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17-3 - brown(twb493 Ch17-h3 chiu(56565 1 This print-out...

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brown (twb493) – Ch17-h3 – chiu – (56565) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of2)10.0points The potential difference from one end of a 1 cm-long wire to the other in a circuit is Δ V = V B V A = 1 . 7 V , as shown in the figure below. A B 1 cm Which end of the wire is at the higher po- tential? 1. B correct 2. A 3. The ends are at the same potential. Explanation: Since V B V A is a positive number, it is clear that B is at the higher potential. 002(part2of2)10.0points What is the magnitude and direction of the field E inside the wire? (Since this is a 1D problem, the sign of your answer will indicate the direction.) Correct answer: 170 V / m. Explanation: It should be clear that E points toward the left, since E points toward the lower potential. For a constant electric field, Δ V = integraldisplay vector E · d vector = vector E · Δ vector ℓ. For a path from A to B, Δ vector = ( Δ x, 0 , 0 ) . Δ V = E x Δ x E x = Δ V Δ x = 1 . 7 V 0 . 01 m = 170 V / m . 003 10.0points In a television picture tube, electrons are boiled out of a very hot metal filament placed near a negative metal plate. These electrons start out nearly at rest and are accelerated toward a positive metal plate. They pass through a hole in the positive plate on their way toward the picture screen, as shown in the diagram. Hot filament + + + + + + + + + + Plate Plate F = e E E v = ? L The high-voltage supply in the television set maintains a potential difference of 17500 V between the two plates, what speed do the electrons reach? Use m e = 9 . 11 × 10 31 kg and q e = 1 . 6 × 10 19 C and assume that this is not relativistic. Correct answer: 7 . 84034 × 10 7 m / s. Explanation: The net energy remains constant through- out the whole process. We can use the follow- ing train of logic:
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brown (twb493) – Ch17-h3 – chiu – (56565) 2 Δ E = 0 Δ U + Δ K = 0 q Δ V + Δ K = 0 ( e V + Δ K = 0 Δ K = e Δ V = (1 . 6 × 10 19 C)(17500 V) = 2 . 8 × 10 15 J . Then Δ K = K f K i = 1 2 m v 2 f 0 v f = radicalbigg K m = radicalBigg 2(2 . 8 × 10 15 J) 9 . 11 × 10 31 kg = 7 . 84034 × 10 7 m / s . 004(part1of2)10.0points What is the maximum possible potential (rel- ative to infinity) of a metal sphere of 20 cm in air? Remember that the breakdown electric field strength for air is roughly 3 × 10 6 N / C and also take 1 4 πǫ 0 = k = 9 × 10 9 N m 2 / C 2 . Correct answer: 6 × 10 5 V. Explanation: The problem indicates that E max at the sur- face of the sphere is 3 × 10 6 N / C. Now, using the equations for electric field and potential for a spherically distributed charge, we have E = 1 4 πǫ 0 Q R 2 Q = (3 × 10 6 N / C)(20 cm) 2 9 × 10 9 N m 2 / C 2 = 1 . 33333 × 10 5 C . Now that we know the charge on the sphere under these circumstances, we can find the potential: V = 1 4 πǫ 0 Q R = (9 × 10 9 N m 2 / C 2 ) parenleftbigg 1 . 33333 × 10 5 C 20 cm parenrightbigg = 6 × 10 5 V .
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