Solution 7 - homework 07 KANUNGO ANIL Due 4:00 am Question...

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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 λ and the corresponding charge per unit length on the outer conductor is λ (same magnitudes but with opposite signs) and λ > 0. Q R 1 R 2 R 3 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 < r 1 < R 2 . 1. E = λ 2 π ǫ 0 r 1 correct 2. E = λ R 1 3 π ǫ 0 r 1 2 3. E = 0 4. E = λ 2 π ǫ 0 R 1 5. E = λ 3 π ǫ 0 r 1 6. E = λ R 1 4 π ǫ 0 r 1 2 7. None of these. 8. E = λ 2 π ǫ 0 r 1 9. E = 2 λ 3 π ǫ 0 r 1 10. E = λ 2 R 1 4 π ǫ 0 r 1 2 Explanation: Pick a cylindrical Gaussian surface with radius r 1 and apply Gauss’s law: E · d A = E · (2 π r 1 ) = Q ǫ 0 E = λ 2 π ǫ 0 r 1 . Question 2, chap 24, sect 3. part 2 of 4 10 points The electric field vector points 1. in the negative ˆ r direction. 2. in the positive ˆ 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 ˆ 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 < r 2 . 1. E = λ 2 π ǫ 0 r 2 2. E = λ 2 R 1 4 π ǫ 0 r 2 2 3. E = λ 3 π ǫ 0 r 2 4. None of these. 5. E = 2 λ 3 π ǫ 0 r 2 6. E = λ R 1 4 π ǫ 0 r 2 2 7. E = λ 2 π ǫ 0 R 1 8. E = λ R 1 3 π ǫ 0 r 2 2
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homework 07 – KANUNGO, ANIL – Due: Sep 18 2007, 4:00 am 2 9. E = 0 correct 10. E = λ 2 π ǫ 0 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 ǫ 0 = 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 π ǫ 0 q integraldisplay R 1 R 2 dr r = 1 2 π ǫ 0 q ln r vextendsingle vextendsingle vextendsingle vextendsingle R 1 R 2 = q 2 π ǫ 0 ln parenleftbigg R 2 R 1 parenrightbigg . Since C = q V , the capacitance is C = q V = 2 π ǫ 0 ln parenleftbigg R 2 R 1 parenrightbigg = 2 π (8 . 85419 × 10 12 c 2 / N · m 2 ) ln parenleftbigg 1 . 9 mm 1 mm parenrightbigg × (100 m) = 8 . 66747 nF .
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