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Ch17-h3-solutions

# Ch17-h3-solutions - pan(zp695 Ch17-h3 chiu(56565 This...

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pan (zp695) – Ch17-h3 – chiu – (56565) 1 This print-out should have 27 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 The potential di f erence From one end oF a 1 cm-long wire to the other in a circuit is Δ V = V B - V A =2 . 1V , as shown in the fgure below. AB 1cm Which end oF the wire is at the higher po- tential? 1. The ends are at the same potential. 2. A 3. B correct Explanation: Since V B - V A is a positive number, it is clear that B is at the higher potential. 002 (part 2 of 2) 10.0 points What is the magnitude and direction oF the feld E inside the wire? (Since this is a 1D problem, the sign oF your answer will indicate the direction.) Correct answer: - 210 V / m. Explanation: It should be clear that E points toward the leFt, since E points toward the lower potential. ±or a constant electric feld, Δ V = - ± ± E · d ± ² = - ± E · Δ ± ² . ±or a path From A to B, Δ ± ² = ± Δ x, 0 , 0 ² . Δ V = - E x Δ x E x = - Δ V Δ x = - 2 . 0 . 01 m = - 210 V / m . 003 10.0 points In a television picture tube, electrons are boiled out oF a very hot metal flament 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 flament - - - - - - - - + + + + + + + + + + Plate Plate F = eE E v =? L The high-voltage supply in the television set maintains a potential di f erence oF 14500 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 Candassumethatth is is not relativistic. Correct answer: 7 . 13674 × 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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pan (zp695) – Ch17-h3 – chiu – (56565) 2 Δ E =0 Δ U + Δ K q Δ V + Δ K ( - e ) Δ V + Δ K Δ K = e Δ V =(1 . 6 × 10 - 19 C)(14500 V) =2 . 32 × 10 - 15 J . Then Δ K = K f - K i = 1 2 mv 2 f - 0 v f = ± 2 Δ K m = ² 2(2 . 32 × 10 - 15 J) 9 . 11 × 10 - 31 kg = 7 . 13674 × 10 7 m / s . 004 (part 1 of 2) 10.0 points What is the maximum possible potential (rel- ative to infnity) oF a metal sphere oF 19 cm in air? Remember that the breakdown electric feld strength For air is roughly 3 × 10 6 N / C and also take 1 4 π± 0 = k =9 × 10 9 Nm 2 / C 2 . Correct answer: 5 . 7 × 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 feld and potential For a spherically distributed charge, we have E = 1 4 0 Q R 2 Q = (3 × 10 6 N / C)(19 cm) 2 9 × 10 9 2 / C 2 =1 . 20333 × 10 - 5 C . Now that we know the charge on the sphere under these circumstances, we can fnd the potential: V = 1 4 0 Q R =(9 × 10 9 2 / C 2 ) ³ 1 . 20333 × 10 - 5 C 19 cm ´ = 5 . 7 × 10 5 V . 005 (part 2 of 2) 10.0 points What is the maximum possible potential oF a metal sphere oF only 1 .
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Ch17-h3-solutions - pan(zp695 Ch17-h3 chiu(56565 This...

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