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E&M Homework Problem

  1. Jun 11, 2008 #1
    1. The problem statement, all variables and given/known data

    A particle of charge 5.03 nC is placed at the origin of an
    xy-coordinate system
    A second particle of charge -2.00 nC is placed on the positive x-axis at x = 4.04 cm.
    A third particle, of charge 6.01 nC is now placed at the point x = 4.04 cm, y = 3.00 cm.


    Find the x-component of the total force exerted on the third charge by the other two.
    Find the y-component of the total force exerted on the third charge by the other two.
    Find the magnitude of the total force acting on the third charge.
    Find the direction of the total force acting on the third charge in radians.

    2. Relevant equations

    Coulomb's Law = k(q1q2)/r^2



    3. The attempt at a solution

    The answers I got were
    Fx = 6.38*10^-5
    Fy = -3.49*10^-5
    Ft =
    Direction was .944 radians

    all of which were wrong. I don't know how to calculate the Ftotal, or what I did wrong.
     
  2. jcsd
  3. Jun 12, 2008 #2

    Dick

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    Science Advisor
    Homework Helper

    Just giving the numerical answers doesn't help to tell what you did wrong, now does it? Can you give a detailed working of finding Fx? That's the easy one, since the -2nC charge doesn't affect Fx.
     
  4. Jun 12, 2008 #3
    Your equation is correct. For the x-component, only the first particle will influence the motion of the third particle along the x-axis; the second particle has the same x-coordinate, and therefore, will only change the motion of the the third particle along the y-axis. To accomplish this, find the force of the first particle on the third particle in terms of magnitude and angle, then use trigonometry to compute the x-component motion. Use a diagram to illustrate this force vector and its components. For the remaining parts, add the remaining force vector to your diagram and use more trigonometry.
     
  5. Jun 12, 2008 #4
    This is what i did to find Fx:

    To find r between Particle 1 (5.03*10^-9) and Particle 3 (6.01*10^-9)
    Sqrt[(3*10^-2)^2 + (4.04*10^-2)^2] = 5.032*10^-2
    To find the F13:
    (9*10^9)*(6.01*10^-9)*[(5.03*10^-9)/((5.0321*10^-2)^2)
    = (5.409*10^1)*(1.9864*10^-6)
    =1.07*10^-4
    Fx:
    Tan(a) = 4.04/3 = 53.40
    (1.07*10^-4)*cos53.40 = 6.38*10^-5
     
  6. Jun 12, 2008 #5
    This is the first problem. Tan(a) = opposite/adjacent = 3/4.04. I recommend that you learn your trigonometric identities as soon as possible as they are fundamental to solving many problems.
     
  7. Jun 12, 2008 #6
    Oh boy. How silly of me. I was able to get the Fx component this way. Thank you so much!

    However, I am still confused on how to find the Fy and the Ftotal.

    Using the same procedure, I got

    6.38*10^-5 for Fy(13) and -7.2*10^-5 for Fy(23)

    To get the Fytotal

    6.38*10^-5 - 7.2*10^-5 = -8.2*10^-6.

    The answer was wrong. What do I need to do instead and how do I find the magnitude of the total force?

    Thank you so much!
     
  8. Jun 12, 2008 #7
    Since you know the angle and magnitude of the force vector of particle 1 on particle 3, you can determine the x-component (which you have recently done) and also the y-component of this force. What does the force vector of particle 2 on particle 3 look like? What is its direction and magnitude? We discussed particle 2 previously. Now, just add the vector components together.
     
  9. Jun 12, 2008 #8
    I was able to find the Fy and Ftotal components. How do I find the angle of the total force in radians.

    I tried:

    Fy/Fx = (-5.7*10^-5)/(8.61*10^-5) = .662

    Cos.662 = .789 radians
    Sin.662 = .614 radians
    Tan.662 = .779 radians

    All of these were wrong. What should I try now?
     
  10. Jun 12, 2008 #9

    Dick

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    You might try understanding what you are doing instead of using random trig functions. Fy/Ft is the sine of the angle. Use an inverse trig function to get the angle. And don't just try all of them. Pick the right one.
     
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