chap23 - 23.1 J 357 m 150 1 m 354 1 C 30 4 C 40 2 1 1 1 2 2...

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Unformatted text preview: 23.1: J 357 . m 150 . 1 m 354 . 1 ) C 30 . 4 )( C 40 . 2 ( 1 1 1 2 2 1 = -- = - = ∆ μ μ k r r q kq U J. 357 .- = ∆- = ⇒ 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 . J) 491 . J 608 . ( 2 2 1 J 608 . m 800 . ) C 10 50 . 7 )( C 10 80 . 2 ( ) s m . 22 )( kg 0015 . ( 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 . J 608 . ) C 10 80 . 7 )( C 10 80 . 2 ( J 608 . 6 6 2 1 = × × = ⇒ = ⇒-- k r r q kq 23.4: . m 373 . J 400 . C) 10 20 . 7 )( C 10 30 . 2 ( J 400 . 6 6 2 1 =- × ×- = ⇒ =- =-- k r r q kq U 23.5: a) . J 199 . m 250 . ) C 10 20 . 1 ( ) C 10 60 . 4 ( 6 6 = × × = =-- k r kQq U s. m 6 . 37 , J 198 . ) iii ( s. m 7 . 36 J, 189 . ) ii ( s. m 6 . 26 kg 10 80 . 2 J) 0994 . ( 2 2 1 J 0994 . J 0994 . m 5 . 1 m 25 . 1 ) C 10 20 . 1 ( C) 10 60 . 4 ( J (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 . C) 10 2 . 1 ( 6 6 m 500 . 2 m 500 . 2 6 2 2 2 = × = = + =- k kq kq kq U 23.7: a) J. 10 60 . 3 ) m 100 . ( ) nC 00 . 2 )( nC 00 . 3 ( ) m 100 . ( ) nC 00 . 2 )( nC 00 . 4 ( ) m 200 . ( 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 . , 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 . , m 074 . 6 . 1 26 60 2 . 6 8 60 2 = ⇒ = +- ⇒-- +- = x x x x x Therefore m 074 . = 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- = ⇒ =- ⇒ = 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 . 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 ; 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 2 1 = × = × = = + + =-- U r r e πε r r r πε e U 23.12: Get closest distance .Get closest distance ....
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chap23 - 23.1 J 357 m 150 1 m 354 1 C 30 4 C 40 2 1 1 1 2 2...

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