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Superposition of electric field

  1. Jul 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Charge q1 = 1.8 10-8 C is placed at the origin. Charges q2 = -1.7 10-8 C and q3 = 2.6 10-8 C are placed at points (0.18 m, 0 m) and (0 m, 0.24 m), respectively, as shown in the figure. Determine the net electrostatic force (magnitude and direction) on charge
    q3.

    question attacted with picture


    find the magnitude and direction



    3. The attempt at a solution

    F(2-3) = k*abs(q2 * q3) / l2 = 4.42*10-5 N
    F(2-3)x = 4.42*10-5 N * cos (36.9)
    F(2-3)y =4.42*10-5 N * sin(36.9)



    F(1-3)y = k*abs(q1 * q3) / l2 = 7.3 * 10-5 N

    and then you add each x and y and get the net magnitude

    but answer to the angle seems to be 54.9 degrees not 36.9 why is it 54.9?
    tan-1(.18/.24 ) = 36.9 degrees not 54.9 nor is 90-36.9 not= 54.9

    thank you
    it would help me a lot if someone wrote a walkthrough of this problem
     

    Attached Files:

  2. jcsd
  3. Jul 5, 2011 #2

    vela

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    You need to take into account the direction of the forces due to the sign of the charges. Does each pair attract or repel?
     
  4. Jul 5, 2011 #3
    2 and 3 attracts so the equation should be positive
    and 1 and 3 repels so the equation should be negative
     
  5. Jul 5, 2011 #4
    yjk91,

    Drawing a force diagram and the resultant vector would be extremely helpful for you to realize your mistake. Remember that you can take the arctan of any unitless value, which applies to force/force. Also, keep in mind SOHCAHTOA. It appears you have mistaken sine for cosine in your attempted solution.
     
  6. Jul 5, 2011 #5

    vela

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    Hyperspaceme's suggestion is an excellent one. You don't even have to draw it terribly accurately. It will guide you in setting up your calculations and give you an idea of what the answer should be.
     
  7. Jul 5, 2011 #6
    but how do you get angle of 54.9?
     
  8. Jul 5, 2011 #7
    yjk91,

    Maybe if I point out what you did wrong, you might understand how to proceed.

    The angle that you solved for with tan-1(.18/.24) is the angle between the force of q1 on q3 and the force of q2 on q3. This is not the resultant vector. The resultant vector will be the force of q1 on q3 and the force of q2 on q3, added component by component. This is what got you the magnitude, but it will also be what gets you the angle.

    Did you solve for the magnitude already? How did you achieve this?
     
  9. Jul 5, 2011 #8
    yeah i got the magnitude by doing this

    F(1-3)y = k(q1 * q3) / l^2 = 7.3 * 10^-5

    F(2-3) = k*abs(q2 * q3) / l^2 = 4.42*10-5 N
    F(2-3)x = 4.42*10-5 N * sin (36.9)
    F(2-3)y =4.42*10-5 N * cos(36.9)


    Fy net = F(1-3) - F(2-3) = 3.8 * 10^-5
    Fx net = F(2-3) = 2.65 * 10^-5 N

    F net = rad(Fx net ^2 + Fy net ^2) = 4.61 e -5
    and this is right
    but i don't know how to get the angle counter clockwise from the x axis
     
  10. Jul 5, 2011 #9
    nvm i got it all i had to do was tan^-1(fy net / fx net)
     
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