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I'm making an Arithmetic Error, Electrostatic force diagrams

  1. Sep 4, 2015 #1
    1. The problem statement, all variables and given/known data
    Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 2.7 cm . Two of the particles have a negative charge: q1 = -6.0 nC and q2 = -12.0 nC. The remaining particle has a positive charge, q3= 8.0 nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?

    2. Relevant equations
    Fcos(θ) = Fx
    Fsin(θ) = Fy
    θ = Tanh(Fy/Fx)
    F = (Fx^2 + Fy^2)^1/2
    F(2on3) = K(q1q2)/r^2)
    3. The attempt at a solution
    Using the program's help function, I have found the components
    (F1on3)x = 2.96x10^-4
    (F1on3)y = 5.13x10^-4

    So the only thing left was to find F(2on3) in the same way I found F(1on3).
    Arranging the problem so that q3 was at the origin, q1 was at the top of the triangle and q2 was along the x-axis, I found this:

    Fy = 0 (sin0=0)
    Fx = (9x10^9)(12x10^-9)(8x10^-9) / (2.7x10^-2)^2
    = 1.19x10^-3

    This makes sense given that it should attract.

    Here's where the problem falls off of the rails...

    I need to find Theta

    So Tanh (5.13x10^-4)/((2.96x10^-4)+(1.19x10^-3)) = 0.006

    I really need this to be in the vicinity of 19degrees. I do not understand what mistake I am making... Does anybody else?
  2. jcsd
  3. Sep 4, 2015 #2


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    You should be careful here with your notation. Tan (θ) is not the same function as tanh (θ).


    If θ = tan-1 ((5.13x10-4)/((2.96x10-4)+(1.19x10-3)) is what you are trying to calculate ...
  4. Sep 5, 2015 #3
    Really? I thought i'd looked it up and found it was the same, math-wise, just different in definition. My mistake I guess...
    Scientific calculators only offer the tanh function though. Is that because they expect you to simply find Tan(Θ) and then find 1/x?
  5. Sep 5, 2015 #4


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    I'm confused here.

    The standard trigonometric functions are abbreviated as sin, cos, tan, cot, sec, and csc. cot = 1/tan; sec = 1/cos; and csc = 1/sin.

    The inverse trig functions are sin-1, cos-1, etc.

    Note that sin-1 ≠ 1 / sin. If y = sin (θ), then sin-1(y) = θ; -1 ≤ y ≤ 1

    The hyperbolic functions sinh, cosh, tanh, coth, sech, and csch are defined:

  6. Sep 5, 2015 #5
    Oh wow, yes I can see now how the hyperbolic functions are so much not the same thing...
    Okay, I need to find tan^-1(Θ) in order to solve this problem and get me the angle between the force components? Sounds pretty straightforward now. I was making a mistake with my calculator because I didn't distinguish the notations.
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