solution_pdf11 - tovar(jdt436 homework 11 Turner(59070 This...

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tovar (jdt436) – homework 11 – Turner – (59070) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points A long coaxial cable consists oF an inner cylin- drical conductor with radius R 1 and an outer cylindrical conductor shell with inner radius R 2 and outer radius R 3 as shown. The ca- ble extends out perpendicular to the plane shown. The charge on the inner conductor per unit length along the cable is λ and the corresponding charge on the outer conductor per unit length is - λ (same in magnitudes but with opposite signs) and λ > 0. Q R 1 R 2 R 3 b - Q ±ind the magnitude oF the electric feld at the point a distance r 1 From the axis oF the inner conductor, where R 1 < r 1 < R 2 . 1. None oF these. 2. E = λ R 1 4 π ǫ 0 r 1 2 3. E = λ 2 R 1 4 π ǫ 0 r 1 2 4. E = λ 2 π ǫ 0 r 1 5. E = λ 3 π ǫ 0 r 1 6. E = 0 7. E = λ 2 π ǫ 0 r 1 correct 8. E = 2 λ 3 π ǫ 0 r 1 9. E = λ R 1 3 π ǫ 0 r 1 2 10. E = λ 2 π ǫ 0 R 1 Explanation: Pick a cylindrical Gaussian surFace with the radius r 1 and apply the Gauss’s law; we obtain E · · 2 π r 1 = Q ǫ 0 E = λ 2 π ǫ 0 r 1 002 (part 2 oF 2) 10.0 points ±or a 100 m length oF coaxial cable with inner radius 0 . 834467 mm and outer radius 2 . 02108 mm. ±ind the capacitance C oF the cable. Correct answer: 6 . 28905 n±. Explanation: Let : = 100 m , R 1 = 0 . 834467 mm , and R 2 = 2 . 02108 mm . We calculate the potential across the capaci- tor 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 l i R 1 R 2 dr r = - 1 2 π ǫ 0 q l ln r v v v v R 1 R 2 = q 2 π ǫ 0 l ln R 2 R 1 . Since C = q V , we obtain the capacitance C = 2 π ǫ 0 l ln p R 2 R 1 P = 2 π (8 . 85419 × 10 12 c 2 / N · m 2 ) ln p 2 . 02108 mm 0 . 834467 mm P × (100 m) = 6 . 28905 n± .
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tovar (jdt436) – homework 11 – Turner – (59070) 2 003 10.0 points Given a spherical capacitor with radius of the inner conducting sphere a and the outer shell b . The outer shell is grounded. The charges are + Q and - Q . A point C is located at r = R 2 , where R = a + b . a A B C + Q - Q b What is the capacitance of this spherical capacitor? 1. C = b k e 2. C = 1 k e ( a + b ) 3. C = b 2 4 k e ( b - a ) 4. C = k e b 5. C = a k e 6. C = b - a 2 k e ln p b a P 7. C = 1 k e p 1 a - 1 b P correct 8. C = k e a 9. C = 1 k e ( a - b ) 10. C = a + b k e Explanation: Δ V = V a - V b = k e Q p 1 a - 1 b P - 0 since V b is grounded. The charge on the inside of the shell doesn’t aFect the grounded potential.
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