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Chapter 23 - 23.1 W U kq 1 q 2 1 r2 1 r1 k 2 40 C 4 30 C 1...

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23.1: J 357 . 0 m 150 . 0 1 m 354 . 0 1 ) C 30 . 4 )( C 40 . 2 ( 1 1 1 2 2 1 = - - = - = μ μ k r r q kq U J. 357 . 0 - = - = U W 23.2: = × + × = - = - = × - = - - - J 10 4 . 5 J 10 9 . 1 J 10 9 . 1 8 8 8 f f i U U U U W J 10 3 . 7 8 - × 23.3: a) . s m 5 . 12 kg 0015 . 0 J) 491 . 0 J 608 . 0 ( 2 2 1 J 608 . 0 m 800 . 0 ) C 10 50 . 7 )( C 10 80 . 2 ( ) s m 0 . 22 )( kg 0015 . 0 ( 2 1 2 1 2 6 6 2 = - = + = = = × × + = + = - - f f f f i i i i v r q kq mv E E k U K E b) At the closest point, the velocity is zero: m. 323 . 0 J 608 . 0 ) C 10 80 . 7 )( C 10 80 . 2 ( J 608 . 0 6 6 2 1 = × × = = - - k r r q kq 23.4: . m 373 . 0 J 400 . 0 C) 10 20 . 7 )( C 10 30 . 2 ( J 400 . 0 6 6 2 1 = - × × - = = - = - - k r r q kq U 23.5: a) . J 199 . 0 m 250 . 0 ) C 10 20 . 1 ( ) C 10 60 . 4 ( 6 6 = × × = = - - k r kQq U s. m 6 . 37 , J 198 . 0 ) iii ( s. m 7 . 36 J, 189 . 0 ) ii ( s. m 6 . 26 kg 10 80 . 2 J) 0994 . 0 ( 2 2 1 J 0994 . 0 J 0994 . 0 m 5 . 0 1 m 25 . 0 1 ) C 10 20 . 1 ( C) 10 60 . 4 ( J 0 (i) b) 4 2 6 6 = = = = = × = = = = - × × + = - + = - - - f f f f f f f f i i f v K v K v mv K k U U K K 23.6: . J 078 . 0 C) 10 2 . 1 ( 6 6 m 500 . 0 2 m 500 . 0 2 6 2 2 2 = × = = + = - k kq kq kq U
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23.7: a) J. 10 60 . 3 ) m 100 . 0 ( ) nC 00 . 2 )( nC 00 . 3 ( ) m 100 . 0 ( ) nC 00 . 2 )( nC 00 . 4 ( ) m 200 . 0 ( nC) 00 . 3 )( nC 00 . 4 ( 7 23 2 1 13 2 1 12 2 1 - × - = - + + - = + + = k r q q r q q r q q k U . 0 , 0 b) 12 3 2 3 1 12 2 1 - + + = = x r q q x q q r q q k U If So solving for x we find: . m 360 . 0 , m 074 . 0 0 6 . 1 26 60 2 . 0 6 8 60 0 2 = = + - - - + - = x x x x x Therefore m 074 . 0 = x since it is the only value between the two charges. 23.8: From Example 23.1, the initial energy i E can be calculated: J. 10 09 . 5 m 10 ) C 10 20 . 3 )( C 10 60 . 1 ( ) s m 10 00 . 3 )( kg 10 11 . 9 ( 2 1 19 10 19 19 2 6 31 - - - - - × - = × × - + × × = + = i i i i E k U K E When velocity equals zero, all energy is electric potential energy, so: . m 10 06 . 9 2 J 10 09 . 5 10 2 19 - - × = - = × - r r e k 23.9: Since the work done is zero, the sum of the work to bring in the two equal charges q must equal the work done in bringing in charge Q . . 2 2 2 q Q d kqQ d kq W W qQ qq - = = - =
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23.10: The work is the potential energy of the combination. J 10 31 . 7 3 2 2 m 10 5 ) C 10 6 . 1 ( ) C Nm 10 0 . 9 ( 2 1 2 2 m 10 5 m 10 5 ) 2 ( ) ( m 10 5 ) ( m 10 2 5 ) 2 ( 19 10 2 19 2 2 9 10 2 10 10 10 - - - - - - - × - = - × × × = - - × = × - + × - + × = + + = ke e e k e ke e ke U U U U e pe p Since U is negative, we want do J 10 31 . 7 19 - × + to separate the particles 23.11: 1 2 2 1 2 2 1 1 so 0 ; U K U K U K U K = = = + = + eV 9.00 J 10 44 . 1 m 10 00 . 8 with , 5 4 1 2 2 1 4 18 1 10 2 0 0 2 1 = × = × = = + + = - - U r r e πε r r r πε e U 23.12: Get closest distance . γ Energy conservation: γ ke mv mv 2 2 2 2 1 2 1 = + m 10 38 . 1 ) s m 10 ( ) kg 10 67 . 1 ( ) C 10 6 . 1 ( ) C Nm 10 9 ( 13 6 27 2 19 2 2 9 2 2 - - - × = × × × = = mv ke γ Maximum force: N 012 . 0 ) m 10 38 . 1 ( ) C 10 6 . 1 ( ) C Nm 10 9 ( 2 13 2 19 2 2 9 2 2 = × × × = = - - γ ke F 23.13: B B A A U K U K + = + s m 42 . 7 2 J 0.00550 V) 800 V (200 C) 10 5.00 ( J 00250 . 0 ) ( so , 6 = = = - × - + = - + = + = + = - m K v V V q K K qV K qV K qV U B B B A A B B B A A It is faster at B; a negative charge gains speed when it moves to higher potential.
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23.14: Taking the origin at the center of the square, the symmetry means that the potential is the same at the two corners not occupied by the C 00 . 5 μ + charges (The work done in moving to either corner from infinity is the same). But this also means that no net work is done is moving from one corner to the other.
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