This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **Imalllﬂa PROPERTIES OF A PARALLEL-PLATE CAPACITOR The plates of a parallel-plate capacitor in vacuum are 5.00 mm
apart and 2.00 m2 in area. A 10.0-kV potential difference is applied
across the capacitor. Compute (a) the capacitance; (b) the charge
on each plate; and (c) the magnitude of the electric field between
the plates. EXAMPLE 24.4 A CYLINDRICAL CAPACITOR Two long, coaxial cylindrical conductors are separated by vacuum
(Fig. 24.6). The inner cylinder has radius r” and linear charge den-
sity +;\. The outer cylinder has inner radius :3 and linear charge
density —I\. Find the capacitance per unit length for this capacitor. 24.6 A long cylindrical capacitor. The linear charge density A
assumed to be positive in this ﬁgure. The magnitude of charge
a length L of either cylinder is M. _XAMPLE 24.6 A CAPACITOR NETWORK Find the equivalent capacitance of the five-capacitor network
shown in Fig. 24.1051. 24.10 (a) A capacitor network between points a and b. {b} The 12-}.LF and 61.1.1: capacitors in series in (a) are replaced
by an equivalent 4-,:LF capacitor. {c} The 3-,u.F‘, ll-pLF, and 4-pF capacitors in parallel in (b) are replaced by an equivalent
lS—pF capacitor. (d) Finally, the 1811.1: and 911.13 capacitors in series in (c) are replaced by an equivalent 6-pF capacitor. (6) a (b) a it) 6! it!) a L.--'... replace these series
capacitors by an equivalent Replace these Series capacitors '=’- ' 9 F replace lheﬁc by an equivalent capacitor T ,u. parallel capacitors by
an equivalent capacitor E? b b b 24.21 -- For the system of capacitors shown in Fig. E2421, a
potential difference of 25 V is maintained across ab. (a) What
is the equivalent capacitance of this system between a and b?
(b) How much charge is stored by this system? (c) How much
charge does the 6.5-nF capacitor store? (d) What is the potential
difference across the 7.5-nF capacitor? Figure £24.21 7.5 nF 18.0 nF 30.0 nF 10.0 nF a HH a 6.5 nF Immnnn TRANSFERRING CHARGE AND ENERGY BETWEEN CAPACITORS We connect a capacitor C; = 8.0 pF to a power supply, charge it
to a potential difference V0 = 120 V, and disconnect the power
supply (Fig. 24.12). Switch S is open. (a) What is the charge Q0 on
C1? (b) What is the energy stored in C I? (c) Capacitor
C2 = 4.0 FF is initially uncharged. We close switch 3. After
charge no longer flows, What is the potential difference across each
capacitor, and what is the charge on each capacitor? ((1) What is the
final energy of the system? 24.12 When the switch S is closed, the charged capacitor C] is
connected to an uncharged capacitor C2. The center part of the
switch is an insulating handle; charge can flow only between the
two upper terminals and between the two lower terminals. _XA|VIPLE 24.9 TWO WAYS TO CALCULATE ENERGY STORED IN A CAPACITOR The spherical capacitor described in Example 24.3 (Section 241)
has charges +Q and —Q on its inner and outer conductors. Find
the electric potential energy stored in the capacitor (a) by using the
capacitance C found in Example 24.3 and (b) by integrating the
electric-ﬁeld energy density n. 24.5 A spherical capacitor.
Inner shell, charge +Q Gaussian surface Outer shell. charge -Q Immmala A SPHERICAL CAPACITOR WITH DIELECTRIC Use Gauss’s law to ﬁnd the capacitance of the spherical capacitor
of Example 24.3 (Section 24.1) if the volume between the shells is
ﬁlled with an insulating oil with dielectric constant X. 24.5 A spherical capacitor.
Inner shell, charge +Q
Gaussian surface Outer shell. charge -Q 24.66 " A parallel—plate ca- FigUle P24.66
pacitor is made from two
plates 12.0 cm on each side
and 4.50 mm apart. Half of
the space between these plates
contains only air, but the other
half is ﬁlled with Plexiglas®
of dielectric constant 3.40 (Fig. P24.66). An [8.0—V battery is
connected across the plates. (a) What is the capacitance of this
combination? (Hint: Can you think of this capacitor as equivalent
to two capacitors in parallel?) (b) How much energy is stored in
the capacitor? (c) If we remove the Plexiglas® but change nothing
else, how much energy will be stored in the capacitor? Imalllm. CURRENT DENSITY AND DRIFT VELOCITY IN A WIRE An lS-gauge copper wire (the size usually used for lamp cords),
with a diameter of 1.02 mm, carries a constant current of 1.67 A
to a ZOO-W lamp. The free-electron density in the wire is
3.5 X 1028 per cubic meter. Find (a) the current density and (b) the
drift speed. Iﬂmllm ELECTRIC FIELD, POTENTIAL DIFFERENCE, AND RESISTANCE IN A WIRE The IB-gauge copper wire of Example 25.! has a cross-sectional
area of 8.20 X 10-7 m2. It carries a current of 1.67 A. Find (a) the
electric-ﬁeld magnitude in the wire; (b) the potential difference
between two points in the wire 50.0 m apart; (c) the resistance of a
50.0-m length of this wire. _ONCEPTUAL EXAMPLE 25.4 A SOURCE IN AN OPEN CIRCUIT Figure 25.16 shows a source (a battery) with emf 5' = 12 V and
internal resistance r = 2 I]. (For comparison, the internal resist-
ance of a commercial 12-V lead storage battery is only a few thou-
sandths of an ohm.) The wires to the left of a and to the right of the
ammeter A are not connected to anything. Determine the respec-
tive readings Vab and I of the idealized voltmeter V and the ideal—
ized am meter A. 25.16 A source of emf in an open circuit.
Vab a
r=20,8=12V 25.18 “ What diameter must a copper wire have if its resistance
is to be the same as that of an equal length of aluminum wire with
diameter 2.14 mm? Imalllm A SOURCE IN A COMPLETE CIRCUIT We add a 4-D, resistor to the battery in Conceptual Example 25.4,
forming a complete circuit (Fig. 25.17). What are the voltmeter
and ammeter readings Verb and i now? 25.17 A source of emf in a complete circuit.
ii: = Vn'b' _ONCEPTUAL EXAMPLE 25.6 USING VOLTMETERS AND AMMETERS We move the voltmeter and ammeter in Example 25.5 to different
positions in the circuit. What are the readings of the ideal volt-
meter and ammeter in the situations shown in (a) Fig. 25.1851 and
(b) Fig. 25.1%? 25.18 Different placements of a voltmeter and an ammeter in a
complete circuit. (a) (b) A SOURCE WITH A SHORT CIRCUIT In the circuit of Example 25-5 we replace the 4-0 resistor with a
zero-resistance conductor. What are the meter readings now? EXAMPLE 25.8 POWER INPUT AND OUTPUT IN A COMPLETE CIRCUIT For the circuit that we analyzed in Example 25.5, find the rates of
energy conversion (chemical to electrical) and energy dissipation
in the battery, the rate of energy dissipation in the 4-0 resistor. and
the battery’s net power output. 25.33 “ The circuit shown Figure £25.33 in Fig. E2533 contains two 1.6 n 160 V
batteries, each with an emf and . an internal resistance, and two a b
resistors. Find (a) the current in 5,0 9 9_0 Q
the circuit (magnitude and di— 1.4 0 8.0 V rection) and (b) the terminal volt-
age Vab of the 16.0-V battery. Imalllm. EQUIVALENT RESISTANCE Find the equivalent resistance of the network in Fig. 26.3a (next
page) and the current in each resistor. The source of emf has neg-
ligible internal resistance. 26.3 Steps in reducing a combination of resistors to a single equivalent resistor and finding the
current in each resistor. S: [8V.r=0 (b) (C) (d) (e) (1‘) _XAMPLE 26.4 CHARGING A BATTERY In the circuit shown in Fig. 26.11, a l2-V power supply with un-
known internal resistance r is connected to a run-down recharge-
able battery with unknowa emf 8 and internal resistance 1 D. and
to an indicator light bulb of resistance 3 ﬂ carrying a current of
2 A. The current through the run-down battery is l A in the direc-
tion shown. Find r, S, and the current I through the power supply. Find power dissipated in each resistor.
26.11 In this circuit a power supply charges a run-down battery and lights a bulb. An assumption has been made about the polarity
of the emf 5 of the battery. ls this assumption correct? _XAMPLE 26.6 A COMPLEX NETWORK Figure 26.12 shows a “bridge" circuit of the type described at the
beginning of this section (see Fig. 26.6b). Find the current in
each resistor and the equivalent resistance of the network of five ”Sis‘o‘s' Find potential difference between a and b. 26.12 A network circuit with several resistors. 26.31 -- Ii‘l the circuit shown in Fig. E26.31 the batteries have
negligible internal resistance and the meters are both idealized.
With the switch 5' open, the voltmeter reads 15.0 V. (a) Find the emf 6' of the battery. (b) What will the ammeter read when the
switch is closed? Figure £26.31 _XAMPLE 26.8 DESIGNING AN AMMETER What shunt resistance is required to make the 1.00-mA, 20.0-0 meter described above into an ammeter with a range of 0 to
50.0 mA? _XAMPLE 26.9 DESIGNING A VOLTMETER What series resistance is required to make the LOO—mA, 20.0-[1
meter described above into a voltmeter with a range of 0 to 10.0 V? Immllml. MEASURING RESISTANCE || Suppose the meters of Example 26.10 are connected to a differ-
ent resistor as shown in Fig. 26.16b, and the readings obtained on
the meters are the same as in Example 26.10. What is the value of
this new resistance R, and what is the power dissipated in the ' a
”5'5“" ' 26.16 Ammeter—voltmeter method for measuring resistance. (a)
o R b c
i.“
RV
(h)
R (1 RV 'earson Education Inc, _XAMPLE 26.10 MEASURING RESISTANCE] The voltmeter in the circuit of Fig. 26.16a reads 12.0 V and the
ammeter reads 0.100 A. The meter resistances are RV = 10,000 I!
(for the voltmeter) and RA = 2.00 0 (for the ammeter). What are h ’ t R dth ' ' ' th ' t ?
t e resrs ance an e power dlSSIPEIBdm e was or 26.16 Ammeter—voltmeter method for measuring resistance. 32016 Pearson Education Inc, 26.22 Discharging a capacitor. (3) Before the switch is closed at time t : 0, the capacitor charge is Q0 and the
current is zero. (b) At time rafter the Iﬂlmllma DISCHARGING A CAPACITOR switch is closed, the capacitor charge is g and the current is t‘. The actual current . . direction is o osite to the direction
The reststor and capacttor of Example 26.12 are reconnected as shown; i is negative. After a long time, shown in Fig. 26.22. The capacitor has an initial charge of 5.0 {LC 4 and 1' both approach zero.
and is discharged by closing the switch at t = 0. (a) At what time (a) Capacitor initially charged
will the charge be 0.50 ,u.C? (b) What is the current at this time? Switch open (b) Discharging the capacitor Switch
closed When the switch is
closed. the charge
on [he capacitor
and the current
both decrease
over lime. 2016 Pearson Education Inc, 26.43 '0 (P In the circuit shown Figure £26.43
in Fig. E26.43 both capacrtors are initially charged to 45.0 V. (a) How H
long after closing the switch S will 15.0 20.0
the potential across each capacitor HF - ill-F be reduced to 10.0 V, and (b) what
will be the current at that time? 30.0 .0. 50.0 .0. Imalilm. MAGNETIC FORCE ON A PROTON A beam of protons (q : 1.6 X 10—19 C) moves at 3.0 X [05 m/s through a uniform 2.0-T magnetic field directed along the positive
z-axis (Fig. 27.10). The velocity of each proton lies in the .tz-plane
and is directed at 30° to the +z~axis. Find the force on a proton. 27.10 Directions of ii and 0 for a proton in a magnetic ﬁeld.
.Y Imalllm MAGNETIC FLUX CALCULATIONS Figure 27.162 is a perspective view of a ﬂat surface with area
3.0 cm2 in a uniform magnetic field E. The magnetic ﬂux through
this surface is +0.90 me. Find the magnitude of the magnetic
field and the direction of the area vector 3. 27.16 (a) Aﬁflat area A in a uniform magnetic field 3. (hi The
area vector A makes a 60° angle with B. (If we had chosen A to
point in the opposite direction, 4'; would have been 120" and the
magnetic flux (1)3 would have been negative.) (a) Perspective view (b) Our sketch of the problem
(edge-on View) (Hz: Dunn-"n Emu—nun" I...- I Imﬁlllm ELECTRON MOTION IN A MAGNETRON A magnetron in a microwave oven emits electromagnetic waves
with frequency f = 2450 MHz. What magnetic ﬁeld strength is
required for electrons to move in circular paths with this frequency? 27.24 "I A beam of protons travel- Figure £27.24
ing at 1.20 km/s enters a uniform mag-
netic field, traveling perpendicular to
the field. The beam exits the magnetic
field, leaving the field in a direction
perpendicular to its original direction
(Fig. E2124). The beam travels a dis-
tance of 1.18 cm while in thefiefd. What
is the magnitude of the magnetic ﬁeld?" Immum. HELICAL PARTICLE MOTION IN A MAGNETIC FIELD In a situation like that shown in Fig. 27.18, the charged particle
is a proton (q = 1.60 X 10—'9 C. m = 1.6? X 10—27 kg) and the
uniform, 0.500-T magnetic ﬁeld is directed along the x-axis. At
t = 0 the proton has velocity components ox = 1.50 X 105 m/s,
0). = 0. and v: = 2.00 X 105 m/s. Only the magnetic force acts
on the proton. (a) At t = 0, find the force on the proton and its
acceleration. (b) Find the radius of the resulting helical path, the
angular speed of the proton. and the pitch of the helix (the distance
traveled along the helix axis per revolution). 16 Ppamnn Fdllralinn lnr‘ 27.18 The general case of a charged
particle moving in a uniform magnetic
field B. The magnetic ﬁeld does no work
on the particle, so its speed and kinetic
energy remain constant. This particle’s motion has components both parallel {vi} and perpendicular (vi 1 to the
magnetic ﬁeld. so it moves in a helical path. Immllm AN elm DEMONSTRATION EXPERIMENT You set out to reproduce Thomson‘s 2/»: experiment with an
accelerating potential of 150 V and a deflecting electric field of
magnitude 6.0 X 10‘5 N/C. (a) How fast do the electrons move?
(b) What magnetic-ﬁeld magnitude will yield zero beam deflec-
tion? (c) With this magnetic field. how will the electron beam
behave if you increase the accelerating potential above 150 V? Imalllm MAGNETIC FORCE ON A STRAIGHT CONDUCTOR A straight horizontal copper rod carries a current of 50.0 A from
west to east in a region between the poles of a large electromag-
net. In this region there is a horizontal magnetic ﬁeld toward the
northeast (that is, 45” north of east) with magnitude 1.20 T.
(21) Find the magnitude and direction of the force on a LOO-m sec-
tion of rod. (b) While keeping the rod horizontal. how should it be
oriented to maximize the magnitude of the force? What is the force magnitude in this case? I 27.29 Our sketch of the copper rod as seen from overhead. B =1.20T N _XAMPLE 27.8 MAGNETIC FORCE ON A CURVED CONDUCTOR [n Fig. 27.30 the magnetic ﬁeld E is uniform and perpendicular
to the plane of the figure, pointing out of the page. The conductor,
carrying current I to the left, has three segments: (1) a straight seg-
ment with length L perpendicular to the plane of the figure, (2) a
semicircle with radius R, and (3) another straight segment with length L parallel to the x-axis. Find the total magnetic force on
this conductor. 27.30 What is the total magnetic force on the conductor? )n1 R D nnnnn Erin Iraiinn Inv- 27.35 H A long wire carry-
ing 4.50 A of current makes
two 90° bends, as shown in
Fig. E2135. The bent part
of the wire passes through a
uniform 0.240-T magnetic
field directed as shown in the figure and confined to a lim- 60.0 cm ited region of space. Find the : ma ° ' ' ' B '
gmtude and directmn of I. o o o . I the force that the magnetic 2:656:51: __ field exerts on the wire. ...

View
Full Document

- Spring '07
- Fuchs
- Capacitance, Charge, Magnetism, Energy, Magnetic Field, Electric charge