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oldhw9solution_pdf

# oldhw9solution_pdf - klarin(sjk772 oldhomework 09...

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klarin (sjk772) – oldhomework 09 – Turner – (56705) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find 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 2 μ C / m 3 . (The volume of a cylinder with radius r and length is V = π r 2 .) R 1 . 9 cm What is the electric field inside the insulator at a distance 1 . 9 cm from the axis (1 . 9 cm < R )? Correct answer: 2145 . 88 N / C. Explanation: Given : ρ = 2 μ C / m 3 = 2 × 10 - 6 C / m 3 , r = 1 . 9 cm = 0 . 019 m , and ǫ 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 field parallel to the axis is negligible. r R The flux leaving the ends of the Gaussian cylinder is negligible, and the only contribu- tion to the flux is from the side of the cylinder. Since the field is perpendicular to this surface, the flux is Φ s = 2 π r ℓ E , and the charge enclosed by the surface is Q enc = π r 2 ℓ ρ . Using Gauss’ law, Φ s = Q enc ǫ 0 2 π r ℓ E = π r 2 ℓ ρ ǫ 0 . Thus E = ρ r 2 ǫ 0 = ( 2 × 10 - 6 C / m 3 ) (0 . 019 m) 2 (8 . 85419 × 10 - 12 C 2 / N / m 2 ) = 2145 . 88 N / C . 002 (part 2 of 3) 10.0 points Determine the absolute value of the potential difference between r 1 and R , where r 1 < R . (For r < R the electric field takes the form E = C r , where C is positive.) 1. | V | = C ( R - r 1 ) r 1 2. | V | = C parenleftbigg 1 r 1 - 1 R parenrightbigg 3. | V | = C radicalBig R 2 - r 2 1 4. | V | = 1 2 C ( R 2 - r 2 1 ) correct 5. | V | = C parenleftbigg 1 r 2 1 - 1 R 2 parenrightbigg 6. | V | = C ( R 2 - r 2 1 ) 7. | V | = C r 1 8. | V | = C r 1 2 9. | V | = 1 2 C ( R - r 1 ) r 1

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klarin (sjk772) – oldhomework 09 – Turner – (56705) 2 10. | V | = C ( R - r 1 ) Explanation: The potential difference 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 = - integraldisplay B A vector E · vector ds = - integraldisplay R r 1 E dr , since E is radial. Δ V = - integraldisplay R r 1 C r dr = - C r 2 2 vextendsingle vextendsingle vextendsingle vextendsingle R r 1 = - C parenleftbigg R 2 2 - r 2 1 2 parenrightbigg . The absolute value of the potential differ- ence is | Δ V | = C parenleftbigg R 2 2 - r 2 1 2 parenrightbigg = 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 correct 2. None of these 3. V r 1 = V R 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 C > 0 and R > r 1 . Thus V B < V 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.
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oldhw9solution_pdf - klarin(sjk772 oldhomework 09...

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