6. Gauss's law 2 -solutions

6. Gauss's law 2 -solutions - gilvin(jg47854 – 6...

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Unformatted text preview: gilvin (jg47854) – 6. Gauss’s law 2 – meyers – (21235) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This homework is due SUNDAY, February 13, at midnight Tucson time. 001 (part 1 of 4) 10.0 points A solid sphere of radius 39 cm has a total pos- itive charge of 34 . 2 μ C uniformly distributed throughout its volume. Calculate the magnitude of the electric field at the center of the sphere. Correct answer: 0 N / C. Explanation: Let : r = 39 cm and Q = 34 . 2 μ C = 3 . 42 × 10 − 5 C . By Gauss’ law, contintegraldisplay vector E · d vector A = q ǫ The enclosed charge is zero, so E = 0 N / C . 002 (part 2 of 4) 10.0 points Calculate the magnitude of the electric field 9 . 75 cm from the center of the sphere. Correct answer: 5 . 05217 × 10 5 N / C. Explanation: Let : r = 9 . 75 cm . E · 4 π r 2 = 4 3 π r 3 ρ ǫ E = r ρ 3 ǫ , where the charge density is ρ = Q 4 3 π r 3 = 3 . 42 × 10 − 5 C 4 3 π (39 cm) 3 = 0 . 00013764 C / m 3 and R is the radius of the sphere. Thus, E = r ρ 3 ǫ = (0 . 0975 m) (0 . 00013764 C / m 3 ) 3 (8 . 85419 × 10 − 12 C 2 / N · m 2 ) = 5 . 05217 × 10 5 N / C . 003 (part 3 of 4) 10.0 points Calculate the magnitude of the electric field 39 cm from the center of the sphere. Correct answer: 2 . 02087 × 10 6 N / C. Explanation: Let : r = 39 cm . The field outside the sphere is the same as that for a point charge. Thus E = Q 4 π r 2 ǫ (1) = 3 . 42 × 10 − 5 C 4 π (0 . 39 m) 2 × 1 8 . 85419 × 10 − 12 C 2 / N · m 2 = 2 . 02087 × 10 6 N / C . 004 (part 4 of 4) 10.0 points Calculate the magnitude of the electric field 68 . 1 cm from the center of the sphere. Correct answer: 6 . 62786 × 10 5 N / C. Explanation: Let : r = 68 . 1 cm . Using Eq. 1, E = Q 4 π r 2 ǫ = 3 . 42 × 10 − 5 C 4 π (0 . 681 m) 2 × 1 8 . 85419 × 10 − 12 C 2 / N · m 2 = 6 . 62786 × 10 5 N / C . gilvin (jg47854) – 6. Gauss’s law 2 – meyers – (21235) 2 005 (part 1 of 5) 10.0 points A solid conducting sphere of radius a is placed inside of a conducting shell which has an inner radius b and an outer radius c . There is a charge q 1 on the sphere and a charge q 2 on the shell. O A P R S a b c r q 1 q 2 Find the electric field at point A , where the distance from the center O to A is d , such that a < d < b . 1. E A = k q 1 a 2 2. E A = q 1 k d 2 3. E A = k q 1 a 4. E A = k q 1 d 5. E A = k q 2 d 2 6. E A = 0 7. E A = q 2 k d 2 8. E A = k q 1 d 2 correct 9. E A = k ( q 1 + q 2 ) d 2 10. E A = k ( q 1 + q 2 ) d Explanation: Taking a Gaussian sphere of radius d from the center and applying Gauss’ Law, E A 4 π d 2 = Q enclosed ǫ = q 1 ǫ , E A = 1 4 π ǫ q 1 d 2 = k q 1 d 2 ....
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This note was uploaded on 04/29/2011 for the course PHYS 202 taught by Professor Doc during the Spring '11 term at North Texas.

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6. Gauss's law 2 -solutions - gilvin(jg47854 – 6...

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