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Unformatted text preview: Arb, Kellen – Midterm 1 – Due: Sep 14 2004, 10:00 pm – Inst: Turner 1 This printout should have 25 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 3) 10 points Consider a long, uniformly charged, cylindri cal insulator of radius R and charge density 1 . 4 μ C / m 3 . (The volume of a cylinder with radius r and length ‘ is V = π r 2 ‘ .) R 1 . 1 cm What is the electric field inside the insulator at a distance 1 . 1 cm from the axis (1 . 1 cm < R )? Correct answer: 869 . 645 N / C. Explanation: Given : ρ = 1 . 4 μ C / m 3 = 1 . 4 × 10 6 C / m 3 , r = 1 . 1 cm = 0 . 011 m , and ² = 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 ‘ E , and the charge enclosed by the surface is Q enc = π r 2 ‘ ρ . Using Gauss’ law, Φ s = Q enc ² 2 π r ‘ E = π r 2 ‘ ρ ² . Thus E = ρ r 2 ² = ( 1 . 4 × 10 6 C / m 3 ) (0 . 011 m) 2(8 . 85419 × 10 12 C 2 / N / m 2 ) = 869 . 645 N / C . 002 (part 2 of 3) 10 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 ) 2.  V  = C µ 1 r 2 1 1 R 2 ¶ 3.  V  = C r 1 2 4.  V  = C q R 2 r 2 1 5.  V  = 1 2 C ( R 2 r 2 1 ) correct 6.  V  = C ( R 2 r 2 1 ) 7.  V  = 1 2 C ( R r 1 ) r 1 8.  V  = C µ 1 r 1 1 R ¶ 9.  V  = C ( R r 1 ) r 1 Arb, Kellen – Midterm 1 – Due: Sep 14 2004, 10:00 pm – Inst: Turner 2 10.  V  = C 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 = Z B A ~ E · ~ ds = Z R r 1 E dr , since E is radial. Δ V = Z R r 1 C r dr = C r 2 2 fl fl fl fl R r 1 = C µ R 2 2 r 2 1 2 ¶ . The absolute value of the potential differ ence is  Δ V  = C µ R 2 2 r 2 1 2 ¶ = 1 2 C ( R 2 r 2 1 ) . 003 (part 3 of 3) 10 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 . Thus V B < V A and the potential is higher at point A where r = r 1 than at point B, where r = R ....
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This note was uploaded on 10/06/2008 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Physics

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