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Coulombs Law with vectors question help, test tomorrow.

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data[/b]

    1. Charge A (-2uC) is 0.10m left of charge B (+3uC), with charge C (+4uC), 0.075m below charge B, forming a right angle triangle with the right angle at B. Find the net electrostatic charge on C

    if its confusing the 0.10 is the opposite of the hypotenuse with the 0.075m being the adjacent.

    i dont how to draw a damn triangle i cant do it with just normal symbols but if oyu dont understand the triangle i will try to explain it further
    2. Relevant equations
    Fe=kq1q2/r^2

    3. The attempt at a solution
    Fe(b on c) = (8.99*10^9)(3*10^-6)(4*10^-6)/ .075^2 = 19.17866666..

    Fe (a on c) = first i found the hypotenuse, (.1^2+.075^2) = .015625m

    then found Fe, K * ( 4*10^-6) (2*10^-6)/.015625^2 = 294.58432

    I found the angle at C, by using pythagoras, tan-1(0.10/.075) = 53.13010235, then found the angle to use to find the x and y component of Fe (a on c) by subtracintg 53.. from 90, which is 36.86089765..

    Fe (aonc x) = 294... * cos36... = 235.667456
    Fe (aonc y) = 294... * sin 36... = 176.750592

    then [itex]\sqrt{235..^2 + 176...^2}[/itex] which equals back to 294.58432.

    I may have done the whole Fe ( a on c) wrong, if i did i dont know where.

    The answer is 16.8N 12.6 degrees W of S
     
    Last edited: Oct 3, 2012
  2. jcsd
  3. Oct 4, 2012 #2

    ehild

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    Homework Helper
    Gold Member

    You made a mistake when calculating the hypotenuse: .015625 is the square of of the hypotenuse, you do not need to square again in the equation for the force. .

    ehild
     
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