Question

Four point charges, each of magnitude 8.27

Four point charges, each of magnitude
8.27

Homework Answers

Answer #1

Apply F = kqq/r2 for each charge pair...

For the charges directly to the side and below the negative charge, we get...

F = (8.98755 X 109)(8.27 X 10-6)2/(.111)2

F = 49.9 N

Then for the charge diagonally across, we need the distance from the Pythagorean Theorem...

r2 = (.111)2 + (.111)2

r = .1569777 m

The the force = (8.98755 X 109)(8.27 X 10-6)2/(.1569777)2

F = 24.9 N

We need the x and y components of this, so multiply by cos(45)

F = 17.64 N

That means there is a net horizontal and vertical force of 17.64 + 49.9 = 67.5 N

The overall force, then, is by the Pythagorean Theorem

F2 = (67.5)2 + (67.5)2

F = 95.5 N

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