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Posted by Mary 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?

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 - drwls, 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 - Mary, 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.

I misread the problem, thinking you wanted the potential zero at the center, that is why all my distances were the same.

Here, the potential at the two corners is zero. Let s be the side length..so the diagonal length from the corner to the center is s*1/sqrt2
0= -3q/s + +2q/s+ X/(s*1/sqrt2)

then q/sqrt2=X the charge at each corner, or X=.707q at each corner

The answer 0.707q is correct but I am still not understanding how to get to that answer.

I have another problem just like this one but with different values for q. When I use the formula above I am coming up with an incorrect answer. Please help. My question is:

A charge of +1q 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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