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EXAM I physics Help

  1. Sep 21, 2006 #1

    Question 7

    i know that the equation is

    Fx = (-)(K)(12)(6) / (4^2) + K (12)^2 / (Square root of 32)) * Cos theta

    Fy = K (12)(6) / 4^2 - K(12)^2 / (square root of 32)) * Sin theta

    I just need to for somone to explain how come for the first variable is negative. I dont understand why the signs are the way they are
  2. jcsd
  3. Sep 21, 2006 #2


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    Hello Alt+F4,

    assume the x-axis "pointing" to the right and the y-axis to the top.

    If you want to find the x-component Fx of the resulting force F on Q1 you'll find:
    • Q2 does not contribute
    • Q4 and Q1 will repel each other since they both are positive charges forcing Q1 in the "negative" x direction, hence the (-) sign
    • Q3 and Q1 will attract each other forcing Q1 in the positive x direction (and negative y direction, but that will be covered by Fy), hence the (+) sign

    You can determine the correct signs for the y-component of the resulting force on the same way.


    Last edited: Sep 21, 2006
  4. Sep 21, 2006 #3
  5. Sep 21, 2006 #4


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  6. Sep 21, 2006 #5
    ya so how do u know wat is leaving and entering, i have my circuit labeled but i dont see it
  7. Sep 21, 2006 #6


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    The small triangles/arrows next to I1, I2 and I3 represent the (randomly!) chosen directions. I say randomly, because if you actually try to analyze a circuit you will know the correct "directions" for the currents after you've done all calculations. It will be represented by the sign of your result.
  8. Sep 22, 2006 #7
  9. Sep 22, 2006 #8
    ya i am still no seeing that cause i had a HW due and the equation was
    I1 = I2+I3
  10. Sep 23, 2006 #9
  11. Sep 23, 2006 #10
    As nazzard said there is no trick here. The current entering a junction must equal the current leaving a junction. So as you can see I1 and I3 are entering the junction. So that means that the sum of those currents must equal the current leaving the junction I2.

    You don't need to really calculate the force. Just draw the force vectors. You know that the two positive charges are going to repel the 12 uC charge and that in the horizontal and vertical directions. So add those two vectors and you will get a force that is along the diagonal. Now for the -12 uC charge you know it will attract the 12 uC charge along the diagonal. So the obvious choice is...

    No it won't be forcing it in the positive x direction. -ve x-direction is the left and the +ve x-direction is on the right. Since the +2 uC charge will be attracting the -2 uC charge towards it, it will be forcing it left. Then you need to add the effect of the -4 uC charge on the -2 uC charge to find the overall direction.
  12. Sep 23, 2006 #11

    I thought stuff goes from postive to negative so -2 should be going dwon
    Last edited: Sep 23, 2006
  13. Sep 23, 2006 #12
    Yes the -2 uC charge will be forced downwards due to the repulsion force from the -4 uC charge and then it will be moved in the positive y and negative x direction due (ie diagonally) to the +2 uC charge. However because the repulsion force is much greater the attractive force from the +2 uC charge the -2 uC charge will still be forced downwards and to the left.
  14. Sep 23, 2006 #13
    got it thanks
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