30
15.58
CHAPTER 15
y
As shown in the sketch, the electric field
at any point on the x-axis consists of two
parts, one due to each of the charges in the
dipole.
r
+q
q
E = E+ E =
E=
ke q
( x a)
2
ke q
r+2
a
u
r
E
a
r+
ke q
( x + a)
x
r2
ke q
2
u
r
E+
( x
Reflection and Refraction of Light
22.16
The angle of incidence is
air
water
4.00 m
2.00 m
1 = tan 1
= 26.6
4.00 m
Therefore, Snells law gives
2.00
m
q2
243
f
q1 q1
n1 sin 1
n2
2 = sin 1
( 1.333 ) sin 26.6
= sin 1
= 36.6
1.00
and the angle
268
CHAPTER 23
Answers to Even Numbered Conceptual Questions
2.
If the finger and the image are at the same distance from you, then they will coincide
regardless of what angle you view them from. However, if one is closer than the other,
they will appear
322
CHAPTER 24
24.38
(a) If d =
1 cm
= 6.67 10 4 cm = 6.67 10 6 m , the highest order of = 500 nm that can
1 500
be observed will be
mmax =
(b) If d =
=
( 6.67 10
6
500 10
m ) ( 1)
9
m
= 13.3 or 13 orders
1 cm
= 6.67 10 5 cm = 6.67 10 7 m , then
15 000
mm
Mirrors and Lenses
23.15
277
The focal length of the mirror may be found from the given object and image distances
as 1 f = 1 p + 1 q , or
f=
pq
( 152 cm ) ( 18.0 cm )
=
= + 16.1 cm
p + q 152 cm + 18.0 cm
For an upright image twice the size of the object,
170
CHAPTER 19
19.65
Note: We solve part (b) before part (a) for this problem.
(b) Since the magnetic force supplies the centripetal acceleration for this particle,
qvB = mv 2 r or the radius of the path is r = mv qB . The speed of the particle may be
wri
304
CHAPTER 24
Answers to Even Numbered Conceptual Questions
2.
The wavelength of light is extremely small in comparison to the dimensions of your hand,
so the diffraction of light around obstacles the size of your hand is totally negligible.
However, sou
Wave Optics
24.17
313
With ncoating > nair and ncoating > nlens , light reflecting at the air-coating boundary
experiences a phase reversal, but light reflecting from the coating-lens boundary does
not. Therefore, the condition for destructive interferenc
356
CHAPTER 25
25.43
A fringe shift occurs when the mirror moves distance 4 . Thus, the distance moved
(length of the bacterium) as 310 shifts occur is
650 10 9 m
5
L = N shifts = 310
= 5.04 10 m = 50.4 m
4
4
25.44
A fringe shift occurs when the mirro
230
CHAPTER 21
R=
(b)
VDC 12 V
=
= 4.0
I DC
3.0 A
2
From Z = R2 + X L = R 2 + ( 2 f L ) , we find
2
( 6.0 ) ( 4.0 )
= 1.2 10 2 H = 12 mH
2 ( 60 Hz )
2
Z 2 R2
L=
=
2 f
21.65
2
(a) The radiation pressure on the perfectly reflecting sail is
p=
2 ( Intensity
128
CHAPTER 18
(d) The power dissipated in each resistor is found from P = ( V ) R with the
following results:
2
( V )ac
2
Pac =
Rac
( V )ed
=
2
Ped =
Red
( V )cd
=
2
Pcd =
18.48
Rcd
=
( 6.0 V )
= 6.0 W
6.0
( 1.8 V )
( 3.0 V )
Pce =
= 0.54 W
Pfd =
= 1.5
286
CHAPTER 23
23.36
We are given that f = +12.5 cm and q = 30.0 cm . Then, the thin lens equation,
111
+ = , gives
pqf
p=
qf
( 30.0 cm ) ( 12.5 cm )
=
= +8.82 cm
q f
30.0 cm 12.5 cm
M=
and the lateral magnification is
q
30.0 cm
=
= + 3.40
p
8.82 cm
Since
176
CHAPTER 20
Answers to Conceptual Questions
2.
Consider the copper tube to be a large set of rings stacked one on top of the other. As the
magnet falls toward or falls away from each ring, a current is induced in the ring. Thus,
there is a current in t
185
Induced Voltages and Inductance
20.25
(a) After the right end of the coil has
entered the field, but the left end
has not, the flux through the area
enclosed by the coil is directed into
the page and is increasing in
magnitude. This increasing flux
in
64
16.61
CHAPTER 16
The charges initially stored on the capacitors are
Q1 = C1 ( V )i = ( 6.0 F ) ( 250 V ) = 1.5 10 3 C
and
Q2 = C2 ( V )i = ( 2.0 F ) ( 250 V ) = 5.0 10 2 C
When the capacitors are connected in parallel, with the negative plate of one co
202
CHAPTER 21
Answers to Even Numbered Conceptual Questions
2.
At resonance, X L = XC . This means that the impedance Z = R 2 + ( X L XC ) reduces to
2
Z = R.
4.
The purpose of the iron core is to increase the flux and to provide a pathway in which
nearl
98
CHAPTER 17
(b) If the minimum acceptable power delivered to the cleaner is Pmin , then Equations
(2) and (3) give the maximum allowable total resistance as
Rt , max = Ri + 2Rc , max =
( V )
4
Pmin P1
=
( V )
2
Pmin P1
so
Rc , max =
2
2
2
1 ( V ) 2
1 (
252
CHAPTER 22
22.41
(a) Snells law can be written as
sin 1 v1
= . At the critical angle of incidence (1 = c ) ,
sin 2 v2
v
the angle of refraction is 90 and Snells law becomes sin c = 1 . At the concrete-air
v2
boundary,
v1
1 343 m s
= 10.7
= sin
Alternating Current Circuits and Electromagnetic Waves
211
(d) When the instantaneous current is a maximum ( i = I max ) , the instantaneous voltage
across the resistor is vR = iR = I max R = VR , max = 1.3 10 2 V . Again, the
instantaneous voltage across
234
CHAPTER 22
Answers to Even Numbered Conceptual Questions
2.
Ceilings are generally painted a light color so they will reflect more light, making the
room brighter. Textured materials are often used on the ceiling to diffuse the reflected
light and red
338
CHAPTER 25
Answers to Conceptual Questions
2.
The objective lens of the microscope must form a real image just inside the focal point of
the eyepiece lens. In order for this to occur, the object must be located just outside the focal
point of the obje
Electric Forces and Electric Fields
(c) Outside the spherical dome, E =
( 8.99 10
E=
15.35
9
ke q
. Thus, at r = 4.0 m ,
r2
N m 2 C2 ) ( 2 .0 10 4 C )
( 4.0 m )
21
2
= 1.1 10 5 N C
For a uniformly charged sphere, the field is strongest at the surface.
Thu
Optical Instruments
347
(b) With the image at the normal near point, the angular magnification is
m = mmax = 1 +
25.20
25.0 cm
25.0 cm
= 1+
= + 7.14
f
4.07 cm
(a) For maximum magnification, the image should be at the normal near point
( q = 25.0 cm ) of t
80
17.23
CHAPTER 17
At 80C,
I=
I = 2.6 10 2 A = 26 mA
or
17.24
V
V
5.0 V
=
=
R
R0 1 + ( T T0 ) ( 200 ) 1 + ( 0.5 10 3 C 1 ) ( 80C 20C )
If R = 41.0 at T = 20C and R = 41.4 at T = 29.0C , then R = R0 1 + ( T T0 ) gives
the temperature coefficient of resis
Electric Forces and Electric Fields
31
(b) Any point on the x-axis in the range x < 0 is located closer to the larger magnitude
charge ( q = 5.00 C ) than the smaller magnitude charge ( q = 4.00 C ) . Thus, the
contribution to the resultant electric field
Direct-Current Circuits
101
20.
Even if the fuse were to burn out or the circuit breaker to trip, there would still be a
current path through the device, and it would not be protected.
22.
When connected in series, all bulbs carry the same current. Thus,
Magnetism
19.47
N
The magnetic field inside a long solenoid is B = 0 nI = 0 I . Thus, the required
L
current is
I=
19.48
161
(1.00 104 T ) ( 0.400 m ) = 3.18 102 A = 31.8 mA
BL
=
0 N ( 4 10 -7 T m A ) ( 1 000 )
(a) From R = L A , the required length of wi
110
CHAPTER 18
18.13
The resistors in the circuit can be combined in the stages shown below to yield an
equivalent resistance of Rad = ( 63 11) .
6.0 W
3.0 W
a 3.0 W
b
I1
c
d
3.0 W
6.0 W
4.0 W
I2
3.0 W
a
I2
2.0 W
I
e
12 W
I12
I1
3.0 W
c
d
b
e
3.0 W
2.0 W
Electrical Energy and Capacitance
65
The final charge on each of the capacitors is
Q1 = C1 ( V ) = ( 4.0 F ) ( 44.4 V ) = 1.8 10 2 C
and
16.63
Q2 = C2 ( V ) = ( 2.0 F ) ( 44.4 V ) = 89 C
(a) V = V1 + V2 + V3 =
keQ 2 keQ keQ
+
x+d
x
xd
x ( x d ) 2 ( x 2 d
User access level feature is used to achieve different access levels for different
authorized users. It is useful for protecting the IP phone from unauthorized configuration,
and popularly used for the Hosted PBX solution. The following describes how to
c
Tested headset list compatible with Yealink IP Phone
Tested wired headset list compatible with Yealink IP Phone
Manufacturer
Model
Wired Headset Connection
SupraPlus (HW251, HW251N, HW261, HW261N)
Plantronics
Entera (HW111N, HW121N)
Practica (SP11, SP12)