Electric Forces and Electric Fields
7
Problem Solutions
15.1
Since the charges have opposite signs, the force is one of attraction .
Its magnitude is
F=
15.2
k e q1 q2
r2
9
9
N m 2 ( 4.5 10 C ) ( 2 .8 10 C )
= 8.99 109
= 1.1 10 8 N
2
2
C
( 3.2 m )
The ele
Optical Instruments
359
(b) When the object distance is p = 33.0 cm , the corrective lens should produce parallel
rays ( q ) . Then the implanted lens will focus the final image on the retina.
From the thin lens equation, the required focal length is f =
Current and Resistance
17.5
The period of the electron in its orbit is T = 2 r v , and the current represented by the
orbiting electron is
I=
=
17.6
Q e v e
=
t T 2 r
( 2 .19 10
6
m s ) ( 1.60 10 19 C )
2 ( 5.29 10 11 m )
= 1.05 10 3 C s = 1.05 mA
The mas
Alternating Current Circuits and Electromagnetic Waves
(c)
2
Z = R2 + X L , so X L = Z 2 R 2 =
and
21.30
(a)
L=
2
2
( 208 ) ( 40.0 ) = 204
XL
204
=
= 0.541 H
2 f 2 ( 60.0 Hz )
X L = 2 fL = 2 ( 60.0 Hz ) ( 0.100 H ) = 37.7
XC =
1
1
=
= 13.3
2 fC 2 ( 60
34
CHAPTER 15
Thus, if R = 1.27 10 3 m , we must have
sin 2 =
ay R
2
v0
=
( 6.90 10
10
m s 2 ) ( 1.27 10 3 m )
( 9 550
m s)
2
= 0.961
giving 2 = 73.9 or 2 = 180 73.9 = 106.1 . Hence, = 37.0 or 53.0
(b) The time of flight for each possible angle of project
24
CHAPTER 15
The total charge enclosed by the cylindrical gaussian surface is Q = A , where is the
charge density on the conducting surface. Hence, Gausss law gives
EA =
15.47
F=
15.48
(a)
k e q1 q2
r2
F=
A
or E =
0
o
9
2
19
k e 2 ( 8.99 10 N m C )( 1.60
Direct-Current Circuits
18.4
a
(a) The current through this series combination is
I=
( V )bc
Rbc
=
b
9.0 W
12 V
= 2.0 A
6.0
Therefore, the terminal potential difference of the
power supply is
105
c
6.0 W
DV
V = I Req = ( 2 .0 A ) ( 9.0 + 6.0 ) = 30 V
(b)
Games of Incomplete Information
A
L TEX le: harsanyi Daniel A Graham <[email protected]>, June 16, 2005
The Harsanyi Setup
n: the number of players.
Di : the set of possible actions or decisions available to the ith player, i = 1, . . . n.
Ti : th
Reflection and Refraction of Light
247
Finally, using Snells law in the third placement gives
sin R =
and
22.27
sin 31.7
n1 sin 26.5
= ( n1 sin 26.5 )
= 0.392
n3
n1 sin 36.7
R = 23.1
When the Sun is 28.0 above the horizon, the angle of
incidence fo
58
16.45
CHAPTER 16
The capacitance of this parallel plate capacitor is
6
2
A
C 2 ( 1.0 10 m )
12
C = 0 = 8.85 10
= 1.1 10 8 F
2
d
N m ( 800 m )
With an electric field strength of E = 3.0 106 N C and a plate separation of d = 800 m ,
the potential differe
A Simple Genetic Algorithm
A
L TEX le: ga-nb-all Daniel A. Graham <[email protected]>, June 30, 2005
For the objective function to be maximized, take a quadratic with one local and one global maximum
in the interval [0,31]
In:
objective = (x 14) 2;
I
Solution by Pretense
A
L TEX le: pretense Daniel A. Graham <[email protected]>, November 15, 2005
This handout describes a trick commonly used to solve for the symmetric Nash equilibrium in games
of incomplete information. To illustrate this method,
Wave Optics
Thus,
335
1
2nt
sin 2 =
(1 sin 2 2 ) = 2nt cos 2
sin 2
cos 2
= 2 nt tan 2
The condition for constructive interference is then
2nt cos 2 = ( m + 1 2 )
(b) When 1 = 30.0 , then
n1 sin 1
1 ( 1.00 ) sin 30.0
= 21.2
= sin
n
1.38
2 = sin
Direct-Current Circuits
18.17
3.00 W
We name the currents I1 , I 2 , and I 3 as shown.
Using Kirchhoffs loop rule on the rightmost
loop gives
I1
I2
5.00
W
+12 .0 V- ( 1.00+3.00 ) I 3
( 5.00 + 1.00 ) I 2 4.00 V = 0
( 2 .00 ) I 3 + ( 3.00 ) I 2 = 4.00 V
or
Reflection and Refraction of Light
265
(b) Observe from the figure above that sin = L d . Thus, the distance the light travels
inside the plastic is d = L sin , and if L = 50.0 cm = 0.500 m , the time required is
t =
22.62
1.20 ( 0.500 m )
d L sin
nL
=
=
Envelopes
A
L TEX le: EnvelopesPuzzle Daniel A. Graham <[email protected]>, June 16, 2005
An honest but mischievous father tells his two risk-neutral sons, Andy and Bill, that he has placed
10n dollars in one envelope and 10n+1 dollars in another env
Mirrors and Lenses
289
(b) The overall magnification is
q q
M = M1 M2 = 1 2
p1 p2
30.0 cm ( 25.0 cm )
=
= + 2.50
15.0 cm ( 20.0 cm )
(c) Since q2 < 0 , the final image is virtual ; and since M > 0 , it is upright
23.43
From the thin lens equati
Electrical Energy and Capacitance
Problem Solutions
16.1
(a) The work done is W = F s cos = ( qE ) s cos , or
W = ( 1.60 10 19 C ) ( 200 N C ) ( 2 .00 10 2 m ) cos 0 = 6.40 10 19 J
(b) The change in the electrical potential energy is
PEe = W = 6.40 10 19
Magnetism
19.32
Since the centripetal acceleration is furnished by the magnetic force acting on the ions,
mv
mv 2
qvB =
or the radius of the path is r =
. Thus, the distance between the impact
r
qB
points (that is, the difference in the diameters of the p
Mirrors and Lenses
23.59
(a) The lens makers equation,
1
1
1
= ( n 1)
f
R1 R2
299
, gives
1
1
1
= ( n 1)
5.00 cm
9.00 cm 11.0 cm
which simplifies to n = 1 +
1
99.0
= 1.99
5.00 11.0 + 9.00
(b) As light passes from left to right through the lens, th
The Envelope Theorem
A
L TEX le: Envelope-nb-all Daniel A. Graham <[email protected]>, June 22, 2005
An example without constraints
Here we consider a simple maximization problem without constraints, with a single choice variable and
a single paramet
Wave Optics
24.44
When light of wavelength passes through a single slit of width a, the first minimum is
observed at angle where
sin =
or = sin 1
a
a
This will have no solution if a < , so the maximum slit width if no minima are to be
seen is a = = 632 .
Current and Resistance
17.5
The period of the electron in its orbit is T = 2 r v , and the current represented by the
orbiting electron is
I=
=
17.6
Q e v e
=
t T 2 r
( 2 .19 10
6
m s ) ( 1.60 10 19 C )
2 ( 5.29 10 11 m )
= 1.05 10 3 C s = 1.05 mA
The mas
Induced Voltages and Inductance
(c) The emf is first maximum at t =
20.36
189
T 0.50 s
=
= 0.13 s
4
4
Treating the coiled telephone cord as a solenoid,
L=
0 N A
2
=
( 4 10
7
2
2
T m A ) ( 70.0 ) ( 1.30 10 2 )
4
0.600 m
= 1.36 10 6 H = 1.36 H
20.37
20.38
Magnetism
19.8
The speed attained by the electron is found from
v=
147
1
mv 2 = q ( V ) , or
2
2 ( 1.60 10 19 C ) ( 2 400 V )
2 e ( V )
=
= 2 .90 107 m s
m
9.11 10 31 kg
(a) Maximum force occurs when the electron enters the region perpendicular to the
fie
Mirrors and Lenses
23.25
281
As parallel rays from the Sun ( object distance, p ) enter the transparent sphere from
air ( n1 = 1.00 ) , the center of curvature of the surface is on the side the light is going
toward (back side). Thus, R > 0 . It is observ
Reflection and Refraction of Light
22.4
239
(a) The time for the light to travel to the stationary mirror and back is
3
2 d 2 ( 35.0 10 m )
t =
=
= 2 .33 10 4 s
8
c
3.00 10 m s
At the lowest angular speed, the octagonal mirror will have rotated 1 8 rev in
19
F qE ( 1.60 10 C ) ( 640 N C )
a= =
=
= 6.12 1010 m s 2
m mp
1.673 10 -27 kg
(a)
(b) t =
1.20 10 6 m s
v
=
= 1.96 10 5 s = 19.6 s
a 6.12 1010 m s 2
x =
(c)
2
v 2 v0
f
(d) KE f =
2
6
10
2
11.8 m
2
1
1
mp v 2 = ( 1.673 10 27 kg )( 1.20 106 m s ) = 1.20 1
300
CHAPTER 23
(b) When q = 3 f , we find p =
(c) In case (a), M =
3 f f
=
3f
4
q
4f
=
= 3
p
4f 3
and in case (b), M =
23.61
( 3 f ) f
q
3 f
=
= +4
p
3f 4
If R1 = 3.00 m and R2 = 6.00 m , the focal length is given by
1 n1
1
1 n1 n2
= 1
+
=
f n2
3.00 m