oldhw9solution_pdf - klarin (sjk772) oldhomework 09 Turner...

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Unformatted text preview: 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 C / m 3 . (The volume of a cylinder with radius r and length is V = 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 : = 2 C / m 3 = 2 10- 6 C / m 3 , r = 1 . 9 cm = 0 . 019 m , and = 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 s = 2 r E , and the charge enclosed by the surface is Q enc = r 2 . Using Gauss law, s = Q enc 2 r E = r 2 . Thus E = r 2 = ( 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 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 V =- integraldisplay B A vector E vector ds =- integraldisplay R r 1 E dr , since E is radial. 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 | 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 = V =- 1 2 C ( R 2- r 2 1 ) < since C > 0 and R > r 1 ....
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oldhw9solution_pdf - klarin (sjk772) oldhomework 09 Turner...

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