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Unformatted text preview: CHAPTER 23 – Electric Potential 1. We find the work done by an external agent from the workenergy principle: W ab = ? K + ? U = 0 + q ( V b – V a ) = ( – 7 . × 10 –6 C)(+ 6.00 V – 0) = – 4.2 × 10 –5 J (done by the field). 2. We find the work done by an external agent from the workenergy principle: W = ? K + ? U = 0 + q ( V b – V a ) = ( 1 . 6 × 10 –19 C)[(– 50 V) – (+ 100 V)] = – 2.40 × 10 –17 J (done by the field). 3. Because the total energy of the electron is conserved, we have ? K + ? U = 0, or ? K = – q ( V b – V a ) = – (– 1.60 × 10 –19 C)(21,000 V) = 3.4 × 10 –15 J . 4. Because the total energy of the electron is conserved, we have ? K + ? U = 0; ? K + q ( V B – V A ) = 0; 16.4 × 10 –16 J + (– 1.60 × 10 –19 C)( V B – V A ) = 0, which gives V B – V A = 1.03 × 10 4 V . Plate B is at the higher potential. 5. We find the potential difference from the workenergy principle: W ab = ? K + ? U = ? K + q ( V b – V a ) 8.00 × 10 –4 J = 2.10 × 10 –4 J + (– 8.10 × 10 –6 C)( V b – V a ), which gives V b – V a = – 72.8 V, or V a – V b = + 72.8 V . 6. For the uniform electric field between two large, parallel plates, we have E = ? V / d ; 1500 V/m = (45 V)/ d , which gives d = 3.0 × 10 –2 m = 3.0 cm . 7. For the uniform electric field between two large, parallel plates, we have E = ? V / d ; 640 V/m = ? V /(11.0 × 10 –3 m), which gives ? V = 7.04 V . 8. For the uniform electric field between two large, parallel plates, we have E = ? V / d = (110 V)/(5.0 × 10 –3 m) = 2.2 × 10 4 V/m . 9. The maximum charge will produce the electric field that causes breakdown in the air: E = Q /4p Å r 2 ; 3 × 10 6 V/m = (9.0 × 10 9 N · m 2 /C 2 ) Q /(0.050 m) 2 , which gives Q = 8 × 10 –7 C = 0.8 µ C . 10. The electric field at the spherical surface is E = Q /4p Å r 2 , while the potential of a sphere, with V = 0 at 8, is V = Q /4p Å r . Thus we have r = V / E , so r min = V / E max = (30,000 V)/(3 × 10 6 V/m) = 1 × 10 –2 m = 1 cm . At this radius the charge is Q = 4p Å r min V = (1 × 10 –2 m)(30,000 V)/(9.0 × 10 9 N · m 2 /C 2 ) = 3 × 10 –8 C . 11. The potential difference between two points in an electric field is found from ? V = – ? E · d ¬ . ( a ) For V BA we have V BA = – E · d A B = – – 300 N/ C i · d y j A B = . ( b ) For V CB we have V CB = – E · d B C = – – 300 N/ C i · d x i B C = 3 N/ C d x 4 m – 3 m = 300 N/ C – 3 m – 4 m = – 2100 V . ( c ) For V CA we have V CA = – E · d A C = – – 300 N/ C i · d x i + d y j A C = 3 N/ C d x 4 m – 3 m = 300 N/ C – 3 m – 4 m = –2100 V . Note that V CA = V CB + V BA . 12. The electric field produced by a large plate is uniform with magnitude σ /2 Å . If we take the potential of the plate to be V , we find the potential a distance x from the plate by integrating: V – V = – E · d x = – E d x x = – ( σ / 2 Å ) x , or V = V – ( σ /2 Å ) x ....
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.
 Spring '09
 Fielding
 Physics, Electric Potential, Energy, Work

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