Chap20Sol

# Chap20Sol - Chapter 20 SOLUTIONS TO PROBLEMS Easy P20.1(a...

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Chapter 20 1 SOLUTIONS TO PROBLEMS Easy P20.1 (a) Energy of the proton-field system is conserved as the proton moves from high to low potential, which can be defined for this problem as moving from 120 V down to 0 V. K i + U i + ! E mech = K f + U f 0 + qV + 0 = 1 2 mv p 2 + 0 1.60 ! 10 " 19 C ( ) 120 V ( ) 1 J 1 V # C \$ % & ' ( ) = 1 2 1.67 ! 10 " 27 kg ( ) v p 2 v p = 1.52 ! 10 5 (b) The electron will gain speed in moving the other way, from V i = 0 to V f = 120 V : K i + U i + ! E mech = K f + U f 0 + 0 + 0 = 1 2 mv e 2 + qV 0 = 1 2 9.11 ! 10 " 31 kg ( ) v e 2 + " 1.60 ! 10 " 19 C ( ) ( ) v e = 6.49 ! 10 6 P20.2 ! V = " 14.0 V and Q = ! N A e = ! 6.02 " 10 23 ( ) 1.60 " 10 ! 19 ( ) = ! 9.63 " 10 4 C ! V = W Q , so W = Q ! V = " 9.63 # 10 4 C ( ) " ( ) = 1.35 MJ P20.4 E = ! V d = 25.0 " 10 3 1.50 " 10 # 2 m = 1.67 " 10 6 =

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2 Electric Potential and Capacitance P20.7 (a) The potential at 1.00 cm is V 1 = k e q r = 8.99 ! 10 9 N " m 2 2 ( ) 1.60 ! 10 # 19 C ( ) 1.00 ! 10 # 2 m = 1.44 ! 10 # 7 V . (b) The potential at 2.00 cm is V 2 = k e q r = 8.99 ! 10 9 N " m 2 C 2 ( ) 1.60 ! 10 # 19 C ( ) 2.00 ! 10 # 2 m = 0.719 ! 10 # 7 V . Thus, the difference in potential between the two points is ! V = V 2 " V 1 = " 7.19 # 10 " 8 V . (c) The approach is the same as above except the charge is ! 1.60 " 10 ! 19 C . This changes the sign of each answer, with its magnitude remaining the same. That is, the potential at 1.00 cm is ! 1.44 " 10 ! 7 V . The potential at 2.00 cm is ! 0.719 " 10 ! 7 V , so ! V = V 2 " V 1 = 7.19 # 10 " 8 V . P20.31 (a) Q = C ! V = 4.00 " 10 # 6 F ( ) 12.0 V ( ) = 4.80 " 10 # 5 C = 48.0 μ C (b) Q = C ! V = 4.00 " 10 # 6 F ( ) 1.50 V ( ) = 6.00 " 10 # 6 C = 6.00 C P20.32 (a) C = Q ! V = 10.0 " 10 # 6 C 10.0 V = 1.00 " 10 # 6 F = 1.00 F (b) ! V = Q C = 100 " 10 # 6 C 1.00 " 10 # 6 F = 100 V P20.39 (a) Capacitors in parallel add. Thus, the equivalent capacitor has a value of C eq = C 1 + C 2 = 5.00 F + 12.0 F = 17.0 F . (b) The potential difference across each branch is the same and equal to the voltage of the battery. ! V = 9.00 V (c) Q 5 = C ! V = 5.00 F ( ) 9.00 V ( ) = 45.0 C and Q 12 = C ! V = 12.0 F ( ) 9.00 V ( ) = 108 C P20.47 (a) U = 1 2 C ! V ( ) 2 = 1 2 3.00 F ( ) 12.0 V ( ) 2 = 216 J (b) U = 1 2 C ! V ( ) 2 = 1 2 3.00 F ( ) 6.00 V ( ) 2 = 54.0 J