Chapter 29

# Chapter 29 - 29.1 Model The mechanical energy of the proton...

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29.1. Model: The mechanical energy of the proton is conserved. A parallel-plate capacitor has a uniform electric field. Visualize: The figure shows the before-and-after pictorial representation. The proton has an initial speed v i = 0 m/s and a final speed v f after traveling a distance d = 2.0 mm. Solve: The proton loses potential energy and gains kinetic energy as it moves toward the negative plate. The potential energy U is defined as U = U 0 + qEx , where x is the distance from the negative plate and U 0 is the potential energy at the negative plate (at x = 0 m). Thus, the change in the potential energy of the proton is Δ U p = U f – U i = ( U 0 + 0 J) – ( U 0 + qEd ) = qEd The change in the kinetic energy of the proton is 22 2 11 1 fi f i f KK K m v m v Δ= − = = The law of conservation of energy is Δ K + Δ U p = 0 J. This means ( ) () 2 1 f 2 19 3 5 f 27 0 J 2 1.60 10 C 50,000 N/C 2.0 10 m 2 1.38 10 m/s 1.67 10 kg mv qEd qEd v m −− + −= × ⇒= = = × × Assess: As described in Section 29.1, the potential energy for a charge q in an electric field E is U = U 0 + qEx , where x is the distance measured from the negative plate. Having U = U 0 at the negative plate (with x = 0 m) is completely arbitrary. We could have taken it to be zero. Note that only Δ U , and not U , has physical consequences.

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29.2. Model: The mechanical energy of the electron is conserved. A parallel-plate capacitor has a uniform electric field. Visualize: The figure shows the before-and-after pictorial representation. The electron has an initial speed v i = 0 m/s and a final speed v f after traveling a distance d = 1.0 mm. Solve: The electron loses potential energy and gains kinetic energy as it moves toward the positive plate. The potential energy U is defined as U = U 0 + qEx , where x is the distance from the negative plate and U 0 is the potential energy at the negative plate (at x = 0 m). Thus, the change in the potential energy of the electron is Δ U e = U f – U i = ( U 0 + qEd ) – ( U 0 + 0 J) = qEd The change in the kinetic energy of the electron is 22 2 11 1 fi f i f KK K m v m v Δ= − = = Now, the law of conservation of mechanical energy gives Δ K + Δ U = 0 J. This means () 2 1 f 2 19 3 6 f 31 0 J 2 1.60 10 C 20,000 N/C 1.0 10 m 2 2.7 10 m/s 9.11 10 kg mv qEd qEd v m −− + = −− × × ⇒= = = × × Assess: Note that Δ U e = qEd is the change in the potential energy of the electron. It is negative because q = e for the electron. Thus, the potential energy becomes more negative as d increases, that is, the potential energy of the electron decreases with an increase in d (or x ).
29.3. Model: The mechanical energy of the proton is conserved. A parallel-plate capacitor has a uniform electric field. Visualize: The figure shows the before-and-after pictorial representation. Solve: The proton loses potential energy and gains kinetic energy as it moves toward the negative plate. The potential energy is defined as U = U 0 + qEx , where x is the distance from the negative plate and U 0 is the potential energy at the negative plate (at x = 0 m). Thus, the change in the potential energy of the proton is ( ) ( ) pfi 0 0 0 J UUU U Uq E d q E d Δ=−= + − + =

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Chapter 29 - 29.1 Model The mechanical energy of the proton...

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