Department of Natural Sciences
Clayton College & State University
January 27, 2002
Physics 1112 – Quiz 2a
Name ___SOLUTION__________________________________

E1 = ke (q1/r12) = (8.99 x 109 N m2/C2)(9.00 x 10 –6 C) (6.00 m)2 = 2250 N/C
E2 = ke (q2/r22) = (8.99 x 109 N m2/C2)(8.00 x 10 –6 C) (4.00 m)2 = 4500 N/C
(E1)x = E1 cos (q1) = (2250 N/C) cos (90o) = 0 N/C
(E1)y = E1 sin (q1) = (2250 N/C) sin (90o) = 2250 N/C
(E2)x = E2 cos (q2) = (4500 N/C) cos (- 90o) = 0 N/C
(E2)y = E2 sin (q2) = (4500 N/C) sin (- 90o) = -4500 N/C
(Etot)x = (E1)x + (E2)x = 0 N/C
(Etot)y = (E1)y + (E2)y = 2250 N/C – 4500 N/C = -2250 N/C
Magnitude : 2250 N/C, directed
downward
F = q E
F = (1.00 x 10 –6 C) (2250 N/C) = 2.25 x 10 –3 N