Gauss's law 2 -solutions

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

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gilvin (jg47854) - 6. Gauss’s law 2 - meyers - (21235) This print-out should have 21 questions. and R is the radius of the sphere. Thus, Multiple-choice questions may continue on 1 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 µC 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 µC = 3.42 × 10 −5 C . By Gauss’ law, E = r ρ 3 ϵ 0 = (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 · dA = q ϵ 0 The enclosed charge is zero, so E = 0 N/C . E = Q 4 π r 2 ϵ 0 = 3.42 × 10 5 C 4 π (0.39 m) 2 (1) 1 × 8.85419 × 10 −12 C 2 /N · m 2 002 (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 . 4 E · 4 π r 2 = 3 π r 3 ρ = 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: ϵ 0 E = r ρ 3 ϵ 0 , where the charge density is ρ = Q 4 3 π r 3 = 3.42 × 10 5 C 4 Let : r = 68.1 cm . Using Eq. 1, E = Q 4 π r 2 ϵ 0 = 3.42 × 10 5 C 4 π (0.681 m) 2 × 1 3 π (39 cm) 3 = 0.00013764 C/m 3 8.85419 × 10 −12 C 2 /N · m 2 = 6.62786 × 10 5 N/C .
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gilvin (jg47854) - 6. Gauss’s law 2 - meyers - (21235) the center and applying Gauss’ Law, 005 (part 1 of 5) 10.0 points 2 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. E A 4 π d 2 = Q enclosed ϵ 0 = q 1 ϵ 0 , 1 q 1 E A = 4 π ϵ 0 d 2 a R P A O S r b q 1 q 2 c = k q 1 d 2 006 (part 2 of 5) 10.0 points Find the electric field at point P where the distance from the center O to P is d, such that b < d < c. 1. E P = k (q 1 + q 2 ) d 2. E P = k (q 1 + q 2 ) d 2 Find the electric field at point A, where the distance from the center O to A is d, such that a < d < b. k q 1 3. E P = 0 correct 4. E P = q 2 k d 2 5. E P = k q 1 1. E A = a 2 2. E A = q 1 k d 2 k q 1 a 2 6. E P = k q 1 a 3. E A = 4. E A = 5. E A = 6. E A = a k q 1 d k q 2 d 2 0 7. E P = k q 1 d 8. E P = q 1 k d 2 9. E P = k q 2 d 2 10. E P = k q 1 d 2 7. E A = q 2 Explanation: k d 2 Taking a Gaussian sphere of radius d from k q 1 the center and applying Gauss’ Law, 8. E A = d 2 correct 9. E A = 10. E A = k (q 1 + q 2 ) E P 4 π d 2 = Q
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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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Gauss's law 2 -solutions - gilvin (jg47854) - 6. Gausss law...

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