Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Calculate the x- and y-components of the net electric field

  1. Sep 6, 2004 #1
    Four charges q1 = q3 = -q and q2 = q4 = +q, where q = 9 µC, are fixed at the corners of a square with sides a = 1.3 m.

    (a) Calculate the x- and y-components of the net electric field at the midpoint M of the bottom side of the square.
    (b) Find the total force exerted on q4 by the charges q1, q2, and q3:
    (c) Find the force on a test charge Q = -0.4 µC placed at the midpoint M' of the top side of the square:

    I figured out that the Y-component of the net electric field at the midpoint M of the bottom side of the square is zero because the fields cancel out leaving only a field with an X component.
    To figure out the X component, I tried to use the law of superposition and add the electric fields like vectors. I used E=F/q and then E=(kQ)/r^2 on each of the charges to the left and right of the midpoint. Then, since the electric fields are moving in opposite directions, I subtracted the two fields to get zero, but apparently the X component of the field isn't zero.

    For part b, I drew the force vector diagram, so I have 3 vectors with one going left, the other going down and another going to the upper right. I don't know what to do next so any help would be appreciated here.

    For part c, it's a similar situation with part A and so the Y component is zero again. This time however, I used coulomb's law and the equation kQq/r^2. I plugged in numbers and got 0.0767. However I believed that the forces on the test charge were in the same direction and so I multiplied by 2 to get 0.1534 . This wasn't right either. Sorry for the long post and any help you can give me would be very awesome.

    Attached Files:

  2. jcsd
  3. Sep 6, 2004 #2
    For part A, the horizontal component is nonzero because one of the charges attracts while the other repels. If each charge is on opposite sides of the point under consideration, then they effectively act in the same direction. For part B, consider breaking each of these vectors down into component form and then summing them component-wise. For part C, consider first that the electric field found in part A is similar to the one acting at the upper midpoint. Now take into account the magnitude of test charge Q to find the force acting on it.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook