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Determining the electric field at point a and b, stuck on finding resultant E!

  1. Jul 1, 2005 #1
    Hello everyone, i'm having problems finding The resultant intensity at point b, i know its the vector sum of E1 and E2. q1 = -6x10^-9; q2 = 6x10^9 they are 12cm apaprt. Determine the electrict feild at A and B, i already figured out A. This is the free body diagram.

    ^ E2
    ....v E1


    The angle that E2 forms is a 37 degree angle.
    E1 is 9cm from q1.
    Point B is 15 cm from q2 diagnoally.

    Okay I figued the feild intensity at B due to q1 is directed downward and is equal to
    E1 = kq1/r^2 = -[(9x10^9)(6x10^-9)]/(9x10^-2m)^2;
    E1 = -.667x10^4 N/C;

    To find E2
    E2 = [(9x10^9)(6x10^-9)]/(15x10^-2m)^2;
    E2 = .240x10^4N/C 37 degree N of W.

    Now this is where I get lost...
    I need to find the X and Y component of E2 which seems easy enough....but i f it up anyways.

    So i figure, you have a triangle that looks like this:

    | 37 degrees

    E2y = E2*sin37;
    E2x = E2*cos37;

    E2x = -.192x10^4 N/C
    E2y = .144x10^4 N/C

    They did the following:

    Ex = -E2x = -(.240x10^4 N/C)*cos37;
    Ex = -.144x10^4 N/C; //is this a misprint? :bugeye:

  2. jcsd
  3. Jul 1, 2005 #2

    Doc Al

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    Looks like a misprint to me. It's obviously wrong. (.240x10^4)*cos37 does not equal .144x10^4 !

    What book are you using?
  4. Jul 1, 2005 #3


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    just aside, in the future you might want to use a scanner to illustrate the diagrams by posting it as an attachment.
  5. Jul 1, 2005 #4
    hah good thinking! The book is called Applied physics, it was published in 1979 :uhh: I'm trying to polish up in physics so i don't screw up physics 212 when i go back in the fall.

    I also had another question about the problem....
    If the problem only gives the distance from the 2 charges, 12cm...how did they figure out:

    #1. The angle 37 degrees
    #2. The distance from point B to charge 2?
  6. Jul 2, 2005 #5

    Doc Al

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    Staff: Mentor

    If all they gave was the distance between the two charges, then you couldn't figure out the angle. But they also give you the fact (I assume) that point B is 9cm above q1. Thus you have two sides of a right triangle: the bottom is 12, the left side is 9. You have all you need to figure out the angle and the hypotenuse of that triangle (which is the distance from B to q2) using a little trig (or the Pythagorean theorem).

    This particular right triangle is a popular one: The sides are in the ratio of 3-4-5, thus it's often called the "3-4-5 right triangle". :smile:
  7. Jul 2, 2005 #6
    Thanks for the info! I really gota work on my trig!
  8. Jul 2, 2005 #7


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    ....and vectors
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