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Electric Force vector components

  1. Jan 18, 2008 #1
    [SOLVED] Electric Force

    1. The problem statement, all variables and given/known data

    What is the force F on the 1 nC charge at the bottom?
    Write your answer as two vector components, separated by a comma. Express each component numerically, in newtons, to three significant figures.

    2. Relevant equations
    F = kq1q2 / r^2
    a^2 + b^2 = c^2
    Sine Law

    3. The attempt at a solution
    As a reference, the "focus" charge is the 1nC at the bottom.
    Note that this appears to be a semicircle. Due to this fact, I conclude that the distance between the focus charge and the -6nC is also 5cm.

    Because both 2nC charges have the same charge, distance, and angle with respect to the origin (Focus), I conclude that they cancel each other out along the X axis. They will both push the focus downwards, however there should be no X axis movement. Since the -6nC charge is along the Y axis, I conclude that this charge will not affect the focus' net X force either.

    Therefore, F_x = 0

    The force given between the 2nC on the left, and the focus, is F = kq1q2 / r^2
    k = 9*10^9
    q1 = 1*10^-9 (Focus)
    q2 = 2*10^-9 (Leftmost charge)
    r = 5cm = 0.05m

    F = 0.0000072

    Due to Sine Law, the X and Y components of this force are equivalent, since the angle is at 45 degrees.
    Sqrt(X^2 + Y^2) = F^2
    Sqrt(2Y^2) = F^2
    2Y^2 = F^4
    Y^2 = (F^4)/2
    Y = Sqrt((F^4)/2)
    Y = 3.6656415536710623664939771731515 * 10^-11

    Since the rightmost charge is identical with respect to the focus, it will put out the same force along the Y axis.

    Now to include the -6nC charge. F = kq1q2 / r^2
    k = 9*10^9
    q1 = 1*10^-9
    q2 = -6*10^-9
    r = 0.05

    F = -0.0000216

    So, the net force on the focus should be:
    = "-6nC" + "2nC:Y" + "2nC:Y"
    = "-6nC" + 2*"2nC:Y"
    = -0.0000216 + 2*3.6656415536710623664939771731515 * 10^-11
    = -0.00002159992668716892657875268
    Rounded to 3 significant digits is 2.16*10^-5

    Expressing answer as asked:

    This answer is incorrect.
    My mistake is where?
    Last edited: Jan 18, 2008
  2. jcsd
  3. Jan 18, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    That should be:
    X^2 + Y^2 = F^2
    Redo this.
  4. Jan 18, 2008 #3
    I just messed up the pythagorean theorem. Holy crap I'm embarassed.

  5. Jan 18, 2008 #4
    X^2 + Y^2 = F^2
    2Y^2 = F^2
    Y^2 = (F^2)/2
    Y = Sqrt((F^2)/2)
    Y = Sqrt((F^2)/2)
    Y = 0.00000509116882454314217568607


    = "-6nC" + 2*"2nC:Y"
    = -0.0000216 + 2*0.00000509116882454314217568607
    = -0.0000114176623509137156486278 * 10^-5
    Rounded to 3 significant digits is -1.14*10^-5

    Expressing answer as asked:

    This answer is correct.
    Thank-you Doc Al :)
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