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Vector addition in electric field

  1. Jan 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Two positive charges = = 2.0 are located at = 0, = 0.30 and = 0, = -0.30 , respectively. Third point charge = 4.0 is located at = 0.40 , = 0.What is the net force ((a)magnitude and (b)direction) on charge exerted by the other two charges?

    2. Relevant equations

    Columbs law (force between two point charges)
    F = 1/4 piEo times abs value of q1*q2 divided by r^2

    3. The attempt at a solution

    first i have to find the force that Q exerts on q1 (F1 on q1)
    i use coulumbs law
    F = k * [q1q2]/[r^2]
    k = 8.988 e 9

    so i plug in q1 and Q and the distance between them and i get....

    F1 on q1 = .29
    i break that down into components x and y

    F1x= .29cos(.40/.50) = .287
    F1y = .29sin(.30/.50) = .00301

    so now i find the force that q2 exerted on q1 (F2 on q1)

    q2 and q1 are both on the vertical axis so there will only be a y component

    so i use cuolumbs law again and now i plug in q1 and q2 and r is the distance between them so..

    F2= k [2.0 e -6][2.0 e -6] / (.60 ^2) = .1

    so now i add the x components and y components up

    there is only one x component and 2 y components sooo

    F1x = .287
    F1y + F2y = .10301

    F = square root of ( .287^2 + .10301^2)

    which is equal to .30 (2 sig figs)

    then to find the angle i would use tan(theta) = Fy/Fx
    which gives me an angle of 19.7 degrees
    It is supposed to be in the +x direction so i think i am supposed to subtract it from 180 and that should give me my angle

    im afraid to type the answer in into masterinphysics because i have only a few attempts left...so i want to make sure im right before i do anything!
  2. jcsd
  3. Jan 13, 2008 #2

    Doc Al

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    Staff: Mentor

    It's hard to understand what you've written (when I compare it to what follows). Please fill in the gaps. What are the charges? Where are they? (I assume those are x and y coordinates?) You need the force on which charge?


    What are the charges?
    Redo these calculations. Also: Direction (and thus signs) counts.
  4. Jan 13, 2008 #3

    opps sorry the question didnt copy right. my fault

    Two positive charges q1 = q2 = 2.0 [tex]\mu[/tex]C are located at coordinate (0,.30) meters and (0,-.30) meters, respectively. The third point charge Q = 4.0 [tex]\mu[/tex]C and is located at (.40,0) meters. What is the net force magnitude and direction on chare q1 exerted by the other two charges.

    i made a lil drawing so that it could help give a visual...its attached

    So first i found the force exerted by Q on q1 which i will call F1
    the diff between q1 and Q is .50 meters

    F1 on q1 = k (2.0 e -6)(4.0 e -6) / (.50^2) = .29 N

    i then had to break that down into x and y components

    i need to use cosine for the x component so... = F1 * cos (.40/.50)
    and then the y component would be = F1 * sin (.30/.50)

    i hope that makes sense now

    Attached Files:

  5. Jan 14, 2008 #4

    Doc Al

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    Much better.
    Good. That's the magnitude of the force.

    F1*cos(theta) = F1*(.40/.50)
    F1*sin(theta) = F1*(.30/.50)

    Those are the magnitudes of the components. What are their signs?
  6. Jan 14, 2008 #5

    :confused: I guess they are positive?
  7. Jan 14, 2008 #6

    Doc Al

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    Staff: Mentor

    Draw yourself a picture of the force that Q exerts on q1. Which way does the vector point?
  8. Jan 14, 2008 #7
    ok so the force that Q exerts on q1 is negative
    and the force that q2 exerts on q1 is positive
  9. Jan 14, 2008 #8
    so F1 and its components are negative?
  10. Jan 14, 2008 #9
    actually no...only the x component is in the negative direction for F1

    the y component of F1 is still positive
  11. Jan 14, 2008 #10

    Doc Al

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    I don't understand what you mean by saying that the force is negative or positive.

    F1 is shown on your diagram in post #3. Which way does its x-component point? Its y-component?
  12. Jan 14, 2008 #11
    its x component points in the - x direction
    the y comp. points in the + y direction
  13. Jan 15, 2008 #12

    Doc Al

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  14. Jan 15, 2008 #13
    ok and the force exerted by q2 on q1 (F2) is also in the positve y direction
    and there is no x component for that force

    so now that i know the components for all the forces do i just add them all up...add the y component of F1 and F2 together?

    then do i just add the final x and y components together?

    and that would be the magnitude of the force exerted on q1 by the other two charges right?
  15. Jan 15, 2008 #14

    Doc Al

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    No. The x and y components are perpendicular to each other--so you can't just add them like numbers. How do you find the magnitude (and direction) of a vector, given its components?
  16. Jan 15, 2008 #15
    F = [tex]\sqrt{}(Fx^2 + Fy^2)[/tex]

    ok i think i got it...Thank you very much for your help!!!!! :smile:
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