UNIT 1
1. 13.3 MC
2. (a) 50.7 million N (uses mass of nucleon in solution)
or
51.3 million N (uses Avagadro's number in the solution)
(b) 144 million or 145 million N North; diagram
3. (5.45)a
4. (a) 4kqQ/a²
(b) towards the negative charge
5. (a) 4.14 cm to right of 1 microC charge
(with the 1 microC charge placed on the left)
(b) sketch of electric field lines
6. (a) kq[(2/(x+a)²)-(1/(x-a)²)]i
(b) graph of E_x vs x
7. 60.75 N/C
8. (a) Q/l (b) along the y axis (up for Q>0)
(c) kQ/(y(sqrt(y²+(l²/4))))
(d) proof
9. For both components, the magnitude turns out
to be k(lambda)/R
10. (2kQ/(pi)a²) to the RIGHT
11. proof
12. -5e
13. 3.09 kN/C
14. (a) 55.6 pm
(b) 5.34 x 10^(-24) N⋅m; sketch
(c) -3.65 x 10^-24 J
15. (a) 693 kg/s (b) 693 kg/s
(c) 347 kg/s (d) 347 kg/s
(e) 575 kg/s
16. 23.9 cm/s
17. (a) S1, 0; S2, 0.081 N⋅m²/C;
S3, 0; S4, 0;
S5, -0.135 N⋅m²/C; S6, 0
(b) -0.478 pC
18. (a) inner, (3q/(4pi))(1/(b^3-a^3))
outer, -(3q/(4pi))(1/(d^3-c^3))
(b) (i) 0
(ii) (kq/r^2)((r^3-a^3)/(b^3-a^3))
(iii) (kq/r^2)
(iv) (kq/r^2)(1-((r^3-c^3)/(d^3-c^3)))
(v) 0
(c) graph
19. (a) 2k(lambda)/r radially outward
(b) 2k(lambda)/r radially outward
(c) graph
(d) inner: -lambda; outer: +lambda
20. (a) (rho)r/(2epsilon_0) radially outward
(b) 2klambda/r
(c) You must show that the expression for
each region gives the same answer.
(d) graph
21. (a) outside the shells: 4 lines pointing inward;
between the shells: 20 lines pointing inward;
inside the inner shell: 8 lines pointing inward
(b) no difference in field lines (except that in the
insulator case, some of the field lines pass
through the shells, while in the conductor case
all field lines stop at the shell surfaces);
+q on inside of inner shell, -(5/2)q on outside;
+(5/2)q on inside of outer shell, -q/2 on outside
22. (a) E = 0
(b) E = kQa/r²
(c) k(Qa + Qb)/r²
(d) (Qa + Qb)/4(pi)b2² on outside of outer shell;
(-Qa)/4(pi)b1² on inside of outer shell;
(+Qa)/4(pi)a2² on outside of inner shell
(e) graph
23. (a) 2.05 x 10^10 excess electrons
(b) 2.90 x 10^18 atoms
(c) 141 million
24. (a) 240 N/C
(b) -6.37 nC/m²
(c) +3.18 nC/m²
25. (a) 3.51 MN/C down (b) 0.79 MN/C down (c) 0.79 MN/C up
(d) top plate: top, -7 microC/m²; bottom, +31 microC/m²
bottom plate: top, -31 microC/m²; bottom, -7 microC/m²
26. (a) on sphere, Q; on shell, Q
(b) for sphere, on outer surface;
for shell, -Q on inner surface, +2Q on outer surface
(c) at a, Q/4(pi)a²; at b, -Q/4(pi)b²; at c, +Q/2(pi)R²
27. (a) (q)sqrt(2k/md)
(b) (2q)sqrt(k/md)
(c) (q)sqrt(2k/md)
(d) zero
28. 4.88 kJ
29. (a) 352 m/s²
(b) 7.50 m/s
Chapter 21
62. (a) top +, middle -, bottom +
You must explain your reasoning.
(b) The field is zero to the right
and left of the central charge
where the lines are least dense.
Draw a picture to show how the
points of zero field are produced
by the three charges.
66. 0.658 pN in the -x direction; attractive
Chapter 22
8. (a) 452 N⋅m²/C
(b) -881 N⋅m²/C
(c) -429 N⋅m²/C
(d) 723 N⋅m²/C
(e) -158 N⋅m²/C
(f) No. You must explain why.
30. (a) 282 kN/C to the left
(b) 395 kN/C to the left
(c) 169 kN/C to the left
54. (a) You may omit part (a).
(b) inside, rho*|x|/epsilon_0 away from center
outside, rho*d/epsilon_0
Chapter 23
12. 12.1 mN
END OF UNIT 1 ANSWERS
-------------------------------------------------------
UNIT 2
1. (a) kq((4/x)-(1/(x-a)))
(b) (4/3)a
(c) kq((4/x²)-(1/(x-a)²))i
(d) 2a
2. (a) k(lambda)ln((y+L)/y)
(b) k(lambda)L/(y+L)y
(c) zero, since V(x) must be a
max at P by symmetry
3. (a) kc(sqrt(L² + y²)-y)
(b) -kc((y/sqrt(L² + y²))-1)
(c) we don't know V(x) for any y
and symmetry cannot help
4. (.113)sigma(R)/epsilon0
5. (a) (kQ/R³)r
(b) (kQ/2R)
(c) center is higher if positive; surface if negative
(d) -kQ/(2R); kQ/R
6. (a) zero J
(b) +823 microJ
(c) -823 microJ
(d) zero V at B; -29400 V at C
(e) +375 microJ; -13400 V at D
(f) K=4.70 fJ; v=1.02e8 m/s (note that this speed, calculated
by Newtonian mechanics, is so large that special relativity
should instead be used in order to correctly calculate the
actual speed)
7. (a) outside c, k2Q/r;
from b to c, k2Q/c constant
from a to b, kQ((2/c)+(1/b)-(1/r))
inside a, kQ((2/c)+(1/b)-(1/a)) constant
(b) graph
8. (a) 288 kV
(b) 0.107 microC from 1 to 2
GBA NOTES THAT THE ORDERING OF PROBLEMS 9-12 WAS CORRECTED
HERE ON 7/13 at 8:30 PM.
9. (a) 3.5 pF
(b) 3.5 pF
(c) 57.1 V
10. After a touch, A always has four times
as much charge as B; so when there is no
more charge transfer on a touch, A will
have charge q and B will have charge q/4.
11. (a) proof
(b) dW = (2kq/a)dq
(c) (k/a)(Q²)
12. (a) 1.41 J
(b) 4.22 J
GBA NOTES THAT THE ORDERING OF PROBLEMS 9-12 WAS CORRECTED
HERE ON 7/13 at 8:30 PM.
13. (a) Q = 1.58 mC; V_4 = 396 V; V_6 = 264 V
(b) V = 317 V; Q_4 = 1.27 mC; Q_6 = 1.90 mC
14. (a) circuit diagram
(b) 12.0 nF
(c) Q1 = 1.20 microC; Q2 = 0.40 microC; Q3 = 0.80 microC
(d) V1 = 60 V; V2 = V3 = 40 V
15. (a) 6900 N/C
(b) 85.5 V and 31.7 V
(c) 22.1 pF
(d) 30.0 nJ and 11.1 nJ
(e) 11.3 kN/C
(f) 702 pC and 1.15 nC
(g) 13.4 pF
(h) 30.0 nJ and 49.0 nJ
16. (a) 725 pF <== this number uses HRW's value of kappa
or for Pyrex, which is 4.7
863 pF <== this number uses Wolfson's value of kappa
for Pyrex, which is 5.6
(b) about 28 kV (more exactly 27.2 kV) <== these numbers
use 14 MV/m for the dielectric strength
of Pyrex (from HRW or Wolfson)
or
about 20 kV (more exactly 19.4 kV) <== these numbers
use 10 MV/m for the dielectric strength
of Pyrex (from Y&F)
17. 66.2 microJ
18. (a) 200 kV/m (b) 200 kV/m
(c) 1.77 microC/m² (d) 4.60 microC/m²
(e) -2.83 microC/m²
19. (a) 7.18
(b) 0.766 microC
(c) 2.30 mm
20. 4.80 mA away from the screen
21. (a) 2.44X10^19 electrons per second
(b) 1.18 MA/m²
(c) 0.0869 mm/s
(d) a unchanged; b decreases by a factor of 4;
c decreases by a factor of 4
(e) 12.9 MA/m²
(f) 1.91 mm/s
(g) 19.9 s
22. (a) 8700 A
(b) 15.1%
(c) 316 kV
23. (a) 871 pJ/m³
(b) change in U is 13600 J, which is
1.56 x 10^19 times bigger than
the available kinetic energy
24. 10.0 V
25. (a) 48.4 V
(b) 57.3 V
(c) 59.9 V
Chapter 23
50. (a) 763 pC; proof of result from 23:49(a)
must be shown here
(b) scale drawing; assign 0 V to r=infinity
(the 100 V circle is at 4.00 cm)
(c) drawing; answers to questions
should be yes
74. (a) i. zero ii. zero iii. 2.25 MV iv. zero
(b) graph, the potential at r=0 is 2.25 MV
(c) graph, the potential at r=0 is -2.25 MV
(d) graph, the potential at r=0 is 2.34 MV
Chapter 24
58. (a) there are (at least) two possible solutions
(b) your answer to this part will depend
on which solution you choose for (a);
be sure to state the voltage across
each good capacitor after the one bad
capacitor shorts out
64. (a) 7.20 mC
(b) 0.654 mC
(c) 0.432 J for (a); .0393 J for (b)
Chapter 25
48. (a) 5 Ohm - 1.11 W; 9 Ohm - 1.99 W
(b) 7.18 W
(c) 4.07 W
(d) proof
60. (a) 45.0 A
(b) 45.0 A
(c) 2.74 V/m
(d) 2.34 V/m
(e) 2.81 V
64. (a) proof
(b) (V_ab)ab/(rho(b-a)r²)
(c) proof; hint - let b = a+L with a >> L
and apply binomial expansion
Chapter 26
14. 25 Ohm - 7.00 A; 20 Ohm - 9.95 A
62. (a) 8.60 V
(b) 3.80 A
(c) 2.00 A towards the bottom of the figure
END OF UNIT 2 ANSWERS
-------------------------------------------------------
UNIT 3
1. (a) 3750 m/s
(b) drawing
2. (a) 2.60 x 10^6 m/s
(b) 0.109 microseconds
(c) 141 keV
(d) 70.3 kV
3. (-11.4 N/C)i - (6 N/C)j + (4.8 N/C)k
4. (a) drawing
(b) free-body diagram
(c) 8.64 mT
5. (a) 20 min
(b) 59.4 m(N⋅m)
6. proof; out of the page
7. proof; out of the page
8. proof
9. (a) 78.5 microT (b) 1.08 microN⋅m
10. 7.14 microN West
11. (a) (mu0)J(|z|) (b) (mu0/2)Jh
12. (a) (pi/3)R²J_0
(b) (mu0/6r)R²J_0
(c) (mu0/2)rJ_0(1-(2r/3R))
13. (a) 4.8 mT
(b) 0.933 mT
(c) zero
14. (a) (mu0)I(sqrt(n²+(1/2piR)²))
(b) INVtan(1/2(pi)nR)
15. (a) Bmax = 174 microT; Bmin = 139 microT <== these answers use
(b) 0.140 microWb (1 Wb = 1 T⋅m²) HRW's value of rho_CU
(c) heating rate is 1.57 W; doesn't get very hot (16.9 nOhms⋅m)
(a) Bmax = 175 microT; Bmin = 140 microT <== these answers use
(b) 0.141 microWb (1 Wb = 1 T⋅m²) Wolfson's value
(c) heating rate is 1.58 W; doesn't get very hot (16.8 nOhms⋅m)
(a) Bmax = 171 microT; Bmin = 137 microT <== these answers use
(b) 0.138 microWb (1 Wb = 1 T⋅m²) Y&F's value
(c) heating rate is 1.54 W; doesn't get very hot (17.2 nOhms⋅m)
16. (a) 0.1 V
(b) 25 mA
(c) 1.25 mN
(d) 2.5 mW by using I²R
(e) 2.5 mW by using F⋅v
17. (a) 0.452 V by using motional EMF
(b) 0.452 V by using Faraday's Law
18. (a) (b-a)hB
(b) (b-a)hB/deltaT CCW
(c) 4(b-a)hB/deltaT CW
(d) zero
(e) zero
(f) (mu_0*Ih/(2pi))ln(b/a)
(g) (mu_0*Ih/(2pi))v*((1/(a+vt))-(1/(b+vt)))
in the CCW direction
19. (a) 3.48 mWb
(b) 15.5 microC
20. (a) 25 times per second
(b) 466 turns
(c) 20 Ohms
(d) 63.7 mWb
(e) 0.01 and 0.03 s (you must explain why)
21. (a) 1.51 Wb
(b) at 90 degrees, 1.20 Wb
at 180 degrees, 0.892 Wb
(c) (posted after due)
(d) (posted after due)
(e) 155 Wb/s
(f) (posted after due)
22. ((mu_0)l/2pi)ln((a+w)/a)
23. (a) 3.21 mV
(b) 10.2 mN/C
(c) 40.1 mA
(d) will be posted after HW is due
(e) 22.9 mH
Chapter 27
76. (a) 0.0299 N⋅m in the +j direction
(b) 0.0173 N⋅m in the -j direction
(c) the answers for the two cases are the same;
you must carefully explain why
(d) 0.0173 J for the field in the +x dir;
0.00463 J for the field in the -z dir
78. (a) on PQ, F=0; on RP, F = 12 N into page;
on QR, F= 12 N out of page
(b) zero
(c) on PQ, tau=0; on RP, tau=0;
on QR, tau = 3.60 N⋅m to right
(d) 3.60 N⋅m to right; the results agree
(e) out of the plane
Chapter 28
10. a, 2.00 microT; b, 0.705 microT;
c, 2.00 microT; d, zero;
e, 0.545 microT; also a figure
36. a, zero; b, -5.0 microT⋅m;
c, +2.5 microT⋅m; d, +5.0 microT⋅m
86. 0.5(mu_0)v(sigma)
Chapter 29
6. (a) 0.0302 V + (0.302 mV/s³)t³
(b) 0.113 mA
26. (a) graph
(b) graph
28. (a) 170 microN/C
(b) 339 microN/C
(c) 530 microN/C
(d) will be posted after this HW is due
77. (a) from a to b
(b) (Rmg)tan(phi)/(L²B²cos(phi))
(c) (mg)tan(phi)/(LB)
(d) Rm²g²tan²(phi)/(L²B²)
(e) same answer as in (d) by a different method
END OF UNIT 3 ANSWERS
-------------------------------------------------------
UNIT 4
1. (a) 7.20 x 10^{11} (V/m)/s
(b) increasing
2. (a) V1 = 25 V; V2 = V4 = 50.0 V; V3 = 0 V
A1 = A3 = 0.500 A; A2 = 0 A
(b) V3 = 75.0 V; all other meters read zero
(c) 5.63 microC; long after S is closed
3. (a) 1.89 pT
(b) graph; at 10 cm, B = 0.565 pT
4. (a) (4/3) A
(b) at R/4 and at 4R
5. proof
6. (a) by integration, U_C = (1/2)C(EMF)² and
energy out from battery = C(EMF)²
(b) by integration, heat from resistor = (1/2)C(EMF)²
7. (a) 238.7 W (b) 154.8 W (c) 393.5 W
8. (a) 18.7 J (b) 5.10 J (c) 13.6 J
9. DiffEq is -I(t)R - L(dI/dt) = 0
10. (a) 894 microC
(b) 20.0 mJ
(c) graph (cos function with amp 2.0 A)
(d) graph (sin function with amp 894 microC)
11. (a) 0.405 J
(b) 31700 times per second
(c) 0.304 J
(d) 7.89 microseconds
(e) 1.06 A
12. (a) n(5.00 microseconds) where n=1,2,3,4,...
(b) (2n+1)(2.50 microseconds) where n=0,1,2,3,4,...
(c) (2n+1)(1.25 microseconds) where n=0,1,2,3,4,...
13. (a) 3.6 mH (b) 1326 Hz (c) 189 microseconds
14. (a) (EMF_max)sin((w_d)t)=L(dI/dt)
(b) I(t)=-(EMF_max/(Lw_d))cos((w_d)t)
(c) 5.22 mA (d) zero
(e) 4.52 mA (f) taking energy (you must explain why)
15. (a) (EMF_max)sin((w_d)t)=Q(t)/C; no
(b) I(t)=(EMF_max*Cw_d)cos((w_d)t)
(c) 39.1 mA (d) zero
(e) -33.9 mA (f) supplying energy (you must explain why)
16. (a) (EMF_max)sin((w_d)t)=I(t)R; no
(b) I(t)=(EMF_max/R)sin((w_d)t)
(c) 0.6 A
(d) 0.6 A
17. (a) 16.6 Ohms (b) 422 Ohms (c) 521 mA
(d) 33.2 Ohms (e) 408 Ohms (f) 539 mA
18. (a) 14400 rad/s
(b) 37.5 V
19. posted after due
20. (a) 2.50 Ohms
(b) 1.93 microF
(c) 50900 rad/s
21. (a) 76.4 mH (b) yes; 17.8 Ohms; resistor would
consume energy, not store it
22. C is 2.86 pF; R is 126 mOhms
23. (a) 0.824 radians
(b) 0.680
(c) 356 Ohms
(d) 161 V
(e) 49.4 W
(f) 49.4 W (posted after due)
(g) zero (posted after due)
(h) 0.665 A
(i) 107 W
24. 563 nJ/m³
25. (a) 4.35 mT
(b) 7.53 J/m³
(c) 1.52 microm³
(d) 11.4 microJ
(e) 3.65 microH
(f) same as (d) by a different method
26. (a) 3.33 microT
(b) 1.67 microT in the negative x direction
(c) 50 m
27. (a) 0.260 pW/m²
(b) 8.66x10^-22 J/m³
(c) 14.0 microV/m
Satellite dishes can process amazingly small signals; of
course, in the first step of the process the signal is
concentrated at the receiver by the parabolic reflector
of the dish (the actual receiver is at the focal point
of the parabolic reflector).
Chapter 26
84. (a) 7.10 J
(b) 3.62 kW
(c) 1.81 kW
Chapter 30
46. (a) EMF(t)=(23.4 V)sin((240pi rad/s)t)
two graphs
(b) 23.4 V; zero
(c) 0.124 A; zero
62. (a) V1 = 33.3 V; V2 = V4 = 16.7 V; V3 = 0 V
A1 = A3 = 0.333 A; A2 = 0 A
(b) V1 = 38.5 V; V2 = 0; V3 = V4 = 11.5 V
A1 = 0.385 A; A2 = 0.153 A; A3 = 0.230 A
68. (a) i1 = 1.50 A; i2 = 1.80 A
(b) 2.40 A
Chapter 31
36. (a) 945 rad/s
(b) 70.6 ohms
(c) resistor - 120 V; capacitor - 450 V;
inductor - 450 V
(d) V4 = 0 V; V5 = 84.9 V
(e) V1 = 27.4 V; V2 = 70.1 V; V3 = 150.4 V;
V4 = 80.3 V; V5 = 84.9 V
(f) V1 = 36.6 V; V2 = 180.6 V; V3 = 104.0 V;
V4 = 76.6 V; V5 = 84.9 V
Chapter 32
18. 15.9 microJ
20. (a) 240 microW
(b) 17.4 V/m
END OF UNIT 4 ANSWERS
-------------------------------------------------------