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Electrostatic problem

  1. Mar 21, 2012 #1
    Hi guys I came upon this problem while solving in a book, there is something I am confused about the way he solved it in the answers.


    (II)Three charged particles are placed at the corners of an
    equilateral triangle of side 1.2m. The charges are 4uC,-8uC,and
    -6uC. Calculate the magnitude and direction of the net force
    on each due to the other two.
    The charge -4uC is on top and -6.00uC is on the right and -8uC is on the left

    I answered this by calculating the magnitude of each charge then getting the net charge by adding the vector components of each.

    F12 = 0.2N;
    F13 = 0.15N;
    F32 = 0.3N;
    F31 = 0.15;
    F23 = 0.3N;
    F21 = 0.2N;

    I concluded the force on charge Q1 should be pointing to Q2 since its larger magnitude so I first calculated the components vectors F12 as negative as so:-

    F12x = -cos(60)*0.2 = -0.1N;
    F12y = -sin(60) * 0.2) = (-(3^1/2) / 10) N


    F13x = cos(60)(0.15) = (3/40)N
    F13y = sin(60)(0.15) = 0.129N


    Now since the force is favored to go to F12 then we can add F12 to F13 we are subtracting anyway thats why I did F13 as positive,since I know that force net should go to F12;

    F1x = -0.3 + 3/40 = -1/40N
    F1y = -(3^1/2) / 10 + 0.129 = -0.044N

    But the way he answered in the book is really different.

    the book answer is as follows:
    F1x = -(0.1997N)cos(60) + (0.1498)cos60 = -2.495 * 10^-2;
    F1y = -(0.1997)sin(60) - (0.1498N)sin60 = -3.027 * 10^-1;

    Notice that he subtracted the y component vector which really doesn't make sense I mean both are attracting the positive proton so it wouldn't make sense to subtract. Did I miss something here? I also notice that 0.1997 F12 is really close just my rounding must have been incorrect. I also didn't include here the calculation for the force relative to point 2,since if I know what I did wrong here same thing apply to point 2 calculation(thats if I did something wrong).


    Thanks alot.
     
  2. jcsd
  3. Mar 21, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    Is that first charge +4 μC or -4 μC? You've used both in your description of the problem.

    Did you draw a diagram first and sketch in vectors for all the various forces, taking into account how the charges of same and different signs interact? Ascertaining the directions of the vectors is key.
     
  4. Mar 21, 2012 #3
    Srry I typed it wrong the first charge is +4uC not -4uC the one on top yeh I drawed all vectors calculated first the magnitude then used logic to see which one it would get attracted more in order to get the net charge as I did above.
     
  5. Mar 21, 2012 #4

    gneill

    User Avatar

    Staff: Mentor

    Look at the forces acting on charge 1. Both have y-components that are pointing downwards (negative), and one has a negative x-component while the other a positive x-component. The math should reflect these component directions.

    attachment.php?attachmentid=45356&stc=1&d=1332371244.gif
     

    Attached Files:

  6. Mar 21, 2012 #5
    I see I made the x-component that way,but I guess I made the y-component minus too.
    I thought since both are pointing downards i'd minus the force of larger component in order to get net accelaration,but since the x-component is the one that has opposite sides that would take care of the math.

    Thanks it totally makes sense now :).
     
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