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Electric potential with charges

  1. Jan 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Two charges are placed at the corners of a square. One charge, +4.0uC, is fixed to one corner and another, -6.0uC is fixed to the opposite corner. WHat charge would need to be placed at hte intersection of the diagonals of the square in order ot make the potential difference zero at each of the two unoccupied corners?

    2. Relevant equations
    V = kq/r

    3. The attempt at a solution
    Given that potential difference (or voltage) is calculated based on the formula above, can I just assume that the radius of this square is infinitely large and therefore V becomes zero and that the charge is some number that is much smaller than infinite?
    I'm not sure if I'm allowed to do this. Please help.
  2. jcsd
  3. Jan 11, 2009 #2


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    Homework Helper

    Hi canadianballa,

    No, I don't think that's what they want you to do here. Instead, let the side length of the square be some unknown distance d. If the side has length d, what is the potential from each of the three charges (4 uC, -6uC, and Q) at one of the unoccupied corners (using your equation)? (You'll need to find the distance, in terms of d, that each of the three charges is from the corner.) If the sum of those three terms has to be zero, then that will give you the unknown charge Q.
  4. Jan 11, 2009 #3
    Let the side of the sqaure be s. Place one known charge on the y axis, the other on the x axis of a coordinate system, and the unknown charge at the intersection of the square diaganols. V = Ed where E is the electric field produced by the point charge and d is the distance from the point charge. A zero potential exists when no net electric field is present. So, use the expression for determining the electric field at a square corner by the point charges of the two known and one unknown charges and solve for the unkown charge by setting the vector sum of the three electric fields equal to zero. Your answer will be in terms of s.
  5. Jan 11, 2009 #4
    Thanks guys for your help. I'll work on solving it now and post the attempt in a bit.
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