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Net electric field and magnitude

  1. Jan 22, 2006 #1
    Three positive charges are arranged in a rectangle. The charge in the bottom left corner is 3.0 nC, in the top right corner is 7.0 nC, and in the bottom right corner it is 1.0 nC. The sides have a length of 0.10 m. Find the magnitude of the electric field at the fourth corner of the rectangle. Answer in units of N/C.

    In the x-direction, the field is from the charge in the top right corner.
    so k* 7 x 10^-9/.10^2 = 6300

    In the y-direction, the field is from the charge in the bottom left corner.
    so k* 3 x 10^-9/ .10^2 =2700

    I know that the last charge needs to be broken up into components, but I'm not really sure how to do that. I know that the diagonal of the rectangle is .141 m, and that would be the distance from the point to the charge, but I always get confused on which angle to use. Help please??
  2. jcsd
  3. Jan 22, 2006 #2


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    You have to think of r as a position vector. Your rectangle is a square, so the angle the diagonal makes with respect to any of the sides is 45 degrees.
  4. Jan 23, 2006 #3
    ok I got that part.. thanks
    The second part says,
    What is the direction of this electric field ( as an angle between -180 and 180 measured from the positive x-axis with counterclockwise positive)? Answer in units of degrees.

    I know that to find the angle, you take the inverse tangent of the x and y components.
    so tan^-1 (3021.1/6620.1) = 24.5 degrees.
    This is wrong. I don't really understand how they are measuring direction from the x-axis. Can someone help me with how to get my answer to fit? Or am I doing it completely wrong?
  5. Jan 23, 2006 #4


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    The Electric Field points AWAY from positive charges ...
    (unlike the gravitational Field, which points TOWARD positive masses).
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