22_InstSolManual_PC

# 22_InstSolManual_PC - ELECTRIC POTENTIAL E XERCISES Section...

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Unformatted text preview: ELECTRIC POTENTIAL E XERCISES Section 22.1 Electric Potential Difference 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 μ = ∆ = = ( Note: Since only magnitudes are needed in this problem, we omitted the subscripts A and B .) 17. I NTERPRET This problem is about the energy gained by an electron as it moves through a potential difference . V ∆ 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. For r ∆ v in the direction of a uniform electric field, Equation 22.2b gives | | (650 N/C)(1.4 m) 910 V. V E r ∆ = ∆ = = v (See note in solution to Exercise 16. Since , dV E dr = - ⋅ r v the potential always decreases in the direction of the electric field.) 21. I NTERPRET This problem is about the work done by the battery to move the charge from the positive terminal to the negative terminal. D EVELOP The work done by the battery is equal to the kinetic energy gained by the charge, and is equal to | |. W q V = ∆ E VALUATE Substituting the values given, we have (3.1 C) (9.0 V) 27.9 J W q V = ∆ = = A SSESS The charged particle gains kinetic energy as it moves toward the negative plate (in the direction of the electric field). The battery is needed to maintain the potential difference between the plates. 22. The energy gained is q V ∆ (see Example 22.1). The proton and singly-ionized helium atom have charge e , so they gain 19 17 100 eV (1.6 10 C)(100 V) 1.6 10 J,-- = × = × while the-particle α has charge 2 e and gains twice this energy....
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22_InstSolManual_PC - ELECTRIC POTENTIAL E XERCISES Section...

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