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Coulomb's Law

  1. Jan 14, 2009 #1
    1. The problem statement, all variables and given/known data
    http://img382.imageshack.us/img382/4100/coulombko9.jpg [Broken]

    2. Relevant equations
    In the place marked X what is the direction of the electric field?

    3. The attempt at a solution

    My solution says that it's upward to the right, am I right?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 15, 2009 #2


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    I take it this is a square?

    Figure first what is the field from the 2 negative charges. By symmetry this will be directed along the diagonal won't it? What is the magnitude then?

    Then figure the field from the larger positive charge which is also along the same diagonal.

    Which magnitude is larger?
  4. Jan 15, 2009 #3
    how do you calculate the magnitude of both?
  5. Jan 15, 2009 #4


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    As LowlyPion said, you have to assume it's a square. Denote the sides by say, d. Now you should be able to calculate the magnitude of the E field strength in terms of d using Coulomb's law.
  6. Jan 15, 2009 #5


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    That only makes me wonder how you arrived at your solution.
    Or was that just a guess? You weren't using Coulomb's Law?
  7. Jan 15, 2009 #6
    Well..I know that Coulomb's Law is F = kq1q2/r^2, am I right? The Coulombs law is only used to find magnitude but not directions...
  8. Jan 15, 2009 #7


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    If there is a force, there is a direction. You've stated the scalar version of the law.

    The vector form ...
  9. Jan 17, 2009 #8
    hmm.. I still haven't solved this problem.. I assume that this problem can be solved by simulating of putting a positive charge on the X location and see where the electric field goes.. am I right? one more thing, why would I want to find the magnitude? The question only asks where the direction is at the position marked X
  10. Jan 17, 2009 #9
    Oops nevermind. I forgot that the diagonal does complicate things a tiny bit.

    X is a positive test charge. Find the magnitude of the diagonal charge and find its components.

    Then just set up force equations in the x and y directions.
    Last edited: Jan 17, 2009
  11. Jan 17, 2009 #10


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    You would want to calculate the magnitude to be certain which direction it is, positive or negative along the diagonal, which through symmetry is what you would easily expect.

    Whether you determine the magnitude as a force with a unit test charge or directly as an E field vector it amounts to the same thing.

    Consider that you have three vectors.

    1 = -kq/a² x-hat
    2 = -kq/a² y-hat
    3 = +2kq/(a*√2)²*cos45 x-hat + 2kq/(a*√2)²*sin45 y-hat

    Now what do the vectors sum to?
  12. Jan 17, 2009 #11
    I assume that the answer is negative of the diagonal
  13. Jan 17, 2009 #12


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    Which direction is negative on the diagonal?
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