homework 10-solutions

homework 10-solutions - howard(cah3459 homework 10...

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howard (cah3459) – homework 10 – Turner – (56705) 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 - 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. E = λ 2 π± 0 R 1 2. E = 2 λ 3 0 r 1 3. E =0 4. E = λ 2 R 1 4 0 r 1 2 5. E = λ 3 0 r 1 6. E = λ R 1 4 0 r 1 2 7. None oF these. 8. E = λ 2 0 r 1 correct 9. E = λ 2 0 r 1 10. E = 1 3 0 r 1 2 Explanation: Pick a cylindrical Gaussian surFace with the radius r 1 and apply the Gauss’s law; we obtain E · ² · 2 π r 1 = Q ± 0 E = λ 2 0 r 1 002 (part 2 oF 2) 10.0 points ±or a 100 m length oF coaxial cable with inner radius 0 . 68339 mm and outer radius 1 . 798 mm. ±ind the capacitance C oF the cable. Correct answer: 5 . 75094 n±. Explanation: Let : ² = 100 m , R 1 . 68339 mm , and R 2 =1 . 798 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 0 q l ± R 1 R 2 dr r = - 1 2 0 q l ln r ² ² ² ² R 1 R 2 = q 2 0 l ln R 2 R 1 . Since C = q V , we obtain the capacitance C = 2 0 l ln ³ R 2 R 1 ´ = 2 π (8 . 85419 × 10 - 12 c 2 / N · m 2 ) ln ³ 1 . 798 mm 0 . 68339 mm ´ × (100 m) = 5 . 75094 n± .
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howard (cah3459) – homework 10 – Turner – (56705) 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 AB C + Q - Q b What is the capacitance of this spherical capacitor? 1. C = 1 k e ( a - b ) 2. C = a k e 3. C = k e b 4. C = a + b k e 5. C = b - a 2 k e ln ± b a ² 6. C = k e a 7. C = b k e 8. C = 1 k e ± 1 a - 1 b ² correct 9. C = b 2 4 k e ( b - a ) 10. C = 1 k e ( a + b ) Explanation: Δ V = V a - V b = k e Q ± 1 a - 1 b ² - 0 since V b is grounded. The charge on the inside of the shell doesn’t aFect the grounded potential.
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This note was uploaded on 10/11/2010 for the course PHY 56715 taught by Professor Turner during the Spring '10 term at University of Texas.

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homework 10-solutions - howard(cah3459 homework 10...

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