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
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 th
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
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
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
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
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
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 decision
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 S
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 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 max
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
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
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 )
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
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 d
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 sinc
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 t
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
po
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)
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 constrai
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 maximu
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
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.
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
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
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 s
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
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