Chapter16 - Electrical Energy and Capacitance 59 8 There are eight different combinations that use all three capacitors in the circuit These

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PROBLEM SOLUTIONS 16.1 (a) Because the electron has a negative charge, it experiences a force in the direction opposite to the fi eld and, when released from rest, will move in the negative x direction. The work done on the electron by the fi eld is WF x q E x xx = ( ) = ( ) =− × ( )( ) ΔΔ 1 60 10 375 3 19 .. C N C 20 10 1 92 10 21 8 × ( ) −− m J . (b) The change in the electric potential energy is the negative of the work done on the particle by the fi eld. Thus, Δ PE W =− = − × 192 10 18 . J continued on next page
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60 Chapter 16 (c) Since the Coulomb force is a conservative force, conservation of energy gives ∆∆ KE PE += 0, or KE m PE PE fe f =− = v 2 20 , and v f e PE m = ( ) = −− × ( ) × 2 2 1 92 10 911 10 18 31 . . J kg = × 205 10 6 . m s in the direction x 16.2 (a) The change in the electric potential energy is the negative of the work done on the particle by the F eld. Thus, PE W qE x qE y qx xy ( ) + ( ) ( ) 0 5 40 10 . ( ) + ( ) −× ( ) =+ × 62 327 32 0 10 5 65 10 C N C m .. 4 J (b) The change in the electrical potential is the change in electric potential energy per unit charge, or V PE q == × 565 10 10 105 4 . J +5.40 C V 6 16.3 The work done by the agent moving the charge out of the cell is WW P E q V e input field C =− − ( )=+ ( ) 160 10 19 . ( ) 90 10 1 4 10 32 0 J C J . 16.4 PE q V q V V ef i = ( ) ( ) , so q PE VV e fi = = × 192 10 320 10 17 . . J +60.0 J C 19 C 16.5 E V d × 25 000 15 10 17 10 2 6 JC m NC . . 16.6 Since potential difference is work per unit charge V W q = , the work done is WqV = ( ) ( ) + ( ) 3 6 10 12 4 3 10 56 C J C J 16.7 (a) E V d × 600 533 10 113 10 3 5 m . . (b) Fq E × ( ) × ( ) 1 60 10 1 13 10 1 80 10 19 5 14 . CN C N (c) WF s =⋅ ( ) ( ) × cos . θ 180 10 533 290 10 14 3 N m cos . 0 4 38 10 17 ° J
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Electrical Energy and Capacitance 61 16.8 (a) Using conservation of energy, ΔΔ KE PE += 0, with KE f = 0 since the particle is “stopped,” we have PE KE m i =− =+ × ( 0 1 2 1 2 911 10 23 1 e kg v . ) × ( ) × 2 85 10 3 70 10 7 2 16 .. ms J The required stopping potential is then Δ Δ V PE q == −× × 370 10 160 10 231 10 16 19 . . . J C 3 231 V kV . (b) Being more massive than electrons, protons traveling at the same initial speed will have more initial kinetic energy and require a greater magnitude stopping potent ial . (c) Since Δ V PE q KE q m q stopping = v 2 2 , the ratio of the stopping potential for a proton to that for an electron having the same initial speed is Δ Δ V V me mm i i p e p e pe = −+ −− v v 2 2 2 2 () 16.9 (a) Use conservation of energy KE PE PE KE PE PE se f i ++ ( ) =++ ( ) or Δ KE PE PE ( ) + ( ) + ( ) = 0 Δ KE = 0 since the block is at rest at both beginning and end. Δ PE kx s ( ) 1 2 0 2 max , where x max is the maximum stretch of the spring. Δ PE W QE x e ( )=− =− ( ) max Thus, 0 1 2 0 2 +− ( ) = kx QE x max max , giving x QE k max CV m 78.0 N m × ( ) × ( ) 2 2 350 10 486 10 64 = × = 436 10 436 2 mc m (b) At equilibrium, Σ FFF k x Q E =+= + = 00 , or eq Therefore, x QE k x eq cm = 1 2 218 max .
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This note was uploaded on 04/23/2010 for the course DS SAD taught by Professor Asd during the Spring '10 term at A.T. Still University.

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Chapter16 - Electrical Energy and Capacitance 59 8 There are eight different combinations that use all three capacitors in the circuit These

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