Tài liệu Physics exercises_solution: Chapter 23 docx

23.1: J357.0 m150.0 1 m354.0 1 )C30.4)(C40.2( 11 12 21                  μμk rr qkqU J.357.0 UW 23.2:   J104.5J109.1J109.1 888 ffi UUUUW J103.7 8  23.3: a) .sm5.12 kg0015.0 J)491.0J608.0(2 2 1 J608.0 m800.0 )C1050.7)(C1080.2( )sm0.22)(kg0015.0( 2 1 21 2 66 2        f f ffi iii v r qkq mvEE k UKE b) At the closest point, the velocity is zero: m.323.0 J608.0 )C1080.7)(C1080.2( J608.0 66 21     k r r qkq 23.4: .m373.0 J400.0 C)1020.7)(C1030.2( J400.0 66 21      k r r qkq U 23.5: a) .J199.0 m250.0 )C1020.1()C1060.4( 66     k r kQq U s.m6.37 ,J198.0)iii( s.m7.36 J,189.0)ii( s.m6.26 kg1080.2 J)0994.0(2 2 1 J0994.0 J0994.0 m5.0 1 m25.0 1 )C1020.1(C)1060.4(J0 (i) b) 4 2 66                   ff ff fff fiif vK vK vmvK k UUKK 23.6: .J078.0C)102.1(66 m500.0 2 m500.0 262 22   kkq kqkq U 23.7: a) J.1060.3 )m100.0( )nC00.2)(nC00.3( )m100.0( )nC00.2)(nC00.4( )m200.0( nC)00.3)(nC00.4( 7 23 21 13 21 12 21                             k r qq r qq r qq kU .0,0b) 12 3231 12 21           xr qq x qq r qq kUIf So solving for x we find: .m360.0,m074.006.12660 2.0 68 600 2    xxx xx Therefore m074.0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.1009.5 m10 )C1020.3)(C1060.1( )sm1000.3)(kg1011.9( 2 1 19 10 1919 2631         i iii E k UKE When velocity equals zero, all energy is electric potential energy, so: .m1006.9 2 J1009.5 10 2 19   r r ek 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 WW qQqq  23.10: The work is the potential energy of the combination. J1031.7 3 2 2 m105 )C106.1()CNm100.9( 21 2 2 m105 m105 )2()( m105 )( m1025 )2( 19 10 219229 10 2 1010 10                                     ke eekekeeke UUUU epep Since U is negative, we want do J1031.7 19  to separate the particles 23.11: 12212211 so 0 ; UKUKUKUK  eV9.00J1044.1 m1000.8with, 5 4 1221 4 18 1 10 2 00 2 1            U r r e πεrrrπε e U 23.12: Get closest distance .γ Energy conservation: γ ke mvmv 2 22 2 1 2 1  m1038.1 )sm10()kg1067.1( )C106.1()CNm109( 13 627 219229 2 2        mv ke γ Maximum force: N012.0 )m1038.1( )C106.1()CNm109( 213 219229 2 2        γ ke F 23.13: BBAA UKUK  sm42.72 J0.00550V)800V(200C)105.00(J00250.0)( so, 6     mKv VVqKK qVKqVKqVU BB BAAB BBAA It is faster at B; a negative charge gains speed when it moves to higher potential. 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 C00.5 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. 23.15: E  points from high potential to low potential, so .and ACAB VVVV  The force on a positive test charge is east, so no work is done on it by the electric force when it moves due south (the force and displacement are perpendicular); . AD VV  23.16: a) J.1050.1 6  KqEdUW b) The initial point was at a higher potential than the latter since any positive charge, when free to move, will move from greater to lesser potential. V.357nC)(4.20J)1050.1( 6   qUV c) C.N1095.5 m)06.0()nC20.4( J1050.1 J1050.1 3 6 6      EqEd 23.17: a) Work done is zero since the motion is along an equipotential, perpendicular to the electric field. b) J105.7)m670.0( m V 104.00nC)0.28( 44          qEdW c) J1006.2)45cos60.2( m V 104.00nC)0.28( 34          qEdW 23.18: Initial energy equals final energy: s.m1089.6 J)102.88J1004.5( kg1011.9 2 2 1 J1004.5 2 1 m0.40 C1000.2( m0.10 C)10(3.00 C)1060.1( J1088.2 m0.25 C)1000.2( m0.25 C)10(3.00 C)1060.1( 2 1 6 1717 31 2 17 2 99 19 17 99 19 2 2 2 1 1 2 2 1 1                                         f fe fef i fe ffii fi v vm vmkE kE vm r keq r keq r keq r keq EE 23.19: a) m.105.2 V90.0 C)1050.2( 3 11      k V kq r r kq V b) m.105.7 V30.0 C)1050.2( 3 11      k V kq r r kq V 23.20: a) C.1033.1 V)(48.0m)250.0( 9  kk rV q r kq V b) V16 m)(0.750 C)1033.1( 9     k V 23.21: a) .V738 m0.05 C1050.6 m05.0 C1040.2 :AAt 99 2 2 1 1                        k r q r q kV A b) .V705 m0.06 C1050.6 m08.0 C1040.2 :BAt 99 2 2 1 1                        k r q r q kV B c) J.108.25V)33(C)1050.2( 89   VqW The negative sign indicates that the work is done on the charge. So the work done by the field is J.1025.8 8  23.22: a) b) . 4 1 2 0 a q πε V  c) Looking at the diagram in (a): 22 00 4 1 2 4 1 2)( xa q πεr q πε xV   d) e) When , 2 4 1 , 0 x q πε Vax  just like a point charge of charge .2q 23.23: a) b) .0 )(    r qk r kq V x c) The potential along the x-axis is always zero, so a graph would be flat. d) If the two charges are interchanged, then the results of (b) and (c) still hold. The potential is zero 23.24: a) . 2 )()( :|| 22 ay kqy ya kq ya kq Vay       . 2 )()( : . 2 )( : 22 22 ay kqa ay kq ya kq Vay ay kqa ay kq ya kq Vay               Note: This can also be written as              |||| ay q ay q kV b) c) . 2 )()( : 2 y kqa ay kq ya kq Vay       d) If the charges are interchanged, then the potential is of the opposite sign. 23.25: a) b) . )( )( 2 : axx axkq ax kq x kq Vax      . )( )3( 2 :0 axx axkq xa kq x kq Vax      . )( )( 2 :0 axx axkq ax kq x kq Vx        Note: This can be also be written as )( || 2 || ax q x q kV   c) The potential is zero at 3./and aax  d) e) For ,: 2 x kq x kqx Vax     which is the same as the potential of a point charge –q. (Note: The two charges must be added with the correct sign.) 23.26:a) . 2 || 12 || 22           ya y kq r kq y kq V b) . 3 3 4 when,0 22 22 2 a yay ya yV    c) d) , 21 : y kq yy kqVay           which is the potential of a point charge q . 23.27: J.104.72C)1060.1()V295( 1719   VqUW But also: s.m1001.1 kg109.11 J)1072.4(2 2 1 7 31 17 2       vmvKW 23.28: a) m.415.0 CN0.12 V98.4  E V d d V E b) C.1030.2 )m415.0()V98.4( 10  kk Vd q d kq V c) The electric field is directed away from q since it is a positive charge. 23.29: a) Point b has a higher potential since it is “upstream” from where the positive charge moves. 0)(||)(||)(  abEVVabEabEVV abba b) C.N800 m3.0 240  V d V E c) J.104.8V)240(C)1020.0( 56   VqUW 23.30:(a) ,0 2  QQ VVV so V is zero nowhere except for infinitely far from the charges. The fields can cancel only between the charges 22 22 2 2)( )( )2( xxd xd Qk x kQ EE QQ    . 21  d x The other root, , 21  d x does not lie between the charges. (b) V can be zero in 2 places, A and B. .oflefttheto 0 )2()( : 30 )2()( : 2 QEE dy yd Qk y Qk Bat dx xd Qk x Qk Aat QQ          12 )( )2( 22     d x xd Qk x kQ (c) Note that E and V are not zero at the same places. 23.31: a) 2211 qVKqVK  ;)( 1221 KKVVq  C10602.1 19 q J;10099.4 182 1e 2 1 1   vmK J10915.2 172 2e 2 1 2   vmK V156 12 21    q KK VV The electron gains kinetic energy when it moves to higher potential. b) Now 0J,10915.2 2 17 1   KK V182 12 21    q KK VV The electron loses kinetic energy when it moves to lower potential [...]... charge is moved AGAINST the electric field 23. 67: From Example 21.10, we have: Ex  Q V   4π ε0 (23. 16) x  ( x  2 1 Qx 2 4πε0 ( x  a 2 )3 / 2 x Q 1/ 2 dx  u 2 3/ 2 a ) 4πε0 ux2  a2 u   1 4πε0 Q x  a2 2  Equation 23. 68: dV  1 Q dl 1 Q dθ 1 1 dq 1 λ dl    V  4πε0 πa a 4πε0 πa 4πε0 4πε0 r 4πε0 a π  0 1 Q Q dθ  4πε0 a πa 0.3 ˆ ˆ j 23. 69: a) S1 and S3 : V13    (  5 xi  3... 360 V) (2.40  109 C)   8.64  107 J     23. 36: a) V  Ed  (480 N C) (3.8  10 2 m)  18.2 V b) The higher potential is at the positive sheet σ c) E   σ  ε0 (480 N C)  4.25  10  9 C m 2 ε0 V 4750 V V d    1.58  10 3 m 6 d E 3.00  10 V m σ b) E   σ  ε0 (3.00  106 V m)  2.66  10 5 C m 2 ε0 23. 37:a) E  σ 47.0  109 C m 2 23. 38: a) E    5311 N C ε0 ε0 b) V  Ed  (5311... balance Energy conservation gives U i  K f kq1q2 kq1q3 kq2 q3 1 1 2    m1v12  m3v3 r12 r13 2 r23 2 v1  v3 , m1  m3 , and q1  q2  q3  q v1   kq 2 m1  1 1 1    r  12 r13 r23     (9  109 Nm 2 C 2 ) (2  10 6 C) 2 0.020 kg  1 1 1    0.08 m  0.16 m  0.08 m   7.5 m s    23. 49: a) WE  K  WF  4.35  105 J  6.50  105 J   2.15  105 J  WE 2.15  105 J    2829... ln(b a) 2πε0 c) Between the cylinders: Vab λ ln(b r )  ln(b r ) V 2πε0 ln (b a) 23. 57: a) (i) V  E   Vab  Vab 1 V  (ln(b r ))  r ln(b a) r ln(b a) r d) The potential difference between the two cylinders is identical to that in part (b) even if the outer cylinder has no charge 23. 58: Using the results of Problem 23. 57, we can calculate the potential difference: Vab 1 E  Vab  E ln (b/a)r... ln(b a ) r  Vab  (2.00  10 4 N C) (ln (0.018 m 145  10 6 m)) 0.012 m  1157 V 23. 59: a) F  Eq  (1.10  103 V m) (1.60  1019 C)  1.76  1016 N, downward 2 b) a  F me  (1.76  10 16 N) (9.11  10 31 kg)  1.93  1014 m s , downward 0.060 m 1 1 2 c) t   9 .23  109 s, y  y0  at 2  (1.93  1014 m s ) (9 .23  10 9 s) 6 6.50  10 m s 2 2  8.22  103 m d) Angle θ  arctan(v y vx )  arctan(at... m s 23. 60: b b a a (a) Use Vab to get λ : V   E  dl   λ λ dr  ln 2πε0 2πε0 λ 2πε0 V ln b a E 2πε0 V ln b a λ V   2πε0 r 2πε0 r r ln b a at outer surface of the wire, r  a = 0.127 mm 2 850 V E  2.65  106 V m 0.000127 m  1.00 cm  ( 2 ) ln ( 0.0127 cm )  2    (b) at the inner surface of the cylinder, r  1.00 cm , which gives E  1.68  104 V m b a 23. 61: a) From Problem 23. 57,... a a  a 2  4x2  a   ( 2  1)    23. 63: a)  2πr dr  2kQ  2 2   x 2  r 2  πR  R kQ dV  R V   dV  0 σ 2 ε0 b) 2kQ R2 x 2 R  0 r dr x r 2 2  r dr x2  r 2 2kQ 1 / 2 (z ) R2 zx2 R2 zx2  2kQ R2 x 2   R2  x    R2  x Ex     V 2kQ  1 σ  x  2   1  1   x R  x2  R2  2ε 0  1  R2 x2      23. 64: a) From Example 23. 12: 2 2 2 2 kQ  a  x  a  kQ  1 ...kq k (3.50  10 9 C) 23. 32: a) V    65.6 V 0.48 m r k (3.50  109 C)  131.3 V 0.240 m c) Since the sphere is metal, its interior is an equipotential, and so the potential inside is 131.3 V b) V  23. 33: a) The electron will exhibit simple harmonic motion for x  a, but will otherwise oscillate between  30.0 cm b) From Example 23. 11, 1  kQ 1   V  kQ  V 2 2 2 2 ... x 0 |  2 0.3 0 5 (0.3) 2  0.225 V 2 S5 is higher 23. 70: From Example 22.9, we have: r kQ dr  kQ r  R : E  2  V   kQ  2   r r  r R  r    kQ kQ r kQr r  R : E  3  V    E d r    E  d r    3  r  dr  R R R R R  r kQ kQ 1 2 kQ kQ kQr 2 V   3 r    R R 2 R R 2R 2R3 V  b) kQ 2R  r2  3 2  R   23. 71: a) Problem 23. 70 shows that Q Q Vr  (3  r 2 R 2 ) for r  R...      x  a : V p  23. 80: Set the alpha particle’s kinetic energy equal to its potential energy: K  U  11.0 Me V  k (2e) (82e) k (164) (1.60  1019 C) 2 r r (11.0  106 eV)(1.60  1019 J eV)  2.15  1014 m 23. 81: a) V  b) EB   kQB kQ A kQB Q 1    Q A  3QB  B  RB RA RA 3 QA 3 V r  r  RB kQB k (QA 3) 3kQA E    3E A  B  3 2 2 2 ( RA 3) EA RB RA 23. 82: a) From Problem 22.57 . m. 323. 0 J608.0 )C1080.7)(C1080.2( J608.0 66 21     k r r qkq 23. 4: .m373.0 J400.0 C)1020.7)(C1030.2( J400.0 66 21      k r r qkq U 23. 5:. 23. 1: J357.0 m150.0 1 m354.0 1 )C30.4)(C40.2( 11 12 21                  μμk rr qkqU J.357.0 UW 23. 2: 