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Unformatted text preview: homework 07 KANUNGO, ANIL Due: Sep 18 2007, 4:00 am 1 Question 1, chap 24, sect 3. part 1 of 4 10 points A long coaxial cable consists of an inner cylindrical conductor with radius R 1 and an outer cylindrical conductor shell with inner radius R 2 and outer radius R 3 as shown. The cable extends out perpendicular to the plane shown. The charge per unit length along the cable on the inner conductor is and the corresponding charge per unit length on the outer conductor is (same magnitudes but with opposite signs) and &gt; 0. Q R 1 R 2 R 3 b Q Find the magnitude of the electric field at the point a distance r 1 from the axis of the inner conductor, where R 1 &lt; r 1 &lt; R 2 . 1. E = 2 r 1 correct 2. E = R 1 3 r 1 2 3. E = 0 4. E = 2 R 1 5. E = 3 r 1 6. E = R 1 4 r 1 2 7. None of these. 8. E = 2 r 1 9. E = 2 3 r 1 10. E = 2 R 1 4 r 1 2 Explanation: Pick a cylindrical Gaussian surface with radius r 1 and apply Gausss law: E dA = E (2 r 1 ) = Q E = 2 r 1 . Question 2, chap 24, sect 3. part 2 of 4 10 points The electric field vector points 1. in the negative r direction. 2. in the positive r direction. correct Explanation: The field points from positive charge to negative change. Since the center conductor is negatively charged, the electric field vector points in the negative r direction. Question 3, chap 24, sect 3. part 3 of 4 10 points Find the magnitude of the electric field at the point a distance r 2 from the axis of the inner conductor, where R 3 &lt; r 2 . 1. E = 2 r 2 2. E = 2 R 1 4 r 2 2 3. E = 3 r 2 4. None of these. 5. E = 2 3 r 2 6. E = R 1 4 r 2 2 7. E = 2 R 1 8. E = R 1 3 r 2 2 homework 07 KANUNGO, ANIL Due: Sep 18 2007, 4:00 am 2 9. E = 0 correct 10. E = 2 r 2 Explanation: Pick a cylindrical Gaussian surface with radius r 2 and apply Gauss law. Because there is no net charge inside the Gaussian surface, the electric field E = 0 . Question 4, chap 24, sect 3. part 4 of 4 10 points For a 100 m length of coaxial cable with in ner radius 1 mm and outer radius 1 . 9 mm, find the capacitance C of the cable. The value of the permittivity of free space is 8 . 85419 10 12 c 2 / N m 2 . Correct answer: 8 . 66747 nF (tolerance 1 %). Explanation: Let : = 100 m , R 1 = 1 mm , R 2 = 1 . 9 mm , and = 8 . 85419 10 12 c 2 / N m 2 . We calculate the potential across the capac itor by integrating E d s . We may choose a path of integration along a radius; i.e., E d s = E dr . V = 1 2 q integraldisplay R 1 R 2 dr r = 1 2 q ln r vextendsingle vextendsingle vextendsingle vextendsingle R 1 R 2 = q 2 ln parenleftbigg R 2 R 1 parenrightbigg ....
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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
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