Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 16.33
Two point charges are held at fixed locations. A +2.0 nC charge is at the origin, and a second –1.0 nC charge is at x = +0.30 m. What is the magnitude of the electric field at x = +0.40 m? (A) 1.1e2 N/C.
(B) 7.9e2 N/C.
(C) 9.0e2 N/C.
(D) 1.0e3 N/C.
Correct answer: (B)
The electric field at x = +0.40 m due to the +2.0 nC charge is directed to the right, and has a magnitude of:
k*|Q_1|/(r_1)^2 = 1.1e2 N/C.
where r_1 = 0.40 m.
The electric field at x = +0.40 m due to the -1.5 nC charge is directed to the left, and has a magnitude of:
k*|Q_2|/(r_2)^2 = 9.0e2 N/C.
where r_2 = 0.10 m.
Thus the total electric field at x = +0.40 m will be the difference of these two magnitudes, as this would be the result of the superposition of two vectors opposite in direction:
k*|Q_2|/(r_2)^2 - k*|Q_1|/(r_1)^2 = 9.0e2 N/C - 1.1e2 N/C = 7.9e2 N/C,
and the direction of the total electric field at x = +0.40 m will point to the right.
Response (A) is k*|Q_1|/(r_1)^2; response (C) is k*|Q_2|/(r_2)^2; response (D) is k*|Q_1|/(r_1)^2 + k*|Q_2|/(r_2)^2.
Student responses
Section 70856
(A) : 3 students
(B) : 7 students
(C) : 1 student
(D) : 0 students
Success level: 63%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.50
No comments:
Post a Comment