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4 Charges on a Square, E Field

  1. Sep 19, 2008 #1
    I've been a reader of this forum last year, but now I've registered and need help understanding electric field.

    1. The problem statement, all variables and given/known data
    Four charges q1 = q3 = -q and q2 = q4 = +q, where q = 7 µC, are fixed at the corners of a square with sides a = 1 m.
    https://online-s.physics.uiuc.edu/c...ourses/phys212/fall08/homework/01/01/cl10.gif
    So 4 charges are on the corners of a square. Top-left and lower-right are -q, and Top-right and lower-left are +q. I need to calculate the x-component of the electric field at the bottom of the square in the middle at point M. The sides of the square are 1m.


    2. Relevant equations
    Gauss' Law


    3. The attempt at a solution
    I attempted to draw where the electric field points and it is to the left. When I attempt the problem though everything cancels out and leaves me with zero.

    The two bottom charges should cancel out (?) leaving two charges at the top. Pyth. thereom finds the radius to the top two corners sq(1.25)=d

    Ex= cos(45)(K)(+Q)/d^2+cos(45)(K)(-q)/d^2 = 0
    What am I doing wrong? This is an old homework question with no chance of getting points, but I'm trying to understand it to prepare for an upcoming test. I'm also kind of confused about Gauss' Law. Do you pretend an electron is at the bottom of the square, calculate the force, then divide by q? The answer also asks for N/C while other problems ask for N. Can anyone give me any general units of the electric field? Any help would be appreciated.
    1. The problem statement, all variables and given/known data
     
  2. jcsd
  3. Sep 19, 2008 #2

    Redbelly98

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    Not so. Think carefully about the direction of the E field due to:
    1. The negative charge that is to the right of the point, and
    2. The positive charge that is to the left of the point.
     
  4. Sep 20, 2008 #3
    The electric field due to the left charge would point right and the field from the right charge would also point right? How would you go about calculating that? When I try to get an answer I always get zero still.
     
  5. Sep 20, 2008 #4

    Redbelly98

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    Since they both point to the right, just add up the two numbers. The direction will still be to the right.
    Since they are both positive numbers, you will not get zero when you add them up.

    (Then there's still the two charges at the top to consider...)
     
  6. Sep 20, 2008 #5
    Thanks a lot for pointing me in the right direction Redbelly98, things finally clicked. Now let's go celebrate a Cubs win!
     
  7. Sep 20, 2008 #6

    Redbelly98

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    You're welcome! And welcome to PF.
     
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