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Find net electrostatic force

  1. Jun 29, 2009 #1
    There are four charges, each with a magnitude of 1.98 μC. Two are positive and two are negative. The charges are fixed to the corners of a 0.288-m square, one to a corner, in such a way that the net force on any charge is directed toward the center of the square. Calculate the magnitude of the net electrostatic force experienced by any charge



    F=k q1*q2/ r^2 --> q1&q2 are absolute values.
    k=8.99*10^9

    So I used the formula above and said F= (8.99*10^9)*(1.98*10^-6)*(1.98*10^-6)/ .288^2
    Then I took that answer and multipled by 2 since it was the same force on the other side. Where did I go wrong?
     
  2. jcsd
  3. Jun 29, 2009 #2

    LowlyPion

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    Draw a careful picture.

    For any charge there will necessarily be 2 that attract and 1 that repels.

    The ones that are attractive must be on the closest corners of the square and the one that repels is on the diagonal. The force is along the diagonal as described, but you must sum the q*E of the 3 E-field components, keeping in mind that the closest ones are perpendicular to each other in their effect on the charge you are calculating.
     
  4. Jun 29, 2009 #3
    It says I have two negative and two positive. I think your post implied that I only have 3 charges instead of 4.
     
  5. Jun 29, 2009 #4

    LowlyPion

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    No, I thought I was implying that there are 3 charges exerting forces on the 1 charge.

    3 + 1 = 4

    The closest 2 are at right angles to the charge you are calculating the force for, whichever it may be.

    These forces are vectors, and need to be added as vectors, which your attempted solution apparently fails to take into account.
     
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