Practice HW 6 solution

Practice HW 6 solution - song (shs546) – Practice HW 6...

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Unformatted text preview: song (shs546) – Practice HW 6 Solutions – weathers – (22202) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Particles A (of mass m and charge Q ) and B (of m and charge 5 Q ) are released from rest with the distance between them equal to . 4778 m. If Q = 16 μ C, what is the kinetic energy of particle B at the instant when the particles are 2 . 4778 m apart? Correct answer: 9 . 71715 J. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , Q = 16 μ C = 1 . 6 × 10 − 5 C , a 1 = 0 . 4778 m , and a 2 = 2 . 4778 m . This is a conservation of energy problem so recall that Δ U + Δ K = 0 At a 2 , since the electrostatic potential energy is inversely proportional to the distance, the system of two particles loses potential energy (compared with U at a 1 ). This loss of poten- tial translates to an increase of kinetic energy. Hence Δ U = k e Q (5 Q ) parenleftbigg 1 a 2- 1 a 1 parenrightbigg = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × (1 . 6 × 10 − 5 C) bracketleftbig 5 (1 . 6 × 10 − 5 C) bracketrightbig × parenleftbigg 1 2 . 4778 m- 1 . 4778 m parenrightbigg = 19 . 4343 J so Δ K =- Δ U = 19 . 4343 J Since the particles were initially at rest Δ K is the kinetic energy of both particles at a 2 . To isolate the kinetic energy of particle B , we note that by conservation of momentum p i = p f 0 = m A v A + m B v B since m A = m B ⇒ | v a | = | v b | . Thus K A = K B so Δ K is shared equally between the two particles. Finally K B = Δ K 2 = 9 . 71715 J . 002 10.0 points Given: V = αx 2 y 2 + β z 2 ( x 2- γ ) + δ y 3 z , where α = 3 V / m 4 , β = 8 . 5 V / m 4 , γ = 6 . 7 m 2 , and δ = 5 . 3 V / m 4 . What is the y component of the electric field E y at (4 . 2 m,- 2 m, 7 . 3 m)? Correct answer:- 252 . 6 V / m. Explanation: Let : α = 3 V / m 4 , β = 8 . 5 V / m 4 , γ = 6 . 7 m 2 , δ = 5 . 3 V / m 4 , and ( x, y, z ) = (4 . 2 m ,- 2 m , 7 . 3 m) . E y =- ∂ V ∂y =- bracketleftbig αx 2 (2 y ) + δ (3 y 2 ) z bracketrightbig =- bracketleftBig (3 V / m 4 ) (4 . 2 m) 2 [2 (- 2 m)] + (5 . 3 V / m 4 ) bracketleftbig 3 (- 2 m) 2 bracketrightbig (7 . 3 m) bracketrightBig =- 252 . 6 V / m . 003 10.0 points A charge of 5 . 98 nC is uniformly distributed along the x-axis from- 4 m to 4 m....
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This note was uploaded on 04/22/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.

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Practice HW 6 solution - song (shs546) – Practice HW 6...

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