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HWCh25Solutions

HWCh25Solutions - Physics 133 P25.2 q 6.41 10 19 Homework...

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Physics 133 Homework solutions Chapter 25 P25.2    K = q V    7.37 × 10 - 17 = q 115 ( 29    q = 6.41 × 10 - 19  C P25.4    W =∆ K =- q V    0 - 1 2 9.11 × 10 - 31  kg ( 29 4.20 × 10 5   m s ( 29 2 =-- 1.60 × 10 - 19  C ( 29 V From which,    V =- 0.502 V . P25.7    U =- 1 2 m v f 2 - v i 2 ( 29 =- 1 2 9.11 × 10 - 31  kg ( 29 1.40 × 10 5   m s ( 29 2 - 3.70 × 10 6   m s ( 29 2 = 6.23 × 10 - 18  J    U = q V :    + 6.23 × 10 - 18 =- 1.60 × 10 - 19 ( 29 V    V =- 38.9 V. The origin is at highest potential. P25.18 (a)    E x = k e q 1 x 2 + k e q 2 x - 2.00 ( 29 2 = 0 becomes    E x = k e + q x 2 + - 2 q x - 2.00 ( 29 2 = 0 . Dividing by     k e ,    2 qx 2 = q x - 2.00 ( 29 2    x 2 + 4.00 x - 4.00 = 0 . Therefore     E = 0 when    x = - 4.00 ± 16.0 + 16.0 2 =- 4.83 m . (Note that the positive root does not correspond to a physically valid situation.) (b)    V = k e q 1 x + k e q 2 2.00 - x = 0 or    V = k e + q x - 2 q 2.00 - x = 0 . Again solving for  x ,    2 qx = q 2.00 - x ( 29 . For     0 x 2.00      V = 0 when    x = 0.667 m and     q x = - 2 q 2 - x . For     x < 0    x =- 2.00 m .

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HWCh25Solutions - Physics 133 P25.2 q 6.41 10 19 Homework...

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