ch16-sol - Chapter 16 16.3 Problem Solutions The work done...

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Chapter 16 Problem Solutions 16.3 The work done by the agent moving the charge out of the cell is ( ) ( ) () 19 3 20 J 1.60 10 C9 0 1 0 1 . 41 0 J C input fiel de WW P E q V −− =− =− −Δ =+ Δ ⎛⎞ + × = × ⎜⎟ ⎝⎠ 16.4 (extra) e fi PE q V q V V Δ= Δ = , so 17 1.92 10 J 3.20 +60.0 J C e PE VV −× == = 10 C q Δ × 16.5 6 2 25 000 J C 1. 71 0 NC 51 0 m V E d Δ = × × 16.11 (a) ( )( ) 92 2 1 9 7 -2 8.99 10 N m C C 1.44 10 V 1.00 10 m e kq V r ×⋅ × = × × (b) ( ) 21 2 1 2 1 9 8 11 C C 0.020 0 m0 . 010 0 m 7.19 10 V ee e kq kq VV V k q rr r r Δ = =−= × × 16.12 12 e qq VVV k =+= + where 1 0.60 m00 . 60 m r = −= . , and Thus, 2 . 30 . m r 29 9 01 C6 . 0 C 2 . 60 . 30 m × + 9 2 2 Nm 3 . 8.99 10 . 0 V C0 V ⋅× = ×
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16.13 (extra) (a) Calling the 2.00 C μ charge 3 q , () 12 8 .00 C0 060 3 22 6 26 6 9 2 6 2.00 10 C Nm 10 . 00 10 C 8.99 10 . 0 . 030 0 m C4 m0 0.060 0 0.030 0 m 2.67 10 V ei e i i kq q Vk rr V −− ⎛⎞ ⎜⎟ == ⎝⎠ × × + + 6 q q r + + + ⋅× + (b) Replacing in part (a) yields 6 2.00 10 C by 2.00 10 C ×− × 6 13 10 V 2.
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This note was uploaded on 10/13/2010 for the course PHYSICS 102 taught by Professor Abdulrazzaq during the Fall '08 term at West Virginia State University.

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ch16-sol - Chapter 16 16.3 Problem Solutions The work done...

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