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# 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 C 90 10 1.4 10 J C input field e W W PE q V = − = − −Δ = + Δ = × + × = × 16.4 (extra) ( ) ( ) e f i PE q V q V V Δ = Δ = , so 17 1.92 10 J 19 3.20 +60.0 JC e f i PE V V × = = = 10 C q Δ × 16.5 6 2 25000 JC 1.7 10 N C 1.5 10 m V E d Δ = = = × × 16.11 (a) ( )( ) 9 2 2 19 7 -2 8.99 10 N m C 1.60 10 C 1.44 10 V 1.00 10 m e k q V r × × = = = × × (b) ( ) ( )( ) 2 1 2 1 2 1 9 2 2 19 8 1 1 1 1 8.99 10 N m C 1.60 10 C 0.0200 m 0.0100 m 7.19 10 V e e e k q k q V V V k q r r r r Δ = = = = × × = − × 16.12 1 2 1 2 1 2 e q q V V V k r r = + = + where 1 0.60 m 0 0.60 m r = = . , and Thus, 2 0.60 m 0.30 m 0.30 m r = = 2 9 9 0 10 C 6.0 10 C 2 .60 m 0.30 m × + 9 2 2 N m 3. 8.99 10 .2 10 V C 0 V × = × = ×

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16.13 (extra) (a) Calling the 2.00 C μ charge 3 q , ( ) ( ) 1 2 1 2 8.00 C 0060 3 2 2 1 2 6 2 6 6 9 2 2 2 6 2.00 10 C N m 10 .00 10 C 8.99 10 . 0 .0300 m C 4 m 0 0.0600 0.0300 m 2.67 10 V e i e i i k q q V k
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ch16-sol - Chapter 16 16.3 Problem Solutions The work done...

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