# chap22 - ELECTRIC POTENTIAL 22 EXERCISES Section 22.1...

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22.1 ELECTRIC POTENTIAL 22 EXERCISES 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. WqV μ =Δ = = ( Note: Since only magnitudes are needed in this problem, we omitted the subscripts A and B .) 17. INTERPRET This problem is about the energy gained by an electron as it moves through a potential difference . V Δ DEVELOP 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 || Uq V . Δ =Δ EVALUATE 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 Ke V −− Δ=−Δ= = × = × ASSESS 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 V Δ == Δ hence AB V Δ = 45 J/15 mC 3 kV. = (Since the work required to move the charge from A to B is positive, and B AA B V VV is positive.) 19. INTERPRET This problem is about conversion of units. DEVELOP 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). =⋅ EVALUATE Combining the two expressions gives1 V 1 J/C 1 N m/C. = It follows that1V /m 1N/C. = ASSESS These are the units for the electric field strength. 20. For in the direction of a uniform electric field, Equation 22.2b gives| . r Δ K | (650 N/C)(1.4 m) 910 V VE r Δ =Δ= = K (See note in solution to Exercise 16. Since , dV E dr = −⋅ G K the potential always decreases in the direction of the electric field.) 21. INTERPRET This problem is about the work done by the battery to move the charge from the positive terminal to the negative terminal. DEVELOP The work done by the battery is equal to the kinetic energy gained by the charge, and is equal to Wq V . EVALUATE Substituting the values given, we have (3.1 C) (9.0 V) 27.9 J =Δ= = ASSESS 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 (see Example 22.1). The proton and singly-ionized helium atom have charge e , so they gain100 while the qV Δ 19 17 eV (1.6 10 C)(100 V) 1.6 10 J, = × -particle α has charge 2 e and gains twice this energy. Section 22.2 Calculating Potential Difference

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22.2 Chapter 22 23. INTERPRET In this problem we are given a uniform electric field and asked to calculate the potential difference between two points. DEVELOP For a uniform field, the potential difference between two points a and b is given by Equation 22.1b: ab b a VV V E r Δ =−= Δ G G where is a vector from a to b .
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## chap22 - ELECTRIC POTENTIAL 22 EXERCISES Section 22.1...

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