Ch 29 Physics for Scientists and Engineers

Ch 29 Physics for Scientists and Engineers - THEELECTRIC...

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THE ELECTRIC POTENTIAL 29.1. Model: The mechanical energy of the proton is conserved. A parallel plate capacitor has a uniform electric field. Visualize: After Before * . v =o . E‘ * I , I x 0 1 .o 2.0 The figure shows the before-and-after pictorial representation. The proton has an initial speed vi = 0 m/s and a final speed vf 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, + qEx, where x is the distance from the negative plate and is the potential energy at the negative plate (at x = 0 m). Thus, the change in the potential energy of the proton is The change in the kinetic energy of the proton is The law of conservation of energy is AK + AUp = 0 J. This means 2(+1.60 x C)(50,000 N / C)(2.0 x m) (1.67 x IO-*’ kg) =1.38x1OS Assess: As described in Section 29.1, the potential energy for a charge 4 in an electric field E is U = + qEx, where x is the distance measured from the negative plate. Having U = Uo at the negative plate (with x = 0 m) is completely arbitrary. We could have taken it to be zero. Note that only AU, and not has physical consequences. 29.2. Model: electric field. Visualize: Before After The mechanical energy of the electron is conserved. A parallel plate capacitor has a uniform + - I=+ * E - + - , I x (-) 0 0.5 1.0 mm 29-1
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29-2 Chapter 29 The figure shows the before-and-after pictorial representation. The electron has an initial speed vi = 0 m/s and a final speed v, after traveling a distance d = 1 .O 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, +qEx, where x is the distance from the negative plate and U,, is the potential energy at the negative plate (at x = 0 m). Thus, the change in the potential energy of the electron is The change in the kinetic energy of the electron is = Kf - K, = +mv: - 3mv' 1 - - 1 2 mv: Now, the law of conservation of mechanical energy gives AK + AU = 0 J. This means + qEd = 0 J (-2)(-1.60 x C)(20,000 N / C)(l.O xIO-~ m) =2.65x106 9.11~10"' kg Assess: Note that AU, = 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: field. Visualize: The mechanical energy of the proton is conserved. A parallel plate capacitor has a uniform electric Charge on each plate is Q 2Q After Before After Before 4 v;=o v,=o *2 2 4 4 ! X X 0 d 0 d 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, + qEx, where x is the distance from the negative plate and is the potential energy at the negative plate (at x = 0 m). Thus, the change in the potential energy of the proton is AUp = -Vi = (U, +O J)-(U, ++Ed)= -qEd The change in the kinetic energy of the proton is AK = - Ki = +mv: -+mv?
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This homework help was uploaded on 04/07/2008 for the course PHSX 211/212 taught by Professor Medvedev during the Spring '08 term at Kansas.

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Ch 29 Physics for Scientists and Engineers - THEELECTRIC...

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