oldhw8_pdf - mcconnell(kam2342 – oldhomework 08 –...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: mcconnell (kam2342) – oldhomework 08 – Turner – (60230) 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 1 . 9 μ C / m 3 . (The volume of a cylinder with radius r and length ℓ is V = π r 2 ℓ .) R 1 . 8 cm What is the electric field inside the insulator at a distance 1 . 8 cm from the axis (1 . 8 cm < R )? Correct answer: 1931 . 29 N / C. Explanation: Given : ρ = 1 . 9 μ C / m 3 = 1 . 9 × 10- 6 C / m 3 , r = 1 . 8 cm = 0 . 018 m , and ǫ = 8 . 85419 × 10- 12 C 2 / N / m 2 . Consider a cylindrical Gaussian surface of radius r and length ℓ much less than the length of the insulator so that the compo- nent of the electric field parallel to the axis is negligible. ℓ 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 ℓE , and the charge enclosed by the surface is Q enc = π r 2 ℓρ. Using Gauss’ law, Φ s = Q enc ǫ 2 π r ℓE = π r 2 ℓρ ǫ . Thus E = ρr 2 ǫ = ( 1 . 9 × 10- 6 C / m 3 ) (0 . 018 m) 2 (8 . 85419 × 10- 12 C 2 / N / m 2 ) = 1931 . 29 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 ( R 2- r 2 1 ) 3. | V | = C radicalBig R 2- r 2 1 4. | V | = 1 2 C ( R 2- r 2 1 ) correct 5. | V | = C r 1 6. | V | = C parenleftbigg 1 r 2 1- 1 R 2 parenrightbigg 7. | V | = 1 2 C ( R- r 1 ) r 1 8. | V | = C r 1 2 9. | V | = C parenleftbigg 1 r 1- 1 R parenrightbigg mcconnell (kam2342) – oldhomework 08 – Turner – (60230) 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. None of these 2. V r 1 = V R 3. V r 1 < V R 4. V r 1 > V R correct 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 ) < since C > 0 and R > r 1 ....
View Full Document

This note was uploaded on 11/22/2010 for the course PHYS 303 taught by Professor Turner during the Spring '10 term at University of Texas.

Page1 / 7

oldhw8_pdf - mcconnell(kam2342 – oldhomework 08 –...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online