Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Conceptual Questions 17.1, 17.4, 17.6
A –12 µC source charge is held at the origin, and a +2.0 µC test charge is at x = +4.0 cm. As it is moved from x = +4.0 cm to x = +5.0 cm, the +2.0 µC test charge moves to a region of __________ potential, while its potential energy __________. (A) lower; decreases.
(B) lower; increases.
(C) higher; decreases.
(D) higher; increases.
Correct answer (highlight to unhide): (D)
The potential of the –12 µC source charge is given by:
V = k·(–12×-6 C)/r,
so increasing r (relocating from x = +4.0 cm to +5.0 cm) would make the potential a smaller negative number, which means moving to a region of higher potential.
The electrical potential energy of the two charges is given by:
EPE = k·(-12×10–6 C)·(+2.0×10–6 C)/r,
so increasing r (moving the +2.0 µC test charge from x = +4.0 cm to +5.0 cm) would make the electric potential energy a smaller negative number, as well, which means it increases.
Student responses
Section 70856
(A) : 2 students
(B) : 5 students
(C) : 0 students
(D) : 4 students
Success level: 36%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.67
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