HOMEWORK #10
Chapter 24
70 ••
24-44.
Find the capacitance of the parallel-plate capacitor shown in Figure
Picture the Problem We can model this parallel-plate capacitor as a combination
of two capacitors C1 and C2 in series with capacitor C3 in parallel.
Express the capacitance of two
series-connected capacitors in
parallel with a third:
Express each of the capacitances C1,
C2, and C3 in terms of the dielectric
constants, plate areas, and plate
separations:
C = C3 + Cs
(1)
where
Cs =
C1C2
C1 + C2
κ1 ∈ 0
C1 =
1
2
C2 =
( 12 A) = κ1 ∈0
d
κ 2 ∈0
1
2
κ 3 ∈0
d
A
d
A
( 12 A) = κ 3 ∈0
A
d
 κ1 ∈0 A  κ 2 ∈0 A 



d  d


Cs =
κ1 ∈ 0 A κ 2 ∈ 0 A
+
d
d
κ κ ∈ A 
= 1 2  0 
κ1 + κ 2  d 
Substitute in equation (1) to obtain:
C=
κ 3 ∈0 A
,
2d
Substitute in equation (2) to obtain:
2d
,
( 12 A) = κ 2 ∈0
d
and
C3 =
(2)
+
κ1κ 2  ∈0 A 


κ1 + κ 2  d 

2κ1κ 2 ∈0 A 

=  κ 3 +

κ1 + κ 2  2d 

76 •• Figure 24-46 shows four capacitors connected in the arrangement
known as a capacitance bridge. The capacitors are initially uncharged. What must
the relation between the four capacitances be so that the potential difference
between points c and d remains zero when a voltage V is applied between points a
and b?
Picture the Problem Note that with V applied between a and b, C1 and C3 are in
series, and so are C2 and C4. Because in a series combination the potential
differences across the two capacitors are inversely proportional to the
capacitances, we can establish proportions involving the capacitances and
potential differences for the left- and right-hand side of the network and then use
the condition that Vc = Vd to eliminate the potential differences and establish the
relationship between the capacitances.
Q
Q
and V3 =
C1
C3
Letting Q represent the charge on
capacitors 1 and 2, relate the
potential differences across the
capacitors to their common charge
and capacitances:
V1 =
Divide the first of these equations
by the second to obtain:
V1 C3
=
V3 C1
(1)
Proceed similarly to obtain:
V2 C4
=
V4 C2
(2)
Divide equation (1) by equation
(2) to obtain:
V1V4 C3C2
=
V3V2 C1C4
(3)
If Vc = Vd then we must have:
V1 = V2 and V3 = V4
Substitute in equation (3) and
rearrange to obtain:
C2C3 = C1C4
92 ••• An air-filled parallel-plate capacitor that has gap-width d has plates
which each have an area A. The capacitor is charged to a potential difference V
and is then removed from the voltage source. A dielectric slab that has a dielectric
constant of 2.00, a thickness d, and an area 12 A is then inserted, as shown in
Figure24-52. Let σ1 be the free charge density at the conductor–dielectric surface,
and let σ2 be the free charge density at the conductor–air surface. (a) Explain why
the electric field must have the same value inside the dielectric as in the free
space between the plates. (b) Show that σ1 = 2σ2. (c) Show that the final
capacitance (after the slab is inserted) is 1.50 times the capacitance when the
capacitor is filled with air. (d) Show that the final potential difference is 23 V .
(e) Show that energy stored after the slab is inserted is only two-thirds of the
energy stored before insertion.
Picture the Problem (b) We can express the electric fields in the dielectric and in
the free space in terms of the charge densities and then use the fact that the
electric field has the same value inside the dielectric as in the free space between
the plates to establish that σ1 = 2σ2. In Parts (c) and (d) we can model the system
as two capacitors in parallel to show that the equivalent capacitance is 3∈0A/(2d)
and then use the definition of capacitance to show that the new potential
difference is 23 V .
(a) The potential difference between the plates is the same for both halves (the
plates are equipotential surfaces). Therefore, E = V/d must be the same
everywhere between the plates.
σ
⇒ σ = κ ∈0 E
κ ∈0
(b) Relate the electric field in each
region to σ and κ :
E=
Express σ1 and σ2:
σ 1 = κ1 ∈0 E1 = 2 ∈0 E1
and
σ 2 = κ 2 ∈0 E2 =∈0 E1
Divide the 1st of these equations
by the 2nd and simplify to obtain:
(c) Model the partially dielectricfilled capacitor as two capacitors
in parallel to obtain:
σ 1 = 2σ 2
Ceq = C1 + C2
where
C1 =
κ ∈0
∈0
( 12 A) = ∈0
A
2d
d
Substitute for C1 and C2 and
simplify to obtain:
Ceq =
(d) Use the definition of capacitance
to relate Vf , Qf , and Cf :
Vf =
A
2d
d
and
C2 =
( 12 A) = κ ∈0
κ ∈0 A ∈0 A
+
=
2∈0 A ∈0 A
+
2d
2d
2d
2d
3∈ 0 A
=
= 1.50C air-filled
2d
Qf
Cf
Because the capacitors are in
parallel:
Qf = Qi = VCi =
Substitute for Qf and Cf and
simplify to obtain:
Vf =
(e) The energy stored after the slab
is inserted is given by:
U f = 12 C f Vf2
Substituting for Cf and Vf and
simplifying yields:
Uf =
V ∈0 A
d
V ∈0 A
V ∈0 A
=
=
Cf d
 3∈ 0 A 

d
 2d 
(
1 3
2 2
Ci )( 23 V ) = 13 CiV 2 =
2
2
3
V
2
3
Ui
The presence of the dielectric slab reduces the potential difference between the
capacitor plates and, hence, the energy stored in the capacitor.
CHAPTER 25
86 ••
In the circuit shown in Figure 25-63, the batteries have negligible
internal resistance. Find (a) the current in each branch of the circuit, (b) the
potential difference between point a and point b, and (c) the power supplied by
each battery.
Picture the Problem Let I2Ω be the current delivered by the 7.00-V battery, I3Ω
the current delivered by the 5-V battery, and I1Ω, directed up, the current through
the 1.00-Ω resistor. We can apply Kirchhoff’s rules to obtain three equations that
we can solve simultaneously for I1, I2, and I3. Knowing the currents in each
branch, we can use Ohm’s law to find the potential difference between points a
and b and the power delivered by both the sources.
(a) Apply Kirchhoff’s junction rule
at junction a:
Apply Kirchhoff’s loop rule to a loop
around the outside of the circuit to
obtain:
Apply Kirchhoff’s loop rule to a loop
around the left-hand branch of the
circuit to obtain:
Solve equations (1), (2), and (3)
simultaneously to obtain:
(b) Apply Ohm’s law to find the
potential difference between points a
and b:
(c) Express the power delivered by
the 7.00-V battery:
(c) Express the power delivered by
the 5.00-V battery:
I 2 Ω = I 3 Ω + I1Ω
7.00 V − (2.00 Ω )I 2 Ω
− (1.00 Ω )I1 Ω = 0
(1)
(2)
7.00 V − (2.00 Ω )I 2 Ω − (3.00 Ω )I 3 Ω
+ 5.00 V = 0
or
(2.00 Ω )I 2 Ω + (3.00 Ω )I 3 Ω = 12.0 V (3)
I 2 Ω = 3.00 A , I 3 Ω = 2.00 A ,
and
I 1 Ω = 1.00 A
Vab = −5.00 V + (3.00 Ω )I 3 Ω
= −5.00 V + (3.00 Ω )(2.00 A )
= 1.00 V
P7 V = ε I 2 Ω = (7.00 V )(3.00 A )
= 21.0 W
P5 V = ε I 3 Ω = (5.00 V )(2.00 A )
= 1 0 .0 W
116 ••• In the circuit shown in Figure 25-78, R1 = 2.00 MΩ, R2 = 5.00 MΩ,
and C = 1.00 µF. The capacitor is initially without charge on either plate. At
t = 0, switch S is closed, and at t = 2.00 s switch S is opened. (a) Sketch a graph
of the voltage across C and the current in R2 between t = 0 and t = 10.0 s. (b) Find
the voltage across the capacitor at t = 2.00 s and at t = 8.00 s.
Picture the Problem We can apply both the loop and junction rules to obtain
equations that we can use to obtain a linear differential equation with constant
coefficients describing the current in R2 as a function of time. We can solve this
differential equation by assuming a solution of an appropriate form,
differentiating this assumed solution and substituting it and its derivative in the
differential equation. Equating coefficients, requiring the solution to hold for all
values of the assumed constants, and invoking an initial condition will allow us to
find the constants in the assumed solution. Once we know how the current varies
with time in R2, we can express the potential difference across it (as well as across
C because they are in parallel). To find the voltage across the capacitor at
t = 8.00 s, we can express the dependence of the voltage on time for a discharging
capacitor (C is discharging after t = 2.00 s) and evaluate this function, with a time
constant differing from that found in (a), at t = 6.00 s. The diagram shows the
circuit shortly after the switch is closed. The directions of the currents in the
resistors and the capacitor have been chosen as shown.
S
I1
ε = 10 V
I3
+
−
Q
1
2
C
R1
a
I2
R2
(a) Apply the junction rule at
junction a to obtain:
I1 = I 2 + I 3
(1)
Apply the loop rule to loop 1 to
obtain:
ε − Q − R1I1 = 0
(2)
Apply the loop rule to loop 2 to
obtain:
Q
− R2 I 2 = 0
C
(3)
C
Differentiate equation (2) with
respect to time to obtain:
dI 1 1 dQ
d 
Q
ε
R
I
R
0
=
−
−
−
−
1
1
1
dt 
C 
dt C dt
dI
1
= − R1 1 − I 3 = 0
dt C
or
dI
1
R1 1 = − I 3
(4)
dt
C
Differentiate equation (3) with
respect to time to obtain:
d Q
dI
 1 dQ
− R2 I 2  =
− R2 2 = 0

dt  C
dt
 C dt
or
dI
1
(5)
R2 2 = I 3
dt C
Using equation (1), substitute for
I3 in equation (5) to obtain:
1
dI 2
( I1 − I 2 )
=
dt
R2C
Solve equation (2) for I1:
I1 =
Substitute for I1 in equation (6) and
simplify to obtain the differential
equation for I2:
ε −Q
R1
=
ε − R2 I 2
R1

dI 2
1  ε − R2 I 2

=
− I 2 
dt
R2C  R1

=
To solve this linear differential
equation with constant coefficients
we can assume a solution of the
form:
C
(6)
ε
 R + R2 
I2
−  1
R1R2C  R1R2C 
I 2 (t ) = a + be −t τ
[
(7)
]
Differentiate I2(t) with respect to
time to obtain:
b
dI 2 d
=
a + be −t τ = − e −t τ
dt
dt
τ
Substitute for I2 and dI2/dt to obtain:
b
− e −t τ =
τ
Equate coefficients of e −t τ to obtain:
τ=
ε
 R + R2 
 a + be −t τ
−  1
R1R2C  R1R2C 
R1R2C
R1 + R2
(
)
ε
Requiring the equation to hold for all
values of a yields:
a=
If I2 is to be zero when t = 0:
0 = a + b ⇒ b = −a = −
Substitute in equation (7) to obtain:
R1 + R2
I 2 (t ) =
ε
R1 + R2
ε
=
−
ε
R1 + R2
ε
R1 + R2
e −t τ
(1 − e )
−t τ
R1 + R2
RRC
where τ = 1 2
R1 + R2
Substitute numerical values and
evaluate τ :
τ=
(2.00 MΩ )(5.00 MΩ )(1.00 µF)
2.00 MΩ + 5.00 MΩ
= 1.43 s
Substitute numerical values and
evaluate I2(t):-
I 2 (t ) =
(
10.0 V
1 − e −t 1.429 s
2.00 MΩ + 5.00 MΩ
(
= (1.429 µA ) 1 − e −t 1.43 s
)
)
Because C and R2 are in parallel, they have a common potential difference given
by:
(
)
(
VC (t ) = V2 (t ) = I 2 (t )R2 = (1.429 µA )(5.00 MΩ ) 1 − e − t 1.429 s = (7.143 V ) 1 − e − t 1.429 s
Evaluate VC at t = 2.00 s:
(
VC (2 s ) = (7.143 V ) 1 − e − 2.00 s 1.429 s
)
)
= 5.381 V
The voltage across the capacitor as a function of time is shown in the following
graph. The current through the 5.00-MΩ resistor R2 follows the same time course,
its value being VC/(5.00 ×106) A.
6
5
4
V C (V) 3
2
1
0
0
2
4
6
8
10
t (s)
(b) The value of VC at t = 2.00 s has
already been determined to be:
When S is opened at t = 2.00 s, C
discharges through R2 with a time
constant given by:
Express the potential difference
across C as a function of time:
Evaluate VC at t = 8.00 s to obtain:
VC (2.00 s ) = 5.381 V = 5.38 V
τ ' = R2C = (5.00 MΩ )(1.00 µF) = 5.00 s
VC (t ) = VC0 e −(t − 2.00 s ) τ '
= (5.381 V )e −(t −2.00 s ) 5.00 s
VC (8.00 s ) = (5.381V )e −6.00 s 5.00 s
= 1.62 V
in good agreement with the graph.
CHAPTER 26
30 ••
A particle has a charge q, a mass m, a linear momentum of magnitude
p and a kinetic energy K. The particle moves in a circular orbit of radius R
r
perpendicular to a uniform magnetic field B . Show that (a) p = BqR and
(b) K = 12 B2 q2 R 2 / m .
Picture the Problem We can use the definition of momentum to express p in
terms of v and apply Newton’s 2nd law to the orbiting particle to express v in
terms of q, B, R, and m. In Part (b) we can express the particle’s kinetic energy in
terms of its momentum and use our result from Part (a) to show that
K = 12 B 2 q 2 R 2 m.
(a) Express the momentum of the
particle:
Apply
∑F
radial
= mac to the orbiting
p = mv
qvB = m
particle to obtain:
(1)
v2
qBR
⇒v =
R
m
Substitute for v in equation (1) to
obtain:
 qBR 
p = m
 = qBR
 m 
(b) Express the kinetic energy of the
orbiting particle as a function of its
momentum:
K=
Substitute our result for p from Part
(a) to obtain:
K=
p2
2m
(qBR )2
2m
=
q2B2R2
2m
52 ••
A rectangular current-carrying 50-turn coil, as shown in Figure 26-36,
is pivoted about the z axis. (a) If the wires in the z = 0 plane make an angle
θ = 37º with the y axis, what angle does the magnetic moment of the coil make
with the unit vector ˆi ? (b) Write an expression for nˆ in terms of the unit vectors
ˆi and jˆ , where nˆ is a unit vector in the direction of the magnetic moment.
(c) What is the magnetic moment of the coil? (d) Find the torque on the coil when
r
there is a uniform magnetic field B = 1.5 T jˆ in the region occupied by the coil.
(e) Find the potential energy of the coil in this field. (The potential energy is zero
when θ = 0.)
Picture the Problem The diagram shows the coil as it would appear from along
the positive z axis. The right-hand rule for determining the direction of n̂ has
been used to establish n̂ as shown. We can use the geometry of this figure to
determine θ and to express the unit normal vector n̂ . The magnetic moment of the
r
r r r
coil is given by µ = NIAnˆ and the torque exerted on the coil by τ = µ × B . Finally,
r r
we can find the potential energy of the coil in this field from U = − µ ⋅ B .
y
I
n̂
37°
θ
x
(a) Noting that θ and the angle
whose measure is 37° have their
right and left sides mutually
perpendicular, we can conclude
that:
θ = 37°
(b) Use the components of n̂ to
express n̂ in terms of iˆ and ĵ :
nˆ = n x iˆ + n y ˆj = cos 37°iˆ − sin 37° ˆj
= 0.799iˆ − 0.602 ˆj
= 0.80iˆ − 0.60 ˆj
r
µ = NIAnˆ
(c) Express the magnetic moment
of the coil:
r
Substitute numerical values and evaluate µ :
(
)(
(0.34 A ⋅ m ) iˆ − (0.25 A ⋅ m ) ˆj
) (
) (
)
r
µ = (50 )(1.75 A ) 48.0 cm 2 0.799iˆ − 0.602 ˆj = 0.335 A ⋅ m 2 iˆ − 0.253 A ⋅ m 2 ˆj
=
2
2
(d) Express the torque exerted on
the coil:
r r r
τ = µ× B
r
r
Substitute for µ and B to obtain:
{(
) (
)}
r
τ = 0.335 A ⋅ m 2 iˆ − 0.253 A ⋅ m 2 ˆj × (1.5 T ) ˆj
( )
( )
= (0.503 N ⋅ m ) iˆ × ˆj − (0.379 N ⋅ m ) ˆj × ˆj =
(e) Express the potential energy of
r r
U = −µ ⋅ B
(0.50 N ⋅ m ) kˆ
the coil in terms of its magnetic
moment and the magnetic field:
r
r
Substitute for µ and B and evaluate U:
{(
) (
)}
U = − 0.335 A ⋅ m 2 iˆ − 0.253 A ⋅ m 2 ˆj ⋅ (1.5 T ) ˆj
= −(0.503 N ⋅ m ) iˆ ⋅ ˆj + (0.379 N ⋅ m ) ˆj ⋅ ˆj = 0.38 J
( )
( )
CHAPTER 28
43 ••
In Figure 28-48, a conducting rod that has a mass m and a negligible
resistance is free to slide without friction along two parallel frictionless rails that
have negligible resistances separated by a distance l and connected by a
resistance R. The rails are attached to a long inclined plane that makes an angle θ
with the horizontal. There is a magnetic field directed upward as shown. (a) Show
that there is a retarding force directed up the incline given by
2 2
2
F = (B l v cos θ )/ R . (b) Show that the terminal speed of the rod is
vt = (mgR sin θ ) / (B l cos θ ).
2 2
2
Picture the Problem The free-body diagram shows the forces acting on the rod
as it slides down the inclined plane. The retarding force is the component of Fm
acting up the incline, i.e., in the −x direction. We can express Fm using the
expression for the force acting on a conductor moving in a magnetic field.
r
Recognizing that only the horizontal component of the rod’s velocity v produces
an induced emf, we can apply the expression for a motional emf in conjunction
with Ohm’s law to find the induced current in the rod. In Part (b) we can apply
Newton’s 2nd law to obtain an expression for dv/dt and set this expression equal to
zero to obtain vt.
y
r
Fn
θ
r
Fm
θ
x
r
mg
(a) Express the retarding force
acting on the rod:
Express the induced emf due to the
motion of the rod in the magnetic
field:
Using Ohm’s law, relate the current
I in the circuit to the induced emf:
F = Fm cosθ
(1)
where
Fm = IlB
and I is the current induced in the rod
as a consequence of its motion in the
magnetic field.
ε = Blv cosθ
I=
ε
R
=
Blv cos θ
R
Substitute in equation (1) to obtain:
 Blv cos θ
F =
R

=
(b) Apply
∑F
x
= max to the rod:

lB cos θ

B 2l 2v
cos 2 θ
R
mg sin θ −
B 2l 2v
dv
cos 2 θ = m
R
dt
and
dv
B 2l 2 v
= g sin θ −
cos 2 θ
dt
mR
When the rod reaches its terminal
speed vt , dv dt = 0 :
Solve for vt to obtain:
B 2 l 2 vt
0 = g sin θ −
cos 2 θ
mR
vt =
mgR sin θ
B 2 l 2 cos 2 θ
Prior to 1960, magnetic field strengths were usually measured by a
72 ••
rotating coil gaussmeter. This device used a small multi-turn coil rotating at a
high speed on an axis perpendicular to the magnetic field. This coil was connected
to an ac voltmeter by means of slip rings, like those shown in Figure 28-61. In one
specific design, the rotating coil has 400 turns and an area of 1.40 cm2. The coil
rotates at 180 rev/min. If the magnetic field strength is 0.450 T, find the
maximum induced emf in the coil and the orientation of the normal to the plane of
the coil relative to the field for which this maximum induced emf occurs.
Picture the Problem We can apply Faraday’s law and the definition of magnetic
flux to derive an expression for the induced emf in the rotating coil gaussmeter.
Use Faraday’s law to express the
ε = − dφ m
induced emf:
dt
Using the definition of magnetic
flux, relate the magnetic flux through
the loop to its angular velocity:
Substitute for φm (t ) and simplify
to obtain:
φm (t ) = NBA cos ωt
ε = − d [NBA cos ωt ]
dt
= − NBAω (− sin ωt )
= NBAω sin ωt = ε max sin ωt
where
ε max = NBAω
Substitute numerical values and evaluate εmax:
ε max = (400)(0.450 T )(1.40 ×10 −4 m 2 )180 rev × 2π rad × 1min  =

min
rev
60 s 
0.475 V
The maximum induced emf occurs at the instant the normal to the plane of the
r
coil is perpendicular to the magnetic field B . At this instant, φm is zero, but ε is
a maximum.