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EXERCISE 16.5 Starting from the initial configuration of three protons, suppose the end two particles are released simultane ously and the middle particle is fixed. Obtain a numerical answer for the speed of the two particles at infinity. (Note that their speeds, by symmetry, must be the same.) ANSWER 1.31 × 107 m/s

EXERCISE 16.5 Starting from the initial configuration of three protons, suppose the end two particles are released simultane ously and the middle particle is fixed. Obtain a numerical answer for the speed of the two particles at infinity. (Note that their speeds, by symmetry, must be the same.) ANSWER 1.31 × 107 m/s

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EXAMPLE 16.5 ELECTRIC POTENTIAL ENERGY AND DYNAMICS GOAL Apply conservation of energy and electrical potential energy to a configuration of charges. 91 PROBLEM Suppose three protons lie on the x-axis, at rest relative to one another at a given instant of time, as in Figure 16.9. If proton g on the right is released while the others are held fixed in place, find a symbolic expression for the proton's speed at infin- ity and evaluate this speed when n = 2.00 fm. (Note: 1 fm = 10-15 m.) Figure 16.9 (Example 16.5) STRATEGY First calculate the initial electric potential energy associated with the system of three particles. There will be three terms, one for each interacting pair. Then calculate the final electric potential energy associated with the system when the proton on the right is arbitrarily far away. Because the electric potential energy falls off as 1/r, two of the terms will vanish. Using conser- vation of energy then yields the speed of the particle in question. SOLUTION ke k,e ke Calculate the electric potential energy associated with the initial configuration of charges: PE, = + "13 T23 To 2r. To k. qı 92 PE, = ke? Calculate the electric potential energy associated with the final configuration of charges: To Write the conservation of energy equation: AKE + APE = KE, - KE, + PE, - PE, = 0 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or dupijRgisdukerwise ka reu, jalieantentER th8 BaYOROCengage Learning. 16.3 | Potentials, Charged Conductors, and Equipotential Surfaces 537 Substitute appropriate terms: k,e? mzv,? – 0+ (k,e² + k,e? ke = 0 ro 2r. (k,e² k,e" = 0 - 3k,e Solve for v, after combining the two remaining potential energy terms: Uy = 3(8.99 x 10° N ·m²/C³)(1.60 × 10-19 C)² (1.67 x 10-27 kg)(2.00 x 10-15 m) Evaluate taking r = 2.00 fm: Vz = 1.44 x 107 m/s REMARKS The difference in the initial and final potential energies yields the energy available for motion. This calculation is somewhat contrived because it would be difficult, although not impossible, to arrange such a configuration of protons; it could conceivably occur by chance inside a star. QUESTION 16.5 If a fourth proton were placed to the right of q3, how many additional potential energy terms would have to be calculated in the initial configuration? EXERCISE 16.5 Starting from the initial configuration of three protons, suppose the end two particles are released simultane- ously and the middle particle is fixed. Obtain a numerical answer for the speed of the two particles at infinity. (Note that their speeds, by symmetry, must be the same.) ANSWER 1.31 × 107 m/s
Transcribed Image Text:EXAMPLE 16.5 ELECTRIC POTENTIAL ENERGY AND DYNAMICS GOAL Apply conservation of energy and electrical potential energy to a configuration of charges. 91 PROBLEM Suppose three protons lie on the x-axis, at rest relative to one another at a given instant of time, as in Figure 16.9. If proton g on the right is released while the others are held fixed in place, find a symbolic expression for the proton's speed at infin- ity and evaluate this speed when n = 2.00 fm. (Note: 1 fm = 10-15 m.) Figure 16.9 (Example 16.5) STRATEGY First calculate the initial electric potential energy associated with the system of three particles. There will be three terms, one for each interacting pair. Then calculate the final electric potential energy associated with the system when the proton on the right is arbitrarily far away. Because the electric potential energy falls off as 1/r, two of the terms will vanish. Using conser- vation of energy then yields the speed of the particle in question. SOLUTION ke k,e ke Calculate the electric potential energy associated with the initial configuration of charges: PE, = + "13 T23 To 2r. To k. qı 92 PE, = ke? Calculate the electric potential energy associated with the final configuration of charges: To Write the conservation of energy equation: AKE + APE = KE, - KE, + PE, - PE, = 0 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or dupijRgisdukerwise ka reu, jalieantentER th8 BaYOROCengage Learning. 16.3 | Potentials, Charged Conductors, and Equipotential Surfaces 537 Substitute appropriate terms: k,e? mzv,? – 0+ (k,e² + k,e? ke = 0 ro 2r. (k,e² k,e" = 0 - 3k,e Solve for v, after combining the two remaining potential energy terms: Uy = 3(8.99 x 10° N ·m²/C³)(1.60 × 10-19 C)² (1.67 x 10-27 kg)(2.00 x 10-15 m) Evaluate taking r = 2.00 fm: Vz = 1.44 x 107 m/s REMARKS The difference in the initial and final potential energies yields the energy available for motion. This calculation is somewhat contrived because it would be difficult, although not impossible, to arrange such a configuration of protons; it could conceivably occur by chance inside a star. QUESTION 16.5 If a fourth proton were placed to the right of q3, how many additional potential energy terms would have to be calculated in the initial configuration? EXERCISE 16.5 Starting from the initial configuration of three protons, suppose the end two particles are released simultane- ously and the middle particle is fixed. Obtain a numerical answer for the speed of the two particles at infinity. (Note that their speeds, by symmetry, must be the same.) ANSWER 1.31 × 107 m/s
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