s21 - Solutions to Homework#2 Chapter 20 r ^ 10 Picture the...

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Solutions to Homework #2 Chapter 20 10. Picture the Problem : A 12.5 µ C charge moves in a uniform electric field ( 29 ˆ 6350 N/C = E x r . Strategy: The change in electric potential energy of a charge that moves against an electric field is given by equation 20-1, 0 U q Ed = . If the charge moves in the same direction as the field, the work done by the field is positive and 0 U W Fd q Ed = - = - = - . The sign changes if the charge moves in the opposite direction as the field. If the charge moves in a direction perpendicular to the field, the work and change in potential energy are both zero. Solution: 1. (a) The charge moves in the same direction as the field: ( 29 ( 29 ( 29 6 0 3 12.5 10 C 6350 N/C 0.0550m 4.37 10 J 4.37 mJ U q Ed - - = - = - = - = - 2. (b) The charges moves in the direction opposite to the field: ( 29 ( 29 ( 29 6 0 3 12.5 10 C 6350N/C 0.0550m 4.37 10 J 4.37 mJ U q Ed - - = = = = 3. (c) The charge moves perpendicular to the field: 0 U = Insight: Note that U changes only along the direction parallel to the field. This is consistent with the statement, “The electric field points downhill on the potential surface,” as described by figure 20-3 and the accompanying text. 22. Picture the Problem : The speed of a proton is changed as it moves through a potential difference. Strategy: Use conservation of energy, with the potential energy of the proton given by , U qV eV = = to determine the potential difference required to change the proton’s speed and kinetic energy. Solution: 1. (a) Set i i f f K U K U + = + and solve for V : ( 29 ( 29 ( 29 ( 29 ( 29 2 2 1 1 i i f f 2 2 2 2 1 1 i f f i 2 2 2 2 2 27 5 i f 19 1.673 10 kg 4.0 10 m/s 0 2 2 1.60 10 C 840 V 0.84 kV mv eV mv eV mv mv e V V e V m v v V e - - + = + - = - = ∆ - - = = = = 2. (b) Find V for when 1 f i 2 v v = : ( 29 ( 29 ( 29 ( 29 2 2 2 2 i i 1 i i 2 2 27 5 19 3 1 1 2 2 4 8 3 1.673 10 kg 4.0 10 m/s 630 V 0.63 kV 8 1.60 10 C mv mv m V v v e e e - - = - = - = = = = 3. (c) Set 1 i i i f 2 K U K U + = + and solve for V : ( 29 ( 29 ( 29 1 i i i f 2 1 i 2 2 27 5 2 i i 19 1.673 10 kg 4.0 10 m/s 0.42 kV 2 4 4 1.60 10 C K eV K eV K e V K mv V e e - - + = + = ∆ = = = = Insight: It requires a smaller potential difference to cut the kinetic energy in half than it does to cut the speed in half because the kinetic energy is proportional to the speed squared; cutting the speed in half reduces K by a factor of four.
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24. Picture the Problem : For the figure shown at right, it is given that q 1 = + Q . Strategy:
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This note was uploaded on 02/10/2012 for the course PHY 102 taught by Professor Korotkova during the Spring '09 term at University of Miami.

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s21 - Solutions to Homework#2 Chapter 20 r ^ 10 Picture the...

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