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Problem 1: Four charges lie at the corners of a square as shown, where the distance from each vertex to the center of the square is a. Three of the charges are positive (+q) and one of the charges is negative (-q). A fifth charge Q is located at the center of the square. Find the force (magnitude and direction) on the charge Q. +q 9

My solution so far:

E = kQ/r2

E = kQ/a2 (^r)

I think tan(45*) should be used, so it should be...

E = kQ/a2 (tan(45*))

Thank you.

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Answer #1

Forces due to two positive q charges cancel out each other. The remaining force is due to +q and-q which are at opposite corners.These two forces add up.

Final F= 2*kqQ/a2

Direction is along the diagonal from +q towards -q charge along the arrow drawn in the figure.

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