Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 16.15
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Solution and grading rubric:
- p:
Correct. Determines that:- the electric field E1 at x = +0.10 m due to the Q1 = 4.0 μC charge at x = +0.20 m points to the right, but is more than the total electric field Etot at x = +0.10 m (that points to the right), thus the electric field E2 at x = +0.10 m due to the Q2 charge at the origin must point to the left, and so the charge Q2 must be negative as well;
- from setting up vector superposition of the two oppositely directed electric fields E1 and E2 at x = +0.10 m, that the Q2 charge at the origin must be –3.8 μC.
- r:
Nearly correct, but includes minor math errors. - t:
Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least recognizes that there is superposition of two electric fields at x = +0.10 m that result in the given Etot there. - v:
Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Misinterprets Etot at x = +0.10 m m to be the electric force exerted by Q1 on Q2, and then solves for Q1 from Coulomb's law. - x:
Implementation of ideas, but credit given for effort rather than merit. No clear attempt at applying electric forces, fields, and vector superposition. - y:
Irrelevant discussion/effectively blank. - z:
Blank.
Sections 30882, 30883
Exam code: finalLd0c
p: 3 students
r: 2 students
t: 8 students
v: 19 students
x: 4 student
y: 3 students
z: 4 student
A sample "p" response (from student 8282):
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A sample "v" response (from student 9224):
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