23_InstructorSolutions

23_InstructorSolutions - ELECTRIC POTENTIAL 23 ra = 0.150 m...

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23-1 E LECTRIC P OTENTIAL 23.1. I DENTIFY : Apply Eq.(23.2) to calculate the work. The electric potential energy of a pair of point charges is given by Eq.(23.9). S ET U P : Let the initial position of 2 q be point a and the final position be point b , as shown in Figure 23.1. 0.150 m a r = 2 2 (0.250 m) (0.250 m) b r = + 0.3536 m b r = Figure 23.1 E XECUTE : a b a b W U U = 6 6 9 2 1 2 0 1 ( 2.40 10 C)( 4.30 10 C) (8.988 10 N m /C ) 4 0.150 m a a q q U r π 2 + × × = = × P 0.6184 J a U = − 6 6 9 2 1 2 0 1 ( 2.40 10 C)( 4.30 10 C) (8.988 10 N m /C ) 4 0.3536 m b b q q U r π 2 + × × = = × P 0.2623 J b U = − 0.6184 J ( 0.2623 J) 0.356 J a b a b W U U = = − − − = − E VALUATE : The attractive force on 2 q is toward the origin, so it does negative work on q 2 when q 2 moves to larger r . 23.2. I DENTIFY : Apply . a b a b W U U = S ET U P : 8 5.4 10 J. a U = + × Solve for . b U E XECUTE : 8 1.9 10 J . a b a b W U U = − × = 8 8 8 1.9 10 J ( 5.4 10 J) 7.3 10 J. b a a b U U W = = × − − × = × E VALUATE : When the electric force does negative work the electrical potential energy increases. 23.3. I DENTIFY : The work needed to assemble the nucleus is the sum of the electrical potential energies of the protons in the nucleus, relative to infinity. S ET U P : The total potential energy is the scalar sum of all the individual potential energies, where each potential energy is 0 0 (1/4 )( / ). U qq r π = P Each charge is e and the charges are equidistant from each other, so the total potential energy is 2 2 2 2 0 0 1 3 . 4 4 e e e e U r r r r π π = + + = P P E XECUTE : Adding the potential energies gives 2 19 2 9 2 2 13 15 0 3 3(1.60 10 C) (9.00 10 N m /C ) 3.46 10 J 2.16 MeV 4 2.00 10 m e U r π × × = = = × = × P E VALUATE : This is a small amount of energy on a macroscopic scale, but on the scale of atoms, 2 MeV is quite a lot of energy. 23
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23-2 Chapter 23 23.4. I DENTIFY : The work required is the change in electrical potential energy. The protons gain speed after being released because their potential energy is converted into kinetic energy. (a) S ET U P : Using the potential energy of a pair of point charges relative to infinity, 0 0 (1/4 )( / ). U qq r π = P we have 2 2 2 1 0 2 1 1 . 4 e e W U U U r r π = Δ = = P E XECUTE : Factoring out the e 2 and substituting numbers gives ( )( ) 2 9 2 2 19 14 15 15 1 1 9.00 10 N m /C 1.60 10 C 7.68 10 J 3.00 10 m 2.00 10 m W = × × = × × × (b) S ET U P : The protons have equal momentum, and since they have equal masses, they will have equal speeds and hence equal kinetic energy. 2 2 1 2 1 2 2 . 2 U K K K mv mv Δ = + = = = E XECUTE : Solving for v gives 14 27 7.68 10 J 1.67 10 kg U v m Δ × = = × = 6.78 × 10 6 m/s E VALUATE : The potential energy may seem small (compared to macroscopic energies), but it is enough to give each proton a speed of nearly 7 million m/s. 23.5. (a) I DENTIFY : Use conservation of energy: other a a b b K U W K U + + = + U for the pair of point charges is given by Eq.(23.9).
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