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Practice MCAT Question - Q20 - Answer!

Two point charges, A and B, with values of 2.0 x 10-4 C and 24.0 x 10-4 C, respectively, are separated by a distance of 6.0 meters. What is the magnitude of the electrostatic force exerted on point A?

A.  2.2 x 10-9 N
B.  1.3 N
C.  20 N
D.  36 N

The correct answer is (C). Coulomb’s law:
F = k(qAqB)/r2
F = (9.00 x 109 N m2/C2) (2.0 x 10-4 C) (-4.0 x 10-4 C) / (6.0 m)2
F = 720 N m2/36 m2 = 20 N

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