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 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 U0 qEx where x is the distance from the negative plate and U0 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 Up Uf Ui U0 0 J U0 qEd qEd K K K f i 1 2 2 mv f 1 2 2 mv i 1 2 2 mv f The law of conservation of energy is K Up 0 J This means v f 2 qEd m 2 1 60 10 19 1 2 2 mv f qEd C 50 000 N C 2 0 10 m 1 67 10 0 J kg 3 27 1 38 10 m s 5 Assess As described in Section 29 1 the potential energy for a charge q in an electric field E is U U0 qEx where x is the distance measured from the negative plate Having U U0 at the negative plate with x 0 m is completely arbitrary We could have taken it to be zero Note that only U and not U has physical consequences 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 vi 0 m s and a final speed vf 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 U0 qEx where x is the distance from the negative plate and U0 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 Ue Uf Ui U0 qEd U0 0 J qEd K K K f i 1 2 2 mv f 1 2 2 mv i 1 2 2 mv f Now the law of conservation of mechanical energy gives K U 0 J This means v f qEd 2 m 2 1 60 10 19 C 20 000 N C 1 0 10 m 3 2 7 10 m s 6 1 2 2 mv f 0 J qEd 9 11 10 31 kg Assess Note that Ue 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 U0 qEx where x is the distance from the negative plate and U0 is the potential energy at the negative plate at x 0 m Thus the change in the potential energy of the proton is U U U p f U 0 i 0 J U qEd 0 qEd The change in the kinetic energy of the proton is K K K f i 1 2 2 mv f 1 2 2 mv i 1 2 2 mv f Applying the law of conservation of energy K Up 0 J we have 2 1 2 2 mv f qEd 0 J 2 v f qEd m When the amount of charge on each plate is doubled then the final velocity of the proton is Dividing these equations 2 v f 2qE d m 2 v f 2 v f E E v f E E v f For a parallel plate capacitor E 0 Q A 0 Therefore v f v f 2 50 000 m s 70 711 m s Q Q Assess The proton s velocity is expected to increase because an increased charge on the capacitor plates leads to a higher electric field between the plates and hence to an increased force on the proton 29 4 Model The mechanical energy of the charged particles is conserved A parallel plate capacitor has a uniform electric field Visualize The figure shows the before and after pictorial representation Solve The potential energy is defined as U U0 qEx where x is the distance from the negative plate and U0 is the potential energy at the negative plate at x 0 m Thus the change in the potential energy of the proton as it moves from the positive plate to the negative plate is U U U p f U 0 i 0 J U eEd 0 eEd This decrease in potential energy appears as an increase in the proton s kinetic energy K K K i f 1 2 m v 2 p f p 1 2 m v 2 p i p 1 2 m v 2 p f p Applying the law of conservation of mechanical energy K Up 0 J we have When the proton is replaced with a helium ion and the same experiment is repeated Dividing the two equations 1 2 mv 2 f p eEd 0 J 2 v f p 2 eEd m p 2 v f ion 2eEd m ion v f ion v f p 1 4 50 000 m s 25 000 m s m p m ion Assess Being a heavier particle the helium ion s velocity is expected to be smaller compared to the proton s velocity 29 5 Model The charges are point charges Visualize Please refer to Figure EX29 5 Solve The electric potential energy of the electron is U electron U 13 U 23 9 9 0 10 N m C 1 12 10 19 19 2 2 1 60 10 J 1 12 10 J 19 2 0 10 m 9 2 1 60 10 C 2 24 10 J 19 0 50 10 m 9 19 C 2 19 1 60 10 2 0 10 m 9 C 2 1 60 10 0 50 10 m 9 19 C 2 29 6 Model The charges are point charges Visualize Please refer to Figure EX29 6 Solve For a system of point charges the potential energy is the sum of the potential energies due to all pairs of charges U elec Kq q i j r ij i j U 12 U 13 U 23 K K K q q 1 2 r 12 q q 1 3 r 13 q …
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