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**Unformatted text preview: **Chapter 16 Electrical Energy and Capacitance Problem Solutions 16.1 (a) The work done is ( ) cos cos W F s qE s θ θ = ⋅ = ⋅ , or ( ) ( ) ( ) 19 2 19 1.60 10 C 200 N C 2.00 10 m cos0 6.40 10 J W--- = × × ° = × (b) The change in the electrical potential energy is 19 6.40 10 J e PE W- Δ = - = - × (c) The change in the electrical potential is 19-19 6.40 10 J 4.00 V 1.60 10 C e PE V q- Δ- × Δ = = = - × 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 = × + × = × 16.5 6 2 25 000 J C 1.7 10 N C 1.5 10 m V E d- Δ = = = × × 16.8 From conservation of energy, ( ) ( ) 2 2 1 or 2 f f q V mv q V v m Δ- = Δ = (a) For the proton, ( ) ( ) 19 5 27 2 1.60 10 C 120 V 1.52 10 m s 1.67 10 kg f v-- ×- = = × × 35 36 &#1; &#1; &#1; &#1; &#1; &#1; CHAPTER 16 (b) For the electron, ( ) ( ) 19 6 31 2 1.60 10 C 120 V 6.49 10 m s 9.11 10 kg f v--- × + = = × × 16.12 1 2 1 2 1 2 e q q V V V k r r = + = + where 1 0.60 m 0.60 m r =- = , and 2 0.60 m 0.30 m 0.30 m. r =- = Thus, 2 9 9 9 2 2 N m 3.0 10 C 6.0 10 C 8.99 10 2.2 10 V C 0.60 m 0.30 m V-- ⋅ × × = × + = × 16.15 (a) 2 9 9 9 2 N m 5.00 10 C 3.00 10 C 8.99 10 103 V C 0.175 m 0.175 m e i i i k q V r-- = ⋅ × × = ×- = ∑ (b) ( )( ) 2 12 9 9 2 7 9 2 5.00 10 C 3.00 10 C N m 8.99 10 3.85 10 J C 0.350 m e i k q q PE r--- = ×- × ⋅ = × = - × The negative sign means that positive work must be done to separate the charges...

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