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# PHY6 - Howard(clh2528 homework#6 shubeita(57885 This...

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Howard (clh2528) – homework #6 – shubeita – (57885) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of6)10.0points A positive test charge is placed at the center of a spherical Gaussian surface. What happens to the net flux through the Gaussian surface for each of the following cases? (a) The surface is replaced by a cube of the same volume whose center is at the same point. 1. the net flux is zero 2. the net flux decreases but is nonvanishing 3. no change correct 4. the net flux increases Explanation: BasicConcepts: Φ S = contintegraldisplay S vector E · d vector A = Q ǫ 0 Solution: As long as the charge remains the same, and it is enclosed by any Gaussian surface, the net flux will remain the same. 002(part2of6)10.0points (b) The sphere is replaced by a cube of one- third the volume centered at the same point. 1. the net flux increases 2. the net flux is zero 3. no change correct 4. the net flux decreases but is nonvanishing Explanation: See part 1. 003(part3of6)10.0points (c) The charge is moved off center in the orig- inal sphere, but remains within the sphere. 1. the net flux increases 2. the net flux decreases but is nonvanishing 3. no change correct 4. the net flux is zero Explanation: See part 1. 004(part4of6)10.0points (d) The charge is moved just outside the orig- inal sphere. 1. the net flux decreases but is nonvanishing 2. the net flux increases 3. no change 4. the net flux is zero correct Explanation: If the charge is moved outside the Gaussian surface, then the amount of enclosed charge is zero, and the net flux is zero. One can imagine the field lines going out from the charge. Any line entering the closed Gaussian surface will also leave, so the net flux is zero. 005(part5of6)10.0points (e) The positive test charge is again placed at the center of a spherical Gaussian surface (as in part a). When a second positive charge is placed near, but outside, the original sphere. 1. flux is zero 2. flux decreases but is nonvanishing 3. flux increases 4. no change correct Explanation: From Part 4 it is clear that an exterior charge does not contribute to the flux, so the flux remains the same.

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Howard (clh2528) – homework #6 – shubeita – (57885) 2 006(part6of6)10.0points (f) A second positive charge is placed inside the original spherical Gaussian surface. 1. flux increases correct 2. flux is zero 3. no change 4. flux decreases but is nonvanishing Explanation: Since the net enclosed charge increases, the flux must increase.
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PHY6 - Howard(clh2528 homework#6 shubeita(57885 This...

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