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Unformatted text preview: brown (twb493) – Ch17h3 – chiu – (56565) 1 This printout should have 27 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points The potential difference from one end of a 1 cmlong 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 (part 2 of 2) 10.0 points 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 ℓ = − vector E · Δ vector ℓ. For a path from A to B, Δ vector ℓ = ( Δ x, , ) . Δ V = − E x Δ x ⇒ E x = − Δ V Δ x = − 1 . 7 V . 01 m = − 170 V / m . 003 10.0 points 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 = eE E v = ? L The highvoltage 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: brown (twb493) – Ch17h3 – chiu – (56565) 2 Δ E = 0 Δ U + Δ K = 0 q Δ V + Δ K = 0 ( − e )Δ 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 mv 2 f − ⇒ v f = radicalbigg 2Δ K m = radicalBigg 2(2 . 8 × 10 − 15 J) 9 . 11 × 10 − 31 kg = 7 . 84034 × 10 7 m / s . 004 (part 1 of 2) 10.0 points 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 = 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 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 ....
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This note was uploaded on 12/01/2011 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Physics

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