Instructor Solutions Manual for Physics by Halliday, Resnick, and Krane
Paul Stanley Beloit College Volume 2
A Note To The 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 restricted to your students. Ihave 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
2 2 vx = v0x + 2ax x,
which are not used inthe text. The worked solutions use only material from the text, so there may be times when the solution here seems unnecessarily convoluted and drawn out. Yes, I know an easier approach existed. But if it was not in the text, I did not use it here. I also tried to avoid reinventing the wheel. There are some exercises and 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, I usually round intermediate answers. As such, some of my 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 considerablymore 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. When two almost equivalent methods of solution exist, often both are presented. You are encouraged torefer 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 email@example.com
The charge transferred is Q = (2.5 × 104 C/s)(20 × 10−6 s) = 5.0 × 10−1 C.
E25-2 Use Eq. 25-4: r= E25-3 Use Eq. 25-4: F = (8.99×109 N·m2 /C2 )(3.12×10−6 C)(1.48×10−6 C) = 2.74N. (0.123 m)2 (8.99×109 N·m2 /C2 )(26.3×10−6 C)(47.1×10−6 C) = 1.40 m (5.66 N)
E25-4 (a) The forces are equal, so m1 a1 = m2 a2 , or m2 = (6.31×10−7 kg)(7.22 m/s2 )/(9.16 m/s2 ) = 4.97×10−7 kg. (b) Use Eq. 25-4: q= (6.31×10−7 kg)(7.22 m/s2 )(3.20×10−3 m)2 = 7.20×10−11 C (8.99×109 N·m2 /C2 )
(a) Use Eq. 25-4, F = 1 q1 q2 1 (21.3 µC)(21.3 µC) = = 1.77 N 2 −12 C2 /N · m2 ) 4π 0 r124π(8.85×10 (1.52 m)2
(b) In part (a) we found F12 ; to solve part (b) we need to ﬁrst ﬁnd F13 . Since q3 = q2 and r13 = r12 , we can immediately conclude that F13 = F12 . We must assess the direction of the force of q3 on q1 ; it will be directed along the line which connects the two charges, and will be directed away from q3 . The diagram below shows the directions.
F 12 F 23 F netF 12
From this diagram we want to ﬁnd the magnitude of the net force on q1 . The cosine law is appropriate here: F net 2 = = = =
2 2 F12 + F13 − 2F12 F13 cos θ, 2 (1.77 N) + (1.77 N)2 − 2(1.77 N)(1.77 N) cos(120◦ ), 9.40 N2 , 3.07 N.
E25-6 Originally F0 = CQ2 = 0.088 N, where C is a constant. When sphere 3 touches 1 the 0 charge on both becomes Q0 /2. When sphere 3 thetouches sphere 2 the charge on each becomes (Q0 + Q0 /2)/2 = 3Q0 /4. The force between sphere 1 and 2 is then F = C(Q0 /2)(3Q0 /4) = (3/8)CQ2 = (3/8)F0 = 0.033 N. 0 E25-7 The forces on q3 are F31 and F32 . These forces are given by the vector form of Coulomb’s Law, Eq. 25-5, F31 F32 = = 1 q3 q1 1 q3 q1 ˆ ˆ , r 2 r31 = 4π 2 31 4π 0 r31 0 (2d) 1 q3 q2 1 q3 q2 ˆ ˆ . r 2 r32 = 4π 2 32 4π 0 r32 0 (d)...
Leer documento completo
Regístrate para leer el documento completo.