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Getting direction of force of two point charges on a third

  1. Feb 10, 2017 #1
    1. The problem statement, all variables and given/known data
    What is the direction of the force F on the -10 nC charge in the figure? Give your answer as an angle measured cw from the +x-axis.
    Express your answer using two significant figures.
    26.P39.jpg
    2. Relevant equations
    ##F = k\frac{Qq}{r^2}##

    3. The attempt at a solution
    I started by getting the components of the net force. The force between the two negative charges will be ##F_1## and the one between the -10 and +15 charges will be ##F_2##.
    ##F_1 = <0, -y> \longrightarrow F_{1x} = k\frac{(5\times10^{-9})(10\times10^{-9})}{0.01^2}##
    ##F_1 = <0, -0.004495>##
    ##F_2 = <-x, y> \longrightarrow F_{2t} = k\frac{(15\times10^{-9})(10\times10^{-9})}{(0.03^2 + 0.01^2)}##
    ##\theta = \tan^{-1}(\frac{1}{3}) \longrightarrow F_2 = <-\cos(\theta)\bullet F_{2t}, \sin(\theta)\bullet F_{2t}>##
    ##F_t = <-0.0012793, -0.0040685>##
    All of this math appears to be correct, because when I calculate the net force (##F_t = \sqrt{x^2 + y^2} \approx 4.3\times10^{-3}##) I get the right answer.
    So, now I just need to find the angle, in degrees, clockwise from the positive x-axis, but whatever I'm doing appears to be wrong. Given these components, the vector should be in the third quadrant. So, I tried calculating the inverse tangent of the y over the x components of the total force (##\tan^{-1}(\frac{y}{x}) \approx 17.45##°), and I should be getting the right angle (I've even drawn it out and checked multiple times) if I subtract this answer from 180°, giving 162.55°, which should be 160° to 2 significant figures. But this is incorrect. What am I missing?
     
    Last edited: Feb 10, 2017
  2. jcsd
  3. Feb 10, 2017 #2

    TSny

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    I haven't checked your entire calculation. But, does F1 point in the negative x direction?
     
  4. Feb 10, 2017 #3
    Yeah, I mixed up a lot of signs and axes in my OP, but it looks like I did so in my work as well - I had the x- and y-components for the final force switched up, hence why I got the right force magnitude but not direction. Thanks for pointing it out! That let me get the right answer (just had to switch the numbers I was putting in the arctan).
     
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