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Unformatted text preview: homework 08 – KANUNGO, ANIL – Due: Sep 20 2007, 4:00 am 1 Question 1, chap 25, sect 2. part 1 of 3 10 points Consider a long, uniformly charged, cylin drical insulator of radius R and charge density 1 . 4 μ C / m 3 . (The volume of a cylinder with radius r and length ℓ is V = π r 2 ℓ .) R 2 . 6 cm What is the electric field inside the insulator at a distance 2 . 6 cm from the axis (2 . 6 cm < R )? Correct answer: 2055 . 52 N / C (tolerance ± 1 %). Explanation: Given : ρ = 1 . 4 μ C / m 3 = 1 . 4 × 10 6 C / m 3 , r = 2 . 6 cm = 0 . 026 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 . 4 × 10 6 C / m 3 ) (0 . 026 m) 2 (8 . 85419 × 10 12 C 2 / N / m 2 ) = 2055 . 52 N / C . Question 2, chap 25, sect 2. part 2 of 3 10 points Determine the absolute value of the po tential 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 parenleftbigg 1 r 2 1 − 1 R 2 parenrightbigg 2.  V  = C ( R 2 − r 2 1 ) 3.  V  = 1 2 C ( R 2 − r 2 1 ) correct 4.  V  = C r 1 5.  V  = C r 1 2 6.  V  = C ( R − r 1 ) 7.  V  = C parenleftbigg 1 r 1 − 1 R parenrightbigg 8.  V  = C ( R − r 1 ) r 1 9.  V  = 1 2 C ( R − r 1 ) r 1 homework 08 – KANUNGO, ANIL – Due: Sep 20 2007, 4:00 am 2 10.  V  = C radicalBig R 2 − r 2 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 ) . Question 3, chap 25, sect 2. part 3 of 3 10 points What is the relationship between the po tentials 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 ) < since C > 0 and R > r 1 ....
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This homework help was uploaded on 04/02/2008 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
 Charge, Work

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