04 - Solutions - Solutions for Homework 4 8. Use the charge...

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Unformatted text preview: Solutions for Homework 4 8. Use the charge per electron and the mass per electron. ! 46 " 10 ! 6 C ( ) 1 electron ! 1.602 " 10 ! 19 C # $ % & ( = 2.871 " 10 14 ) 2.9 " 10 14 electrons 2.871 " 10 14 e ! ( ) 9.109 " 10 ! 31 kg 1 e ! # $ % & ( = 2.6 " 10 ! 16 kg 12. Let the right be the positive direction on the line of charges. Use the fact that like charges repel and unlike charges attract to determine the direction of the forces. In the following expressions, k = 8.988 ! 10 9 N " m 2 C 2 . r F + 75 = ! k 75 C ( ) 48 C ( ) 0.35m ( ) 2 i + k 75 C ( ) 85 C ( ) 0.70m ( ) 2 i = ! 147.2N i " ! 150N i r F + 48 = k 75 C ( ) 48 C ( ) 0.35m ( ) 2 i + k 48 C ( ) 85 C ( ) 0.35m ( ) 2 i = 563.5N i " 560N i r F ! 85 = ! k 85 C ( ) 75 C ( ) 0.70m ( ) 2 i ! k 85 C ( ) 48 C ( ) 0.35m ( ) 2 i = ! 416.3N i " ! 420N i 14. ( a ) If the force is repulsive, both charges must be positive since the total charge is positive. Call the total charge Q . Q 1 + Q 2 = Q F = kQ 1 Q 2 d 2 = kQ 1 Q ! Q 1 ( ) d 2 " Q 1 2 ! QQ 1 + Fd 2 k = Q 1 = Q Q 2 ! 4 Fd 2 k 2 = Q Q 2 ! 4 Fd 2 k 2 = 1 2 90.0 # 10 ! 6 C ( ) 90.0 # 10 ! 6 C ( ) 2 ! 4 12.0N ( ) 1.16m ( ) 2 8.988 # 10 9 N $ m 2 C 2 ( ) % & ( ) * * * = 60.1 ! 10 " 6 C , 29.9 ! 10 " 6 C ( b ) If the force is attractive, then the charges are of opposite sign. The value used for F must then be negative. Other than that, the solution method is the same as for part ( a ). Q 1 + Q 2 = Q F = kQ 1 Q 2 d 2 = kQ 1 Q ! Q 1 ( ) d 2 " Q 1 2 ! QQ 1 + Fd 2 k = Q 1 = Q Q 2 ! 4 Fd 2 k 2 = Q Q 2 ! 4 Fd 2 k 2 = 1 2 90.0 # 10 ! 6 C ( ) 90.0 # 10 ! 6 C ( ) 2 ! 4 ! 12.0N ( ) 1.16m ( ) 2 8.988 # 10 9 N $ m 2 C 2 ( ) % & (...
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This note was uploaded on 11/19/2010 for the course LECTURE 1 taught by Professor Yildiz during the Fall '10 term at University of California, Berkeley.

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04 - Solutions - Solutions for Homework 4 8. Use the charge...

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