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Homework Help: Electric field with three charges

  1. Sep 2, 2006 #1
    Here is the problem which I need to solve:

    The following picture shows a system consisting of three charges, q1 = +4.07 μC, q2 = +4.07 μC, and q3 = -4.07 μC, at the vertices of an equilateral triangle of side d = 2.29 cm. Find the magnitude of the electric field at a point halfway between the charges q1 and q2. Also find the magnitude of the electric field at the point halfway between the charges q2 and q3.


    I pretty much already got down that I have to use x and y components, and got that E1 and E2 cancel each other out (using the equation E = q/r^2). I just pretty much need to figure out how to get the x and y components for E3, and I think I'll be able to finish up the problem on my own. Thanks.
  2. jcsd
  3. Sep 3, 2006 #2


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    Yeah, you're right on track. Can you figure out the angle so that you can split E3 into x & y components? (Hint: You know it's an equilateral triangle)
    Last edited: Sep 3, 2006
  4. Sep 3, 2006 #3
    This is one thing I was stuck at...60 degrees or 30 degrees (to cut the angle in half to go directly in between E2 and E3)? I'm leaning toward 30 degrees, am I right?
  5. Sep 3, 2006 #4
    Hi flipstyle,
    you have an equilateral triangle, so if you trace a line which joint q3 and the midpoint between q1 q2, the lengh of this segment will be d√3/2.
    A test charge placed at halfway between q1 and q2 will be affected only by q3 since the influence of q1 and q2 cancel each other. So the electric field will be along the line joining the midpoint and q3.
  6. Sep 3, 2006 #5
    a test charge halfway between q2 and q3,call this point M, will experience a repulsion from q2, attraction from q3 with the same intensity. so the electric field at M will have a component along the line joining M and q3.

    But the same test charge at M will also experience a repulsion from q1 that you can calculate the same maner as you have done for the first question. This repulsive force will be along the line q1 M.
    This gives you to perpendicular component of the electric force at M.
    From this it should be easy for you to continue.
  7. Sep 3, 2006 #6
    Thanks for all the help/tips...I figured it out.
    Last edited: Sep 3, 2006
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