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Physics 12 electromagnetism

  1. Sep 5, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-9-5_3-44-35.png
    2. Relevant equations
    Let e be electric field.

    e=kq^2/r^2
    Fe=qe

    3. The attempt at a solution
    I have no idea what the steps to solving this are, I am looking for a quick outline on the first few steps only.

    I calculated the magnitude of each point's electric field at 0.144.
     
  2. jcsd
  3. Sep 5, 2015 #2

    blue_leaf77

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    You need to show us the steps with which you arrived at that answer.
     
  4. Sep 5, 2015 #3
    e=k(2x10^-6)^2 / (0.5)^2
    e=(8.99x10^9)(2x10^-6)^2 / (0.5)^2
    e=0.144
     
  5. Sep 5, 2015 #4

    blue_leaf77

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    Force is a physical quantity, it has unit. What's the unit you are using there? In particular how did you have 0.5^2 in the denominator?
     
  6. Sep 5, 2015 #5
    0.5 is the radius, in meters, (it is given as 5 cm in the question). And e is in newtons per coulomb, so the strength of the electric field is 0.144 N/C.
     
  7. Sep 5, 2015 #6

    blue_leaf77

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    5 cm = 0.05 m, not 0.5 m. Anyway what you calculated there is just the E field from one charge, what about the other charge?
     
  8. Sep 5, 2015 #7
    My apologies,

    e=k(2x10^-6)^2 / (0.05)^2
    e=(8.99x10^9)(2x10^-6)^2 / (0.05)^2
    e=14.4 N/C

    The charges are all identical in magnitude, so the electric fields all have the same force of 14.4. Therefore there is this electric field strength at each corner of the triangle.
     
  9. Sep 5, 2015 #8

    blue_leaf77

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    This is just the magnitude of force due to a single charge, however remember that force is a vector quantity. At the same time, there is also the other charge to be considered - the force experience by any charge is the resultant of forces due to the other two charges. The presence of more than one forces makes it necessary to consider the force completely as vectors.
    No, it's not. Again, you need to consider the other two charges and the vector nature of forces due to them.
     
  10. Sep 5, 2015 #9
    Ok, so breaking down the forces:

    Let's assume that we are finding the forces on the top point of the triangle (P). Note that the angles of the triangle are all 60 degrees.

    Therefore the x components of the bottom two electric field forces would be 14.4cos60, however they cancel out as the x components travel in the same direction.
    The y component would be 14.4sin60.

    ey=14.4sin60
    =12.47

    Both the charges will have a y component going upwards, so adding the y components we get a net y component of 24.9 N. Now this value, I have question about: Is this the net electric field in the y component or the net force in the y component.


    The answer in the book is indeed 25N.
     
  11. Sep 5, 2015 #10

    blue_leaf77

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    It's the net force. In your calculation, there is ##q_1 q_2## right?
     
  12. Sep 5, 2015 #11
    Yes, there are 3 charges, all with the same charge.
     
  13. Sep 5, 2015 #12
    I did not use a calculation with q1,q2 in the same expression because I calculated the electric field produced by each.
     
  14. Sep 5, 2015 #13

    blue_leaf77

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    I mean you calculate
    $$
    F = k\frac{q_1q_2}{r^2}
    $$
    with ##q_1=q_2##, this is the formula for electrostatic force felt by charge ##q_1## due to charge ##q_2## (or vice versa). While the electric field is due to a charge ##q## is
    $$
    E = k \frac{q}{r^2}
    $$
    I should have noticed this earlier, what you calculated is the force, the formula you used is the formula for electrostatic force, not electric field. So, the unit should be N (Newton), not N/C.
     
  15. Sep 5, 2015 #14
    Ok I think i amde a mistake as well, I was trying to find the magnitude of the electric field, which is e=kq/r^2, the charge is NOT squared, as I had calculated.

    Therefore,

    e=kq/r^2
    e=(8.99x10^9)(2x10^-6)/(0.05)^2
    e=7192000 N/C

    This is the electric field produced by each charge
     
  16. Sep 5, 2015 #15

    blue_leaf77

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    Alright then, things are already clarified.
     
  17. Sep 5, 2015 #16
    So using this electric field magnitude, I break it into component.

    Xcomponent=7192000cos60
    The X components cancel out

    Ycomponent=7192000sin60
    =6228454

    They y components add up for a net electric field of (2x6228454) in the y direction.
    So enet=12346909 N/C.

    To find force exerted i Use Fe=qe

    Fe=(2x10^-6)(12346909)
    Fe=25 N

    That is the answer, so I was using the wrong equation. Thank you very much for your help.
     
  18. Sep 5, 2015 #17
    However, because you brought it up, when using the formula for electric force=kq1q2/r^2, does this equal the force exerted on the third point. And if so, why must we still use components to get the answer as opposed to just multiplying by 2, as each point produces the same force?
     
  19. Sep 5, 2015 #18

    blue_leaf77

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    Just multiplying by 2 with what? The y component or the force magnitude?
    Force is a vector quantity, you can't add them traditionally like you would add ordinary numbers. I thought you have known this when you explain how the x components cancel leaving out the y component only.
     
  20. Sep 5, 2015 #19
    Yes, I figured it out, thanks again!
     
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