Fa06Soln5 - Physics 7B Fall, 2006 Homework 5 Solutions 21.8...

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Physics 7B Fall, 2006 Homework 5 Solutions 21.8 The number of excess electrons is N = Q /(– e ) = (– 40 × 10 –6 C)/(– 1.60 × 10 –19 C/electrons) = 2.5 × 10 14 electrons . The mass increase is Δ m = Nm e = (2.5 × 10 14 electrons)(9.11 × 10 –31 kg/electron) = 2.3 × 10 –16 kg . 21.10 Using the symbols in the figure, we find the magnitudes of the three individual forces: F 12 = F 21 = kQ 1 Q 2 / r 12 2 = kQ 1 Q 2 / L 2 = (9.0 × 10 9 N · m 2 /C 2 )(70 × 10 –6 C)(48 × 10 –6 C)/(0.35 m) 2 = 2.47 × 10 2 N. F 13 = F 31 = kQ 1 Q 3 / r 13 2 = kQ 1 Q 3 /(2 L ) 2 = (9.0 × 10 9 N · m 2 /C 2 )(70 × 10 –6 C)(80 × 10 –6 C)/[2(0.35 m)] 2 = 1.03 × 10 2 N. F 23 = F 32 = kQ 2 Q 3 / r 23 2 = kQ 2 Q 3 / L 2 = (9.0 × 10 9 N · m 2 /C 2 )(48 × 10 –6 C)(80 × 10 –6 C)/(0.35 m) 2 = 2.82 × 10 2 N. The directions of the forces are determined from the signs of the charges and are indicated on the diagram. For the net forces, we get F 1 = F 13 F 12 = 1.03 × 10 2 N – 2.47 × 10 2 N = – 1.4 × 10 2 N (left). F 2 = F 21 + F 23 = 2.47 × 10 2 N + 2.82 × 10 2 N = + 5.3 × 10 2 N (right). F
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This note was uploaded on 09/09/2008 for the course PHYSICS 7B taught by Professor Packard during the Fall '08 term at University of California, Berkeley.

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Fa06Soln5 - Physics 7B Fall, 2006 Homework 5 Solutions 21.8...

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