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Midterm 1-solutions - Version 063 Midterm 1 Shih(58250 This...

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Version 063 – Midterm 1 – Shih – (58250) 1 This print-out should have 26 questions. Multiple-choice 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 μ C. The surrounding conductor is a cylindrical shell and has an inner diameter of 7 . 048 mm and a charge of - 11 . 04 μ 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 μ C , = 60 m , a = 2 . 939 mm , and b = 7 . 048 mm . The charge per unit length is λ Q . V = - integraldisplay b a vector E · dvectors = - 2 k e λ 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 Δ θ is labeled Δ q . x y - - - - - - - - - - - - - - - - - - Δ θ θ r x y I II III IV B A O Δ q is given by 1. Δ q = Q 2 π 2. Δ q = Q Δ θ π correct 3. None of these 4. Δ q = π Q 5. Δ q = Q 6. Δ q = 2 Q π 7. Δ q = 2 π Q 8. Δ q = Q π
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Version 063 – Midterm 1 – Shih – (58250) 2 9. Δ q = Q Δ θ 2 π 10. Δ q = 2 Q Δ θ π Explanation: The angle of a semicircle is π , thus the charge on a small segment with angle Δ θ is Δ q = Q Δ θ π . 003 (part 2 of 3) 10.0 points The magnitude of the x -component of the electric field at the center, due to Δ q , is given by 1. Δ E x = k | Δ q | (cos θ ) r 2. Δ E x = k | Δ q | sin θ r 2 3. Δ E x = k | Δ q | (cos θ ) r 2 4. Δ E x = k | Δ q | (sin θ ) r 5. Δ E x = k | Δ q | cos θ r 2 correct 6. Δ E x = k | Δ q | r 2 7. Δ E x = k | Δ q | (sin θ ) r 2 8. Δ E x = k | Δ q | sin θ r 9. Δ E x = k | Δ q | cos θ r 10. Δ E x = k | Δ q | r 2 Explanation: Negative charge attracts a positive test charge. At O , Δ E points toward Δ q . Accord- ing to the sketch, the vector Δ E x is pointing along the negative x axis. The magnitude of the Δ E x is given by Δ E x = Δ E cos θ = k | Δ q | r 2 cos θ . 004 (part 3 of 3) 10.0 points Determine the magnitude of the electric field at O . The total charge is - 26 . 1 μ C, the radius of the semicircle is 18 cm, and the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 1. 717473.0 2. 52476.7 3. 62049.0 4. 939376.0 5. 1151900.0 6. 1217810.0 7. 102868.0 8. 39658.6 9. 1064060.0 10. 4609110.0 Correct answer: 4 . 60911 × 10 6 N / C. Explanation: Let : Q = - 26 . 1 μ C , r = 18 cm , and k = 8 . 98755 × 10 9 N · m 2 / C 2 .
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