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Find the direction and the magnitude of the total electric field

  1. Jul 21, 2010 #1
    1. The problem statement, all variables and given/known data

    1. Two point-charges of charges +q and +q are held at the corners of an isosceles rectangle
    triangle, as shown in figure below. The absolute value of q is 1.414 μC. The distance d as shown in the figure is 0.5 m. The angle at A is 90. The gravitational forces are negligible.

    a) Find the direction and the magnitude of the total electric field at
    the apex A. (Use the coordinate system shown in figure by x and y.)

    I don't have the drawing up but if you can imagine a triangle with the top angle equaling 90 degrees and the two bottom angles having charges of +q. A line descending from point A along the y axis dissects the bottom portion of the triangle, divides it in, each side being distance d.

    I know, using Summation that Etotal = E1 + E2 and that E1 and E2 are the same because they have the same charge.

    I know the equation for the magnitude of an electric field is E=k(q/r^2)

    To find r I used the equation .5/sin 45 = x/sin 90, ending with an r value of .707m

    So I know that E = 9x10^9 ((1.414 x 10^-6)/(.707^2))

    Using that equation I end with a value of 25,459V/m for E1. Using my previous equation, I find that Etotal = 25,459 + 25,459 = 50,918 v/m.

    That is incorrect, can someone tell me where I slipped up? Thanks.
     
  2. jcsd
  3. Jul 21, 2010 #2

    berkeman

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    Staff: Mentor

    First, "isosceles rectangle triangle" is a funny typo. Second, I really did try, but without a figure, your text is too hard for me to follow and check your math. Can you please post a figure so we can check your vector math?
     
  4. Jul 21, 2010 #3
  5. Jul 21, 2010 #4

    berkeman

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    Staff: Mentor

    That's a big help. Now show us your calc for the summation of the y components of the E field. The x components cancel obviously.
     
  6. Jul 21, 2010 #5
    Oh wow, I totally ignored the Y component. I realized the X would cancel out.

    Assuming that my other calculation of E = 9x10^9 ((1.414 x 10^-6)/(.707^2)) is correct, then E1 and E2 would be cos45(25,459 + 25,459) or 36004 V/m

    Does that look correct?

    Thanks!
     
  7. Jul 21, 2010 #6
    I have to go to bed soon, but if anyone can help me out, I'd be very grateful!
     
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