YF_ISM_23

YF_ISM_23 - ELECTRIC POTENTIAL 23 ra = 0.150 m rb = (0.250...

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23-1 E LECTRIC P OTENTIAL 23.1. IDENTIFY: Apply Eq.(23.2) to calculate the work. The electric potential energy of a pair of point charges is given by Eq.(23.9). SET UP: 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 = 22 (0.250 m) (0.250 m) b r =+ 0.3536 m b r = Figure 23.1 EXECUTE: ab a b WU U =− 66 92 12 0 1 ( 2.40 10 C)( 4.30 10 C) (8.988 10 N m /C ) 40 . 1 5 0 m a a qq U r π −− 2 −× == × P 0.6184 J a U 0 1 . 3 5 3 6 m b b U r 2 × P 0.2623 J b U 0.6184 J ( 0.2623 J) 0.356 J a b U = − − EVALUATE: 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. IDENTIFY: Apply . a b U SET UP: 8 5.4 10 J. a U × Solve for . b U EXECUTE: 8 1.9 10 J . a b U × = 88 8 1 . 91 0 J (5 . 41 J ) 7 . 31 J . baa b UUW EVALUATE: When the electric force does negative work the electrical potential energy increases. 23.3. IDENTIFY: The work needed to assemble the nucleus is the sum of the electrical potential energies of the protons in the nucleus, relative to infinity. SET UP: The total potential energy is the scalar sum of all the individual potential energies, where each potential energy is 00 (1/ 4 )( / ). Uq q r = P Each charge is e and the charges are equidistant from each other, so the total potential energy is 222 2 13 . 44 eee e U rrr r ππ ⎛⎞ + = ⎜⎟ ⎝⎠ PP EXECUTE: Adding the potential energies gives 21 9 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 42 . 0 0 1 0 m e U r ×× = × = × P EVALUATE: 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. IDENTIFY: 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) SET UP: Using the potential energy of a pair of point charges relative to infinity, 00 (1/ 4 )( / ). Uq q r π = P we have 22 21 02 1 1 . 4 ee WU U U rr ⎛⎞ = = ⎜⎟ ⎝⎠ P EXECUTE: Factoring out the e 2 and substituting numbers gives () ( ) 2 92 2 1 9 14 15 15 11 9.00 10 N m /C 1.60 10 C 7.68 10 J 3.00 10 m 2.00 10 m W −− =×⋅ × = × ×× (b) SET UP: The protons have equal momentum, and since they have equal masses, they will have equal speeds and hence equal kinetic energy. 12 1 . 2 U K K K mv mv Δ= + = = = EXECUTE: Solving for v gives 14 27 7.68 10 J 1.67 10 kg U v m Δ× == × = 6.78 × 10 6 m/s EVALUATE: 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) IDENTIFY: Use conservation of energy: other aa bb KUW KU ++ =+ U for the pair of point charges is given by Eq.(23.9). SET UP: 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. Figure 23.5a EXECUTE: Only the electric force does work, so other 0 W = and 0 1 .
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YF_ISM_23 - ELECTRIC POTENTIAL 23 ra = 0.150 m rb = (0.250...

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