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Electric field strength in a square.

  1. Jun 4, 2006 #1
    Hi guys,

    Suppose I have a point O in the centre of the square (of side length X), where the top 2 corners of the square carries a positive charge each and the bottom left corner a negative charge and the bottom right corner a positive charge, what would the magnitude and direction of the magnetic field strength at point O be? I figured that it would be towards M, but how do I find it's magnitude?

    The given answer is q/(pi)(Enot)(x^2). I don't realy understand why is the 4 missing?

    Thanks!
     
  2. jcsd
  3. Jun 4, 2006 #2

    siddharth

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    First find the electric field due to each point charge, and then find the vector sum.
    One more thing, you do mean electric field, don't you? Why would there be a magnetic field?
     
  4. Jun 4, 2006 #3
    Thanks for the help. I apologise I did mean electric field was thinking of some other question at the same time! :wink:
     
  5. Jun 4, 2006 #4
    Hi Siddarth

    Wodner if you could help me out further here.. after some doing I realised I couldn't get rid of the 4, but it would instead be 2.

    Here's my working, the electric field strength due to the top left and bottom right corner would be 0 as they cancel out (am I correct?). But the field strength due to the top right and bottom left corner would be 2q/4(pi)(Enot)(x^2)?

    Thanks
     
  6. Jun 4, 2006 #5

    siddharth

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    Yeah, if the charges are the same.

    Not quite, what's the distance of the center of the square O from each vertex? (It isn't x)

    Also, I don't get the given answer. Are you sure the charges are equal?
     
  7. Jun 4, 2006 #6
    Thanks Siddarth for the assistance.. I think I got the answer..

    Yes the question states that 4 equal charges except for the bottom left is a negatively charged one compared to the rest of the positively charged ones.

    Here's my complete working, I'll name the sides J, K L and M for ident purposes starting from the top left corner in a clockwise direction. So M is the point that carries the -ve charge. E field due J and L cancels each other. The (distance)^2 between K and O I calculated to be (x^2)/2 which is also the (distance)^2 between O and M. So by vector addition, E field due K and M should be in the same direction I got [q/4(pi)(Enot)(x^2)/2] + [q/4(pi)(Enot)(x^2)/2] which gives the answer to be q/(pi)(Enot0)(x^2) after cancelling out the numbers.

    Thanks once again.
     
  8. Jun 4, 2006 #7

    siddharth

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    Yeah, that's right.
     
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