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Posted by on Monday, September 3, 2007 at 12:40pm.

A charge of +2q is fixed to one corner of a square, while a charge of -3q is fixed to the diagonally opposite corner. Expressed in terms of q, what charge should be fixed to the center of the square, so the potential is zero at each of the two empty corners?

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Physics - bobpursley, Sunday, September 2, 2007 at 11:10am
Potential is a scalar, so you can just add all four contributions.

Vt= -3q/r + 2q/r + 2X/r=0

solve for charge X

when i use the formula i can up with +1q and the system is saying that the answer is incorrect.

  • Physics please clarify - , Monday, September 3, 2007 at 1:17pm

    The equation you should be solving is
    2q/a -3q/a + (sqrt 2)x/a = 0

    Where a is the length of a side of the square. The corners with charges are a distance "a" away from the empty corners, but the center charge x is closer.

    Multiplying bith sides by a gives
    -q + sqrt 2 * x = 0
    x = 0.707 q

  • Physics please clarify - , Monday, September 3, 2007 at 4:39pm

    Can you please explain how you get to x=0.707q.

    I got how you got up to -q+sgrt2*X=0 but i keep coming up with a different answer than 0.707q.

  • Physics please clarify - , Thursday, January 22, 2009 at 10:27pm

    solve the equation on paper to make it clear. you will end up doing -1/-(sqrt 2)

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