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Finding the Force on One Point Charge Due to Multiple Point Charges

  1. Dec 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Two +4 mC charges and two -4mC charges are arranged on a circle of diameter "d" as shown in the figure below. The magnitude and direction of the force on the +4mC charge at the top of the circle is:

    a) 6.39*10^6 N radially inward
    b) 1.2 * 10^7 N radially inward
    c) 2.4 * 10^7 N radially inward
    d) 6.39 *10^6 N radially outward
    e) 2.4 * 10^7 radially outward

    2. Relevant equations

    F = k q1q2 / r^2

    3. The attempt at a solution

    15 cm = .15 m
    4 mC = 4*10^-3 C

    F2,1 = (8.99*10^9 Nm^2/C^2)(-4*10^-3 C)(4*10^3 C) / (0.075 m^2)

    F2,1 = -2.557*10^7 N (2) = -5.114*10^7 N

    I multiplied by "2" because there are two -4mC charges.

    F3,1 = (8.99*10^9 Nm^2/C^2)(-4*10^-3C)(4*10^-3C) / (0.15 m)^2

    F3,1 = 6.392*10^6 N

    -5.114 * 10^7 N + 6.392*10^6 N = -4.4748 * 10^7 N
     

    Attached Files:

    Last edited: Dec 8, 2012
  2. jcsd
  3. Dec 8, 2012 #2

    gneill

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    Staff: Mentor

    Your figure fails to specify the precise positions of the charges on the circle, what their numerical designations are, and what signs their charges have. Okay, perhaps we might be able to figure out a workable arrangement by teasing out the information from your equations, but why should we have to work so hard?

    For F3,1 you have specified a distance of 0.15 m, or 150 cm, which is 10x the diameter of the circle. That can't be right by the premise of the charges being arranged on a 15 cm diameter circle.

    I don't see where you've accounted for the vector components of the forces in your calculations or annotations.
     
  4. Dec 8, 2012 #3
    Sorry, I was trying to mimic how it did on an exam. I see where I left charges - I will update with the correct attachment. The question in the exam didn't specify exact positions which I why I didn't.

    Relative to the +4 mC at the top of the circle, the two -4 mC will create a vector pointing radially inward and the +4 mC charge at the bottom will create a vector pointing radially outward. I thought the forces for the -4 mC charges would point in the same direction so I multiplied those by two. Concerning the +4 mC charge at the bottom I knew that pointed in the opposite direction so I took the difference.
     
  5. Dec 8, 2012 #4

    gneill

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    Staff: Mentor

    Where does the "(0.075 m^2)" in the F2,1 equation come from? Was this a given in the problem statement? Presumably you might locate the actual x,y position of the charges from that and then determine the components of the forces.
     
  6. Dec 8, 2012 #5
    The 0.075 m^2 came from me taking 0.15 m^2 / 2. I'm looking at the answer key now and what the instructor has is:

    F = (8.99 * 10^9 Nm^2/C^2)(0.004 C)^2 / 2(0.075 m^2)

    That's not the entire answer but that's the only part where I see 0.075 m.
     
  7. Dec 8, 2012 #6

    gneill

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    Hmm. I don't understand the 2(0.075 m^2) in the denominator; it's not triggering any "aha!" intuitions from what's been presented so far (maybe I'm just having a slow Saturday night :smile:). Was there some particular geometrical arrangement in the original figure that would prompt using ##\sqrt{2}R## for the distance?
     
  8. Dec 8, 2012 #7
    I should have just posted this initially. I hope the attachment helps.
     

    Attached Files:

  9. Dec 8, 2012 #8

    gneill

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    Yes, that helps. It shows the -4mC charges on the "equator" of the circle, so the distance to them from the top charge is given by ##\sqrt{(15cm/2)^2 + (15cm/2)^2} = \sqrt{2(7.5 cm)^2}## which yields a distance-squared of 2 x (0.0075m)2. It also tells you how the force between the charges will decompose into components since it reveals the angles involved (equilateral triangle).
     
  10. Dec 8, 2012 #9
    Does the square root come from the Pythagorean Theorem? Also, I don't understand what he did with the 12.8*10^6 N.
     
  11. Dec 8, 2012 #10

    gneill

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    Yes, of course. The distance is hypotenuse of the triangle with two sides that have lengths equal to the radius of the circle (7.5 cm).
    He found the vertical component of the force.
     
  12. Dec 8, 2012 #11
    Wow, it took me a little bit but now I understand the first part (the square root and the hypotenuse). Now I'm confused as to how he went from the vertical component of the force to the 9.0 * 10^6 N? Did he subtract the horizontal component so all he'd have left to compare are vertical components?
     
  13. Dec 8, 2012 #12

    gneill

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    Staff: Mentor

    The magnitude of the force along the line joining one of the negative charges to the top positive charge is 12.8*10^6 N. The vertical component of that force IS 9.0*10^6 N. Take a look at the geometry; What's the angle involved? How do you decompose a vector into horizontal and vertical components?

    By symmetry, the horizontal components due to the two negative charges cancel each other since they are equal and oppositely directed. The vertical components reinforce (add).
     
  14. Dec 9, 2012 #13
    I see now. sin 45° * 12.8*10^6 N = 9.0 *10^6 N
     
  15. Dec 9, 2012 #14

    gneill

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    Yup. Or cos(45°), depending upon which of the triangle corners you're working with. With an angle of 45° the sine is equal to the cosine.
     
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