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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 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 C . 10. The electric field at the spherical surface is E = Q /4p r 2 , while the potential of a sphere, with V = 0 at 8, is V = Q /4p 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 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 /2 . 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 = ( / 2 ) x , or V = V ( /2 ) x ....
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 Spring '09
 Fielding
 Physics, Electric Potential, Energy, Work

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