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Unformatted text preview: Version 063 Midterm 1 Shih (58250) 1 This printout should have 26 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A 60 m length of coaxial cable has a solid cylindrical wire inner conductor with a di ameter of 2 . 939 mm and carries a charge of 11 . 04 C. The surrounding conductor is a cylindrical shell and has an inner diameter of 7 . 048 mm and a charge of 11 . 04 C. Assume the region between the conductors is air. The Coulomb constant is 8 . 98755 10 9 N m 2 / C 2 . What is the capacitance of this cable? 1. 2.21279 2. 1.71512 3. 3.81622 4. 2.41954 5. 2.49337 6. 2.78068 7. 1.09339 8. 1.1855 9. 0.910748 10. 4.34197 Correct answer: 3 . 81622 nF. Explanation: Let : k e = 8 . 98755 10 9 N m 2 / C 2 , Q = 11 . 04 C , = 60 m , a = 2 . 939 mm , and b = 7 . 048 mm . The charge per unit length is Q . V = integraldisplay b a vector E dvectors = 2 k e integraldisplay b a dr r = 2 k e Q ln parenleftbigg b a parenrightbigg . The capacitance of a cylindrical capacitor is given by C Q V = 2 k e 1 ln parenleftbigg b a parenrightbigg = 60 m 2 (8 . 98755 10 9 N m 2 / C 2 ) 1 ln parenleftbigg 7 . 048 mm 2 . 939 mm parenrightbigg parenleftbigg 1 10 9 nF 1 F parenrightbigg = 3 . 81622 nF 002 (part 1 of 3) 10.0 points Consider the setup shown in the figure be low, where the arc is a semicircle with radius r . The total charge Q is negative, and dis tributed uniformly on the semicircle. The charge on a small segment with angle is labeled q . x y r x y I II III IV B A O q is given by 1. q = Q 2 2. q = Q correct 3. None of these 4. q = Q 5. q = Q 6. q = 2 Q 7. q = 2 Q 8. q = Q Version 063 Midterm 1 Shih (58250) 2 9. q = Q 2 10. q = 2 Q Explanation: The angle of a semicircle is , thus the charge on a small segment with angle is q = Q . 003 (part 2 of 3) 10.0 points The magnitude of the xcomponent of the electric field at the center, due to q , is given by 1. E x = k  q  (cos ) r 2. E x = k  q  sin r 2 3. E x = k  q  (cos ) r 2 4. E x = k  q  (sin ) r 5. E x = k  q  cos r 2 correct 6. E x = k  q  r 2 7. E x = k  q  (sin ) r 2 8. E x = k  q  sin r 9. E x = k  q  cos r 10. E x = k  q  r 2 Explanation: Negative charge attracts a positive test charge. At O , E points toward q . Accord ing to the sketch, the vector E x is pointing along the negative x axis. The magnitude of the E x is given by E x = E cos = k  q  r 2 cos ....
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 Spring '08
 Turner
 Physics

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