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Homework Help: How to decide a force vector?

  1. Dec 16, 2012 #1
    1. The problem statement, all variables and given/known data


    Q1 = -2 C
    Q2 = 3 C

    Decide the forcevector F

    2. Relevant equations

    F = (k*Q1*Q2)/r^2

    k = 8,99*10^9 Nm^2/C^2

    3. The attempt at a solution

    Q1, Q2 = squarerot(5+7) = 9,24 mm = 0,924 cm = 0,00924 m

    F = (8,99*10^9*-2*3)/0,00924 = -5,838 * 10^12 N

    Uniformity gives

    F/Fx = 0,924/0,7

    Fx = F/(0,924/0,7) = -4.423 N

    F/Fy = 0,924/0,5

    Fy = F/(0,924/0,5) = -3,159 N

    F = (-4,423 N, -3,159 N) ?
    Last edited: Dec 16, 2012
  2. jcsd
  3. Dec 16, 2012 #2
    A few questions for you :
    Have you drawn a diagram for this with the direction of the forces on each charge?
    What force are you trying to calculate?
    What are the positions (or separation) of the 2 charges?
  4. Dec 16, 2012 #3


    User Avatar

    Staff: Mentor

    For some reason your image is not showing up, so here's a copy:
    Question: Which force vector do you want? The one acting on Q1 or the one acting on Q2? In other words, F21 or F12?
    That distance doesn't look right. Better check it.
    It looks like you're losing a good number of powers of ten from the force. The components should be around the same order of magnitude as the force they add up to; your F (currently) has an order of magnitude of 1012.

    Attached Files:

  5. Dec 16, 2012 #4
    Updated picture with direction of the force vector I need (acting on Q2):


    I'm trying to calculate the vectorforce F

    Q1, Q2 = squarerot(5^2+7^2) = 8,602 mm would be the distance between the 2 charges
    Last edited: Dec 16, 2012
  6. Dec 16, 2012 #5
    You have to decide which force to calculate, the force on Q1 or Q2?
    Q2 is the easiest.
    Start by working out the magnitude of the force first and then you can do the components.
  7. Dec 16, 2012 #6


    User Avatar

    Staff: Mentor

    Vectors have a direction. While the forces acting on the two charges will have the same magnitude, their directions will be opposite -- equal and opposite forces thanks to Newton's third law. So you need to decide which force vector you want to calculate (they differ only in direction).

    Here's a picture to make things clear:
    That looks better.

    EDIT: Added diagram.

    Attached Files:

    Last edited: Dec 16, 2012
  8. Dec 16, 2012 #7
    Okay, let's say I want to calculate the force on Q2.
    To get the magnitude (length) isn't it the same as F = squarerot(5^2+7^2) = 8,602 mm ?

    After that I use the uniformity

    F/Fx = F/0,7

    F/Fy = F/0,5

    Not sure how I know what axis (x or y) should be negative or positive?
  9. Dec 16, 2012 #8
    The magnitude of the force is given by Coulomb's law.
    F = (k*Q1*Q2)/r^2

    You can see the direction of the force vector on gneill's diagram.
    If your x-axis is to the right, and your y-axis is straight up, then both components will be positive.
    You can use your "uniformity" expressions will a small change :
    F/Fx = 8.6/7
    F/Fy = 8.6/5

    ( or you can use trigonometry - it'll come out to the same ).
  10. Dec 16, 2012 #9

    I see, so:

    F = -7,290*10^-14

    F/Fx = 8,6/0,7

    Fx = (-7,290*10^-14)/(8,6/0,7) = -5,934*10^-15 N

    F/Fy = 8,6/0,5

    Fy = (-7,290*10^-14)/(8,6/0,5) = -4,238*10^-15 N

    F = (5,934*10^-15 N, 4,238*10^-15 N )
  11. Dec 16, 2012 #10
    Almost there :)

    F = 7.290*10^14 (not -14)
  12. Dec 16, 2012 #11
    Ah, I may be a little too fast sometimes.

    Thank you both for great help :smile:
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