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
Check The Sidebar For Another Practice MCAT Question!! ----->
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
