Ch 22 Practice Pr soln

# Ch 22 Practice Pr soln - Chapter 22 Practice Pr solns 16 W...

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Chapter 22 Practice Pr solns: 16. The potential difference and the work per unit charge, done by an external agent, are equal in magnitude, so (50 C)(12 V) 600 J. W q V μ = ∆ = = ( Note: Since only magnitudes are needed in this problem, we omitted the subscripts A and B .) 17. D EVELOP We assume that the electron is initially at rest. When released from the negative plate, it moves toward the positive plate, and the kinetic energy gained is | | . U q V = E VALUATE As the electron moves from the negative side to the positive side (i.e., against the direction of the electric field), the kinetic energy it gains is 19 17 |( ) | 120 eV (1.6 10 C)(120 V) 1.92 10 J K e V - - = - = = × = × A SSESS Moving a negative charge through a positive potential difference is like going downhill; potential energy decreases. However, the kinetic energy of the electron is increased. 18. The work done by an external agent equals the potential energy change, 45 J , AB AB U q V = = hence AB V = 45 J/15 mC 3 kV. = (Since the work required to move the charge from A to B is positive, and B A AB V V V is positive.) 19. I NTERPRET This problem is about conversion of units. D EVELOP By definition, 1 volt 1 joule/coulomb = (1 V 1 J/C). = On the other hand, 1 joule 1 newton-meter = (1 J 1 N m). = E VALUATE Combining the two expressions gives 1 V 1 J/C 1 N m/C. = = It follows that 1 V/m 1 N/C. = A SSESS These are the units for the electric field strength. 20.

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Ch 22 Practice Pr soln - Chapter 22 Practice Pr solns 16 W...

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