Sep 11, 2016 #6

    kuruman

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    Your drawing shows only two charges. There are three. Draw all three in the positions given by the problem. Then just draw the forces without calculating any numbers. This will help you visualize what's going on.
     
  8. Sep 11, 2016 #7
    I attempted to draw the diagram, but I do not understand anything else about the problem. I know that both of the charged particles (1 and 2) exert force on particle 3 in mirrored directions, which causes Fx to cancel but Fy to remain. However, in order to calculate F, you still need to know what Fx is?
     
  9. Sep 11, 2016 #8

    kuruman

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    What do mean by Fx? If you mean the x-component of the net force on charge 3, that is zero. You said so yourself.
     
  10. Sep 11, 2016 #9

    kuruman

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    Charges 1 and 2 exert separate forces on charge 3. Can you draw these two separate forces?
     
  11. Sep 11, 2016 #10
    Wouldn't the charges exerted on Particle 3 from 1 and 2 just be opposites but equal? For example, if Particle 1 exerted 1 N of force to the right, than particle 2 exerts 1 N to the left.
     
  12. Sep 11, 2016 #11

    kuruman

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    Charge 1 exerts a force to the right and up. Charge 2 exerts a force to the left and up. The right and left cancel but the up and up add. In other words, the net force on charge 3 has zero x-component and non-zero y-component. It's the y-component of the net force that you are asked to find.
     
  13. Sep 11, 2016 #12
    From this point is where I am lost then....
     
  14. Sep 11, 2016 #13

    kuruman

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    Like I said, draw a picture of the three charges on the axes as given by the problem. Then draw two arrows on charge 3 representing the forces from the other two charges. Find expressions, not numbers, for the y-components of these arrows and add them together. That will give you an expression for the net force on charge 3.
     
  15. Sep 11, 2016 #14
    I feel like this does not help me find the answer... I still don't understand exactly what this will do for me. It just winds up being the original net force I started with and the x forces canceling....
     
  16. Sep 12, 2016 #15

    kuruman

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    If you find an expression for the net force on charge 3, it will be a function of y, the distance of charge 3 from the origin because different distances give different values for the net force. You set that expression equal to the force on charge 3 , which is given to you, and then solve for the particular value of y that matches the given force. You cannot do this problem without doing algebra.
     
  17. Sep 12, 2016 #16
    Wouldn't the expression for net force just be SQRT((Fx)^2 + (Fy)^2)? Fx would become 0, Fy would be 3.45x10^-5?
     
  18. Sep 12, 2016 #17

    kuruman

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    I do not understand where the 3.45x10^-5 comes from. It is irrelevant. The expression for the magnitude of the net force is indeed ## F_{net} = \sqrt {F_x ^2+F_y ^2} ##. It is also true that ## F_x = 0 ##, but it is not true that ## F_y = 3.45 \times 10^5 ##. ## F_y ## is the sum of two terms, that is the y-component of the force due to charge 1 added to the y-component of the force due to charge 2. You cannot set anything equal to 6.9x10-5N unless you find an algebraic expression for the net force that has in it the distance of charge 3 (what you are looking for) from the origin.

    I am starting to believe that your resistance to what I asked you to do as a starting point, namely to draw the forces acting on charge 3, may be arising from a lack of understanding of what I am really asking you to do or from your inability to do what I am asking because you don't know how to draw vectors and determine their components. Look at post #13. Is there anything that you do not understand or do not know how to do? If so, ask me and I will try to clarify it. I gave you a starting point and a road map to the solution (post #15). Follow it even though you may not be able to see the solution from where you are right now.
     
  19. Sep 12, 2016 #18
    I do not understand what you are trying to say in post #15. The reason why I said Fy = 3.45 x 10^-5 is because the problem states that particle 3 experiences a force from particle 1 and 2 equal to 6.9 × 10^−5 N ȷ^. I assumed that j^ represented force acting in the y direction, and got 3.45 x 10^-5 by taking half of the original force acting in the y direction so that I could find the amount of force a single particle exerts on particle 3 and figure out the distance from that point.
     
  20. Sep 12, 2016 #19

    kuruman

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    OK, look at the figure that I attached. It is what I asked you to do as a starting point. O marks the origin of the axes and x1, x2 are the positions of charges q1, q2 on the x-axis. Charge 3 is at distance y on the y-axis. Study the drawing. F1 and F2 are the forces from each charge. Can you find an expression for the net force, i.e. the vector sum of F1 and F2? Clearly, your expression must involve y because different values of y give different values for the net force.
    3charges.png
     
  21. Sep 12, 2016 #20
    Ok, so I had something somewhat similar drawn when I first attempted the problem a few days ago. However, breaking it down is giving me a few issues. F1 and F2 represent the total force of particle 1 and particle 2. When you take both of the x components of F1 and F2, they will cancel out because they are equal but opposite. However, all that would be left would be the sum of the y force, which would be the 6.9 x 10^-5, correct? Which means that wouldn't just be the y component of the force, but also the net force acting on particle 3? As for an expression, the only one I can think of is just using the vector sum equation in this case, but considering the x components cancel, there isn't much to it.
     
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