Department of Natural Sciences

Clayton College & State University

 

January 27, 2002

 

Physics 1112 – Quiz 2b

 

 

Name  _____SOLUTION________________________________

 

 

  1. Two point charges lie along the y axis. A charge of q1 = - 4.00 mC is at y = 6.00 m, and a charge of q2 =  3.00 mC is at y = -4.00 m.

 

    1. Find the electric field at the origin. (Specify both magnitude and direction.)

 

 

E1 = ke (q1/r12) = (8.99 x 109 N m2/C2)(4.00 x 10 –6 C) (6.00 m)2  = 990 N/C

 

E2 = ke (q2/r22) = (8.99 x 109 N m2/C2)(3.00 x 10 –6 C) (4.00 m)2  = 1680 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) = 990 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) = 1680 N/C

 

(Etot)x = (E1)x + (E2)x = 0 N/C

(Etot)y = (E1)y + (E2)y = 990 N/C + 1680 N/C = 2680N/C

 

Magnitude : 2680 N/C, directed upward

 

 

    1. Another charge q3 = - 1.00 mC is now placed at the origin. What is the magnitude of the electric force which acts on this charge?

 

F = q E

F = (1.00 x 10 –6 C) (2680 N/C) = 2.68 x 10 –3  N