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oldhomework 09-solutions

# oldhomework 09-solutions - howard(cah3459 oldhomework 09...

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howard (cah3459) – oldhomework 09 – Turner – (56705) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 3) 10.0 points Consider a long, uniFormly charged, cylindri- cal insulator oF radius R and charge density 1 . 1 μ C / m 3 . (The volume oF a cylinder with radius r and length ! is V = π r 2 ! .) R 1 cm What is the electric feld inside the insulator at a distance 1 cm From the axis (1 cm <R )? Correct answer: 621 . 175 N / C. Explanation: Given : ρ =1 . 1 μ C / m 3 . 1 × 10 - 6 C / m 3 , r = 1 cm = 0 . 01 m , and 0 =8 . 85419 × 10 - 12 C 2 / N / m 2 . Consider a cylindrical Gaussian surFace oF radius r and length ! much less than the length oF the insulator so that the compo- nent oF the electric feld parallel to the axis is negligible. ! r R The ±ux leaving the ends oF the Gaussian cylinder is negligible, and the only contribu- tion to the ±ux is From the side oF the cylinder. Since the feld is perpendicular to this surFace, the ±ux is Φ s =2 π r ! E , and the charge enclosed by the surFace is Q enc = 2 !ρ . Using Gauss’ law, Φ s = Q enc 0 2 π r ! E = 2 0 . Thus E = ρ r 2 0 = ( 1 . 1 × 10 - 6 C / m 3 ) (0 . 01 m) 2 (8 . 85419 × 10 - 12 C 2 / N / m 2 ) = 621 . 175 N / C . 002 (part 2 oF 3) 10.0 points Determine the absolute value oF the potential di²erence between r 1 and R , where r 1 . (³or r < R the electric feld takes the Form E = Cr , where C is positive.) 1. | V | = C ± 1 r 1 - 1 R ² 2. | V | = C ± 1 r 2 1 - 1 R 2 ² 3. | V | = C ³ R 2 - r 2 1 4. | V | = C ( R 2 - r 2 1 ) 5. | V | = 1 6. | V | = C ( R - r 1 ) r 1 7. | V | = C ( R - r 1 ) 8. | V | = 1 2 C ( R 2 - r 2 1 ) correct 9. | V | = C r 1 2

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howard (cah3459) – oldhomework 09 – Turner – (56705) 2 10. | V | = 1 2 C ( R - r 1 ) r 1 Explanation: The potential diference between a point A inside the cylinder a distance r 1 From the axis to a point B a distance R From the axis is Δ V = - ± B A " E · " ds = - ± R r 1 E dr , since E is radial. Δ V = - ± R r 1 C r dr = - C r 2 2 ² ² ² ² R r 1 = - C ³ R 2 2 - r 2 1 2 ´ . The absolute value oF the potential difer- ence is | Δ V | = C ³ R 2 2 - r 2 1 2 ´ = 1 2 C ( R 2 - r 2 1 ) . 003 (part 3 oF 3) 10.0 points What is the relationship between the poten- tials V r 1 and V R ? 1. V r 1 <V R 2. None oF these 3. V r 1 >V R correct 4. V r 1 = V R Explanation: Since C> 0 and R > r 1 , From Part 2, V B - V A = Δ V = - 1 2 C ( R 2 - r 2 1 ) < 0 since 0 and R > r 1 . Thus V B A and the potential is higher at point A where r = r 1 than at point B, where r = R. Intuitive Reasoning: The natural ten- dency For a positive charge is to move From A to B, so A has a higher potential. 004 (part 1 oF 3) 10.0 points Consider two concentric spherical conducting shells. The smaller shell has radius a =0 .
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oldhomework 09-solutions - howard(cah3459 oldhomework 09...

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