Physics exercises_solution: Chapter 23

Chia sẻ: Thanh An | Ngày: | Loại File: PDF | Số trang:36

0
66
lượt xem
7
download

Physics exercises_solution: Chapter 23

Mô tả tài liệu
  Download Vui lòng tải xuống để xem tài liệu đầy đủ

Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 23

Chủ đề:
Lưu

Nội dung Text: Physics exercises_solution: Chapter 23

  1. 1 1  1 1  23.1: U  kq1q2     k (2.40 μC)(  4.30 μC) r r     0.357 J  2 1  0.354m 0.150m   W   U   0.357 J. 23.2: W   1.9  10 8 J   U  U i  U f  U f  1.9  108 J  5.4  10 8 J  7.3  108 J 23.3: a) 1 k (2.80  10 6 C)(7.50  10 6 C) Ei  K i  U i  (0.0015 kg )(22.0 m s) 2   0.608 J 2 0.800 m 1 2 kq1q2 2(0.608 J  0.491 J) Ei  E f  mv f   vf   12.5 m s . 2 rf 0.0015 kg b) At the closest point, the velocity is zero: kq q k (2.80  10 6 C)(7.80  10 6 C)  0.608 J  1 2  r   0.323 m. r 0.608 J kq1q 2  k (2.30  10 6 C)(7.20  10 6 C) 23.4: U   0.400 J  r  0.373 m. r  0.400 J kQq k (4.60  106 C) (1.20  106 C) 23.5: a) U    0.199 J. r 0.250 m b) (i) K f  K i  U i  U f  1 1   0 J  k (4.60  10 6 C) (1.20  10 6 C)  0.25 m 0.5 m   0.0994 J     1 2 2(0.0994 J)  K f  0.0994 J  mv f  v f   26.6 m s. 2 2.80  10 4 kg (ii) K f  0.189 J, v f  36.7 m s. (iii) K f  0.198 J, v f  37.6 m s. kq 2 2kq 2 23.6: U    6kq 2  6k (1.2  10 6 C) 2  0.078 J. 0.500m 0.500 m
  2. 23.7: a)  (4.00 nC)(3.00 nC) (4.00 nC)(2.00 nC)     qq qq q1q2   (0.200 m) (0.100m)  U  k 1 2  1 2  r  k  12 r13 r23    (3.00 nC)(2.00 nC)      (0.100 m)    3.60  10 7 J. q q qq q q  b) If U  0, 0  k  1 2  1 3  2 3 . So solving for x we find:  r  12 x r12  x   8 6 0   60    60 x 2  26 x  1.6  0  x  0.074 m, 0.360 m. Therefore x 0.2  x x  0.074 m since it is the only value between the two charges. 23.8: From Example 23.1, the initial energy Ei can be calculated: 1 Ei  K i  U i  (9.11  10 31 kg)(3.00  106 m s) 2 2 k (  1.60  1019 C)(3.20  1019 C)  10 10 m  Ei   5.09  10 19 J. When velocity equals zero, all energy is electric potential energy, so: k 2e 2  5.09  10 19 J    r  9.06  10 10 m. r 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. kq 2 2kqQ q Wqq  WqQ     Q . d d 2
  3. 23.10: The work is the potential energy of the combination. U  U p   U pe  U e  ke(2e) ke(  e) k (  e)(2e)    5 2  10 m 5  10 m 5  1010 m 10 10 ke 2  2   10   1  2 5  10 m  2  (9.0  10 Nm C ) (1.6  1019 C) 2  2 9 2 2   10   3 5  10 m  2  19   7.31  10 J Since U is negative, we want do  7.31  10 19 J to separate the particles 23.11: K1  U1  K 2  U 2 ; K1  U 2  0 so K 2  U1 e2  1 2 2  1 5e 2 U1      , with r  8.00  10 10 m 4πε0  r r r  4πε0 r U1  1.44  10 18 J  9.00 eV 1 1 ke2 23.12: Get closest distance γ. Energy conservation: mv 2  mv 2  2 2 γ ke 2 (9  10 9 Nm 2 C 2 )(1.6  10 19 C) 2 γ   1.38  10 13 m mv 2 (1.67  10  27 kg )(10 6 m s) Maximum force: ke 2 F 2 γ (9  109 Nm 2 C 2 ) (1.6  1019 C) 2  (1.38  1013 m) 2  0.012 N 23.13: K A  U A  K B  U B U  qV , so K A  qV A  K B  qVB K B  K A  q (V A  VB )  0.00250 J  (  5.00  10 6 C) (200 V  800 V)  0.00550 J vB  2 K B m  7.42 m s It is faster at B; a negative charge gains speed when it moves to higher potential.
  4. 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  5.00 μC 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 VB  V A and VC  VA . 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); VD  V A . 23.16: a) W   U  qEd  K  1.50  106 J. 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   U q  (1.50  106 J) (4.20 nC)  357 V. 6 1.50  106 J c) qEd  1.50  10 JE  5.95  103 N C. (4.20 nC)(0.06 m) 23.17: a) Work done is zero since the motion is along an equipotential, perpendicular to the electric field.  V b) W  qEd  (28.0 nC) 4.00  104 (0.670 m)  7.5  10 4 J  m  V c) W  qEd  (28.0 nC) 4.00  104 (  2.60 cos 45)   2.06  10 3 J  m
  5. 23.18: Initial energy equals final energy: keq1 keq2 keq1 keq2 1 2 Ei  E f       me v f r1i r2i r1 f r2 f 2  (3.00  10 9 C) (2.00  10 9 C)  Ei  k (  1.60  10 19 C)       2.88  10 17 J   0.25 m 0.25 m   (3.00  10 C) (2.00  10 C  1 9 9 E f  k (  1.60  10 19 C)     me v 2  2 f  0.10 m 0.40 m  1 2   5.04  10 17 J  me v f 2 2  vf  31 (5.04  10 17 J  2.88  10 17 J) 9.11  10 kg  6.89  106 m s. kq kq k (2.50  10 11 C) 23.19: a) V  r   2.5  10 3 m. r V 90.0 V kq kq k (2.50  10 11 C) b) V  r   7.5  10 3 m. r V 30.0 V kq rV (0.250 m)(48.0 V) 23.20: a) V  q   1.33  10 9 C. r k k 9 k (1.33  10 C) b) V   16 V (0.750 m) q q   2.40  10 9 C  6.50  10 9 C  23.21: a) At A : V A  k  1  2   k  r     738 V.  1 r2     0.05 m 0.05 m   q q   2.40  10 9 C  6.50  10 9 C  b) At B : VB  k  1  2   k  r     705 V.  1 r2     0.08 m 0.06 m   c) W  qV  (2.50  10 9 C)(  33 V)   8.25  10 8 J. The negative sign indicates that the work is done on the charge. So the work done by the field is 8.25  10 8 J.
  6. 23.22: a) 1 q b) V  2 . 4πε0 a 1 q 1 q c) Looking at the diagram in (a): V ( x)  2 2 4πε0 r 4πε0 a 2  x2 d) 1 2q e) When x  a, V  , just like a point charge of charge  2q. 4πε0 x 23.23: a) kq k (  q ) b) V x    0. r r 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
  7. kq kq 2kqy 23.24: a) | y |  a : V    2 . ( a  y ) (a  y ) y  a 2 kq kq  2kqa y  a :V    2 . (a  y ) y  a y  a 2  kq kq 2kqa y   a :V    2 . (a  y ) ( y  a) y  a 2  q q  Note: This can also be written as V  k  | y  a | | y  a |     b) kq kq  2kqa c) y  a : V    . (a  y ) ( y  a) y2 d) If the charges are interchanged, then the potential is of the opposite sign.
  8. 23.25: a) kq 2kq  kq( x  a) b) x  a :V    . x xa x( x  a) kq 2kq kq(3x  a) 0  x  a :V    . x ax x( x  a)  kq 2kq kq( x  a) x  0 :V    . x xa x( x  a ) Note: This can be also be written as V  k ( |q|  x 2q | x  a| ) c) The potential is zero at x   a and a / 3. d)  kqx  kq e) For x  a : V   , which is the same as the potential of a point charge x2 x –q. (Note: The two charges must be added with the correct sign.)
  9. kq 2kq  1 2  23.26:a) V    kq   . | y| r | y | a y  2 2   a y 2 2 a b) V  0, when y 2   3y 2  a 2  y   . 4 3 c) 1 2 kq d) y  a : V  kq     , which is the potential of a point charge  q .  y y   y 23.27: W   U   Vq  (295 V) (1.60  1019 C)  4.72  1017 J. But also: 1 2 2(4.72  10 17 J) W  K  mv  v  31  1.01  10 7 m s. 2 9.11  10 kg V V 4.98 V 23.28: a) E  d    0.415 m. d E 12.0 N C kq Vd (4.98 V ) (0.415 m) b) V  q   2.30  10 10 C. d k k 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. Va  Vb  E (b  a)   | E | (b  a)  Vb  Va  | E | (b  a )  0 V 240 V b) E    800 N C. d 0.3 m c) W   U   qV   (0.20  106 C)(  240 V)   4.8  105 J.
  10. 23.30:(a) V  VQ  V2Q  0, so V is zero nowhere except for infinitely far from the charges. The fields can cancel only between the charges kQ k (2Q) E Q  E 2Q  2   (d  x) 2  2 x 2 x (d  x) 2 x  1d 2 . The other root, x  1d 2 , does not lie between the charges. (b) V can be zero in 2 places, A and B. k (  Q) k (2Q) at A :  0xd 3 x dx k (  Q) k (2Q) at B :  0 y d y dy EQ  E2Q to the left of  Q. kQ k (2Q) d  x x 2 (d  x) 2 2 1 (c) Note that E and V are not zero at the same places. 23.31: a) K1  qV1  K 2  qV2 q(V1  V2 )  K 2  K1; q   1.602  10 19 C K1  1 me v12  4.099  10 18 J; 2 K 2  1 me v2  2.915  10 17 J 2 2 K 2  K1 V1  V2    156 V q The electron gains kinetic energy when it moves to higher potential. b) Now K 1  2.915  10 17 J, K 2  0 K  K1 V1  V2  2   182 V q The electron loses kinetic energy when it moves to lower potential
  11. kq k (3.50  10 9 C) 23.32: a) V    65.6 V. r 0.48 m k (3.50  109 C) b) V   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. 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, kQ 1 1  V  V  kQ   x a 2 2 a x a  2 2    1 1   ΔV  k (24.0  10 9 C)     0.150 m (0.300 m) 2  (0.150 m) 2     796 V 1 19 But W   qV  mv 2  v  2 (1.60  10 10 C) kg V)  1.67  107 m s. 9.11  31 (796 2 23.34: Energy is conserved: 1 2 (1.67  1027 kg ) (1500 m s) 2 mv  qV  V   0.0117 V. 2 2(1.60  1019 C) But: λ  2πε0 V   2πε0 V  V  ln(r0 r )  r0  r exp    r  r0 exp   2πε0  λ   λ   2πε0 (0.0117 V)   r  (0.180 m)exp    5.00  10 12 C / m   0.158 m.    V 360 V 23.35: a) E    8000 N C. d 0.0450 m b) F  Eq  (8000 N C) (2.40  109 C)  1.92  105 N. c) W  Fd  (1.92  105 N) (0.0450 m)  8.64  107 J. d) U  Vq  (  360 V) (2.40  109 C)   8.64  107 J.
  12. 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 V 4750 V 23.37:a) E  d    1.58  10 3 m. d E 3.00  10 V m 6 σ b) E   σ  ε0 (3.00  106 V m)  2.66  10 5 C m 2 . ε0 σ 47.0  109 C m 2 23.38: a) E    5311 N C. ε0 ε0 b) V  Ed  (5311 N / C) (0.0220 m)  117 V. c) The electric field stays the same if the separation of the plates doubles, while the potential between the plates doubles. 23.39: a) The electric field outside the shell is the same as for a point charge at the center of the shell, so the potential outside the shell is the same as for a point charge: q V for r  R. 4πε0 r The electric field is zero inside the shell, so no work is done on a test charge as it moves inside the shell and all points inside the shell are at the same potential as the q surface of the shell: V  for r  R. 4πε0 R kq RV (0.15 m) (  1200 V) b) V  so q     20 nC R k k c) No, the amount of charge on the sphere is very small.
  13. 23.40: For points outside this spherical charge distribution the field is the same as if all the charge were concentrated at the center. Therefore q E 4πε0 r 2 and (3800 N C) (0.200 m) 2 q  4πε0 Er 2   1.69  108 C 9  109 N.m 2 / C 2 Since the field is directed inward, the charge must be negative. The potential of a point charge, taking  as zero, is q (9  109 N.m 2 / C 2 ) (  1.69  10 8 C) V    760 V 4πε0 r 0.200 m at the surface of the sphere. Since the charge all resides on the surface of a conductor, the field inside the sphere due to this symmetrical distribution is zero. No work is therefore done in moving a test charge from just inside the surface to the center, and the potential at the center must also be  760 V. 23.41: a) E   V . V  Ex    ( Axy  Bx 2  Cy )   Ay  2 Bx. x x V  Ey    ( Axy  Bx 2  Cy )   Ax  C. y y V  Ez     ( Axy  Bx 2  Cy )  0. z z 2B C 2B   C  b)  Ay  2 Bx  0  y  x,  Ax  C  0  x   so y  .  A A A  A   2 BC   C  2 BC  2 , E  0 at  , , z A  A A2 
  14. 23.42: a) E   V V   kQ  kQx kQx Ex       3 . x x  x  y  z 2 2 2  (x  y  z ) 2 2 2 32 r   kQy kQz Similarly, E y  3 and Ez  3 . r r kQ  xi yˆ zk  kQ ˆ j ˆ b) So from (a), E  2     ˆ r , which agrees with Equation (21.7). r  r  r r  r2  23.43: a) There is no dependence of the potential on x or y, and so it has no components in those directions. However, there is z dependence: V ˆ E  V  E z     C  E   Ck , for 0  z  d . z  and E  0, for z  d , since the potential is constant there. (b) Infinite parallel plates of opposite charge could create this electric field, where the surface charge is σ   Cε0 .
  15. 23.44: a) kq kq 1 1 (i) r  ra : V    kq   . r r  ra rb  a b  kq kq 1 1  (ii) ra  r  rb : V    kq   . r r  r rb  b  (iii) r  rb : V  0, since outside a sphere the potential is the same as for point charge. Therefore we have the identical potential to two oppositely charged point charges at the same location. These potentials cancel. 1 q q 1 1 1 b) Va     and Vb  0  Vab  r  q   4πε0  a rb  4πε0  ra rb    V q  1 1  1 q Vab 1 c) ra  r  rb : E       r r   . r 4πε0 r  b  4πε0 r 2  1 1  r2    r r   a b  d) From Equation (24.23): E  0, since V is zero outside the spheres. e) If the outer charge is different, then outside the outer sphere the potential is no 1 q 1 Q 1 (q  Q) longer zero but is V    . All potentials inside the outer 4πε0 r 4πε0 r 4πε0 r 1 Q shell are just shifted by an amount V   . Therefore relative potentials within the 4πε0 rb shells are not affected. Thus (b) and (c) do not change. However, now that the potential does vary outside the spheres, there is an electric field there: V   kq  kQ  kq  Q E      2 1  . r r  r r  r   q 
  16. 1 1 23.45: a) Vab  kq     500 V r r   a b  500 V q  7.62  1010 C.  1 1  k  0.012 m  0.096 m     b) c) The equipotentials are closest when the electric field is largest. V   kQ  a 2  x 2  a   23.46: a) E x     ln   x x 2a  a 2  x 2  a      kQ      Ex    x ln( a  x  a)  x ln( a  x  a) 2 2 2 2 2a    x(a 2  x 2 ) 1 2 x(a 2  x 2 ) 1/ 2  kQ kQ   2   2a  a x a 2 a  x  a  x a2  x2 2 2 (2aλ ) 1 λ  Ex   . 4πε0 xa 1  x a 2 2 2πε0 x 1  x 2 a 2 b) The potential was evaluated at y and z equal to zero, and thus shows no dependence on them. However, the electric field depends upon the derivative of the potential and the potential could still have a functional dependence on the variables y and z, and hence E y and E z may be non-zero.
  17. 23.47: a) Equipotentials and electric field lines of two large parallel plates are shown above. b) The electric field lines and the equipotential lines are mutually perpendicular.
  18. 23.48: (a) F  m1a  F12  F13 kq1q2 kq1q 3 2  2  m1a r12 r13 q1  q2  q3  q  1 1 ma kq 2  2  2  r   12 r13   1 1  (0.02 kg)a  (9  109 Nm 2 C 2 ) (2.0  10 6 C) 2  2  2  (0.08 m) (0.16 m)  2 a  352 m s (b) Maximum speed occurs at “infinity”. The center charge does not move since the forces on it balance. Energy conservation gives U i  K f . kq1q2 kq1q3 kq2 q3 1 1    m1v12  m3v3 .2 r12 r13 r23 2 2 v1  v3 , m1  m3 , and q1  q2  q3  q kq 2  1 1 1  v1    r    m1  12 r13 r23  (9  109 Nm 2 C 2 ) (2  10 6 C) 2  1 1 1     0.08 m  0.16 m  0.08 m   7.5 m s  0.020 kg   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 b) WE   qV  V     2829 V. So the initial point is q 7.60  109 C –2829 V with respect to the final point. V 2829 V V c) E    3.54  10 4 . d 0.08 m m
  19. mv 2 ke2 ke2 23.50: a)  2 v . r r mr 1 2 1 ke 2 1 b) K  mv    U. 2 2 r 2 1 1 ke 2 1 k (1.60  10 19 C) 2 c) E  K  U  U    11   2.17  1018 J   13.6 eV. 2 2 r 2 5.29  10 m 4/3 23.51: a) V  Cx 4 / 3  C  (240 V) (0.0130 m) 4/3  7.85  10 4 V m . V 4 4 4/3  V  b) E     Cx1/ 3   (7.85  10 4 V m ) x1/ 3    1.05  105 4/3 x1/ 3  V m , x 3 3  m  toward cathode. c) F   eE  ((1.05  105 ) (0.00650)1/ 3 V m) (1.60  10 19 C)  3.14  10 15 N, toward anode.
  20. 23.52: From Problem 22.51, the electric field of a sphere with radius R and q distributed qr q uniformly over its volume is E  3 for r  R and E  for r  R 4πε0 R 4πε0 r 2 b Va  Vb   Edr. Take b at infinity and V  0. Let point a be a distance r  R from a the center of the sphere. R qr  q q  r2  Vr   dr   dr  3  2   r 4πε R 0 3 R 4πε r 0 2 8πε0 R  R   Set q   2e to get Vr for the sphere. The work done by the attractive force of the sphere when one electron is removed from r  d to  is 2e 2  d2  3  2  Wsphere   eVr   8πε0 R   R  The total work done by the attractive force of the sphere when both electrons are removed is twice this, 2Wsphere . The work done by the repulsive force of the two electrons e2 is Wee  The total work done by the electrical forces is 2Wsphere  Wee . The 4πε0 (2d ) energy required to remove the two electrons is the negative of this, e2  R d2  3    2πε 0 R   4d R 2  We can check this result in the special case of d = R, when the electrons initially sit on the surface of the sphere. The potential due to the sphere is the same as for a point charge  2e at the center of the sphere. Wq  b  U a  U b   2e 2  e2 e2  1   7e 2 U b  0. U a  2     4πε R  4πε (2 R) 4πε R  2   0  0 0  4  8πε0 R The work done by the electric forces when the electrons are removed is  7e 2 8πε0 R and the energy required to remove them is 7e 2 8πε0 R . Setting d =R in our general expression yields this same result.
Đồng bộ tài khoản