Instructor Solutions Manual for Physics by Halliday, Resnick, and Krane Paul Stanley Beloit College Volume 2
A Note To To The Instructor Instructor... ...
The solutions here are somewhat brief, as they are designed for the instructor, not for the student. Check with the publishers before electronically posting any part of these solutions; website, ftp, or server access must be be restricted to your students. I have been somewhat casual about subscripts whenever it is obvious that a problem is one dimensional, or that the choice of the coordinate system is irrelevant to the numerical solution. Although this does not change the validity of the answer, it will sometimes obfuscate the approach if viewed by a novice. There are some traditional formula, such as vx2 = v = v 02x + 2a 2 ax x, which are not used in the text. The worked solutions use only material from the text, so there may be times times when the solution solution here seems unnecessari unnecessarily ly convoluted convoluted and drawn out. Yes, I know know an easier approach existed. But if it was not in the text, I did not use it here. I also tried tried to avoid avoid reinven reinventing ting the wheel. There are some exercises exercises and problems problems in the text which build upon previous exercises and problems. Instead of rederiving expressions, I simply refer you to the previous solution. I adopt a different approach for rounding of significant figures than previous authors; in particular, ular, I usually round intermedi intermediate ate answers. answers. As such, some of my answers answers will differ from those in the back of the book. Exercises and Problems which are enclosed in a box also appear in the Student’s Solution Manual with considerably more detail and, when appropriate, include discussion on any physical implications of the answer. These student solutions carefully discuss the steps required for solving problems, point out the relevant equation numbers, or even specify where in the text additional information can be found. found. When two two almost equivalen equivalentt methods of solution solution exist, often both are presente presented. d. You are encouraged to refer students to the Student’s Solution Manual for these exercises and problems. However, the material from the Student’s Solution Manual must not be copied. Paul Stanley Beloit College
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1
E25-1
The charge transferred is Q = (2. (2.5
E25-2
Use Eq. 25-4: 25-4: r =
E25-3
× 104 C/s)(20 × 10−6 s) = 5.5.0 × 10−1 C.
(8. (8.99 109 N m2 /C2 )(26. )(26.3 10−6 C)(47. C)(47.1 10−6 C) = 1. 1 .40 m (5. (5.66N)
×
·
×
×
Use Eq. 25-4: 25-4: (8. (8.99 109 N m2 /C2 )(3. )(3.12 10−6 C)(1. C)(1.48 10−6 C) F = = 2. 2 .74 N. (0. (0.123 123 m)2
×
E25-4
·
×
×
(a) The forces forces are are equal, equal, so m so m 1 a1 = m 2 a2 , or m2 = (6. (6.31 10−7 kg)(7. kg)(7.22 m/s2 )/(9. (9.16 m/s2 ) = 4.97 10−7 kg. kg.
×
×
(b) Use Eq. 25-4: q = =
E25-5
(6. (6.31 10−7 kg)(7. kg)(7.22 m/s2 )(3. )(3.20 10−3 m)2 = 7. 7 .20 10−11 C (8. (8.99 109 N m2 /C2 )
×
×
×
·
×
(a) Use Eq. 25-4, 25-4, F =
1 q 1 q 2 1 (21. (21.3 µC)(21. C)(21.3 µC) = = 1. 1 .77 N 2 − 12 2 2 4π 0 r12 4π(8. (8.85 10 C /N m ) (1. (1.52m)2
×
·
(b) In part (a) we found F 12 Sincee q 3 = q 2 and 12 ; to solve part (b) we need to first find F 13 13 . Sinc r13 = r = r 12 , we can immediately conclude that F 13 13 = F 12 12 . We must assess the direction of the force of q 3 on q 1 ; it will be directed along the line which connects the two charges, and will b e directed away from q from q 3 . The diagram below shows the directions. F 23
F
θ
12
F 23
F
F net
12
From this diagram we want to find the magnitude of the net force on q 1 . The The cosin cosinee law law is appropriate here: F net 2
F net
= = = =
2 2 F 12 + F 13 2F 12 12 F 13 13 cos θ, (1. (1.77N)2 + (1. (1.77N)2 2(1. 2(1.77N)(1. 77N)(1.77 N) cos(12 cos(1200◦ ), 9.40 N2 , 3.07 N.
−
−
2
E25-6 Originally F 0 = CQ 20 = 0.088N, where where C C is a consta constant nt.. When When sphere sphere 3 touch touches es 1 the
charge on both becomes Q0 /2. When When sphere sphere 3 the touches touches sphere sphere 2 the charge charge on each each becomes becomes (Q0 + Q0 /2)/ 2)/2 = 3Q0 /4. The force between sphere 1 and 2 is then F = C ( C (Q0 /2)(3Q 2)(3Q0 /4) = (3/ (3/8)CQ 8)CQ 20 = (3/ (3/8)F 8)F 0 = 0.033N 033N.. F31 and F32 . These forces are given by the vector form of Coulomb’s The forces forces on q on q 3 are Law, Eq. 25-5, E25-7
31 F
=
32 F
=
1 q 3 q 1 1 q 3 q 1 ˆr31 = ˆr31 , 2 4π0 r31 4π0 (2d (2d)2 1 q 3 q 2 1 q 3 q 2 ˆr32 = ˆr32 . 2 4π0 r32 4π0 (d)2
These two forces are the only forces which act on q 3 , so in order to have q 3 in equilibrium the forces must be equal in magnitude, but opposite in direction. In short, 31 F 1 q 3 q 1 ˆr31 4π0 (2d (2d)2 q 1 ˆr31 4
= = =
−F 32, − 4π1 0 q (d3q )22 ˆr32, − q 12 ˆr32.
Note that ˆ that ˆr31 and ˆ and ˆr32 both point in the same direction and are both of unit length. We then get q 1 =
−4q 2.
The horizontal and vertical vertical contributions from the upper left charge and lower lower right charge are straightforward to find. The contributions from the upper left charge require slightly more work. The diagonal distance is 2a; the component componentss will be weight weighted ed by cos 45 ◦ = 2/2. The diagona diagonall charge will contribute E25-8
√
√
F x F y
√
√
√
√
=
1 (q )(2q )(2q ) 2ˆ 2 q 2 ˆ i = i, 4π0 ( 2a)2 2 8π0 a2
=
1 (q )(2q )(2q ) 2 ˆ 2 q 2 ˆ j = j. 4π0 ( 2a)2 2 8π 0 a2
√ √
(a) The horizontal component of the net force is then F x
√
1 (2q (2q )(2q )(2q ) ˆ 2 q 2 ˆ i+ i, = 4π 0 a2 8π0 a2 4 + 2/2 q 2 ˆ i, = 4π 0 a2 = (4. (4.707)(8. 707)(8.99 109 N m2 /C2 )(1. )(1.13 10−6 C)2 /(0. (0.152 m)2ˆi = 2.34 N ˆi.
√
×
·
×
(b) The vertical component of the net force is then F y
√
1 (q )(2q )(2q ) ˆ 2 q 2 ˆ j j, = + 4π0 a2 8π 0 a2 2 + 2/2 q 2 ˆ j, = 8π0 a2 = ( 1.293)(8. 293)(8.99 109 N m2 /C2 )(1. )(1.13 10−6 C)2 /(0. (0.152 m)2 jˆ =
− −
−
√
×
·
×
3
−0.642N jˆ.
