Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Multiple-Choice Question 17.5, Problems 17.57, 17.58
[10 points.] A parallel plate capacitor is connected to a battery that supplies a constant potential difference. While the battery is still attached, the parallel plates are separated a little more. As the parallel plates are separated, will the charge increase, decrease, or remain constant? Explain your reasoning using the properties of capacitors.
Solution and grading rubric:
- p = 10/10:
Correct. As capacitance depends inversely on plate separation distance, increasing d will decrease C. Since the capacitor is still connected to a battery, the potential difference will remain constant, and due to C = Q/delta(V), as C decreases so will the charge Q. - r = 8/10:
As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. At least has complete/correct explanation for why C will decrease. - t = 6/10:
Nearly correct, but argument has conceptual errors, or is incomplete. - v = 4/10:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. - x = 2/10:
Implementation/application of ideas, but credit given for effort rather than merit. - y = 1/10:
Irrelevant discussion/effectively blank. - z = 0/10:
Blank.
Grading distribution:
Section 31988
p: 4 students
r: 4 students
t: 0 students
v: 5 students
x: 0 students
y: 0 students
z: 0 students
A sample "p" response (from student 1930):
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