tailieunhanh - Physics exercises_solution: Chapter 23

Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 23 | AU kqxq2 -- C C i-------------------- 1 JT -At J. W 8J -AU U -Uf Uf 10 SJ 10 SJ 1 J J 8J a E K U 1 00015 kg 22 0 m s C 7 50xlC C 608 2 m r 1 2 kqAq2 2 J - J c . Et Ef mvf - -E- v p---------------1 m s. f 2 f rf f kg 1 b At the closest point the velocity is zero 1 1 . . 2 80 x 10- C 7 80 x 10 6C _ Q m r J C -0400J r -t 230xl0-6C 720xl0 C 0373m r J a t OxlO C O C _ Q r m b i Kf K U -Uf 0 J 1 x 10 C x 10 C -- 0 0994 J mJ Kf J -mv i v I 2 0 0994 J m s 7 2 f f kg ii Kf J vf s. iii Kf J vf 37 6m s. kq2 2kq2 m 6kq2 6 C 2 J. a nC nC nC nC U k qq qq qq k r r r V r12 r13 r23 J m V nC nC m J x 10 7 J. b IfU 0 0 kI qq I. So solving for x we find V ri2 X ri2 XJ 0 60 8------6 60x2 26x 0 x m m. Therefore x x x m since it is the only value between the two charges. From Example the initial energy Et can be calculated Et Ki Ut 2 x 10 31kg x 106 m s 2 k x 10 19C x 10 19C 10 10 m Et x 10 19 J. When velocity equals zero all energy is electric potential energy so x 10 19 J 22 - r x 10 10 m. r 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. w w q q Wqq WqQ d d Q 2. The work is the potential energy of the combination. U 1 Upe U _ ke 2e ke - e k - e 2e S 2 x 10-10m 5 x 10-10m 5 x 10-10m _--- i -1-21 5 x 10-10m 2 _ x 109 Nm2 C2 x 10-19 C 2 2 __ 31 5 x 10- 10m J1 _ x 10-19 J Since U is negative we want do x 10 19 J to separate the particles K1 U1 K2 U2 K1 U2 0 so K2 U1 U1 -e f- - - 1 5e with r x 10 1 m 4n r r r 4n r U1 x 10 18 J eV Get closest distance y. Energy conservation mv2 .