Chapter 23 HW Solutions - PHYS 241 Ch. 23 Homework...

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Unformatted text preview: PHYS 241 Ch. 23 Homework Solutions 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 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 -- 2 + - = = P 0.6184 J a U = - 6 6 9 2 1 2 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 -- 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 . 23.3.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 (1/4 )( / ). U qq r = P Each charge is e and the charges are equidistant from each other, so the total potential energy is 2 2 2 2 1 3 . 4 4 e e e e U r r r r = + + = P P E XECUTE : Adding the potential energies gives 2 19 2 9 2 2 13 15 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 --- = = = = 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. 23.5.(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). S ET U P : Let point a be where q 2 is 0.800 m from q 1 and point b be where q 2 is 0.400 m from q 1 , as shown in Figure 23.5a. 1 Figure 23.5a E XECUTE : Only the electric force does work, so other W = and 1 2 1 . 4 q q U r = P 2 3 2 1 1 2 2 (1.50 10 kg)(22.0 m/s) 0.3630 J a a K mv- = = = 6 6 9 2 2 1 2 1 ( 2.80 10 C)( 7.80 10 C) (8.988 10 N m /C ) 0.2454 J 4 0.800 m a a q q U r --- - = = = + P 2 1 2 b b K mv = 6 6 9 2 2 1 2 1 ( 2.80 10 C)( 7.80 10 C) (8.988 10 N m /C ) 0.4907 J 4 0.400 m b b q q U r --- - = = = + P The conservation of energy equation then gives ( ) b a a b K K U U = +- 2 1 2 0.3630 J (0.2454 J 0.4907 J) 0.1177 J b mv = + +- = 3 2(0.1177 J) 12.5 m/s 1.50 10 kg b v- = = E VALUATE : The potential energy increases when the two positively charged spheres get closer together, so the kinetic energy and speed decrease....
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This note was uploaded on 03/18/2010 for the course PHYS 241 taught by Professor Milsom during the Fall '08 term at University of Arizona- Tucson.

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Chapter 23 HW Solutions - PHYS 241 Ch. 23 Homework...

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