23.32 See Exercise 32 Section 23.6 Energy Stored in a charged Capacitor 33 How much energy is stored in one cubic meter of air due to an electric field of magnitude 100 V/m?. b Does the
Trang 123.6 Exercises 803
Fig 23.24 See Exercise (20)
V
Δ ° B
Q
Q
V
Δ °
B
(22) The two capacitors of exercise 21 are now connected in series to the same battery (i.e with a potential differenceV = 9 V) (a) Find the equivalent capacitance
of the combination (b) Find the charge on each capacitor (c) Find the potential difference across each capacitor
(23) For the combination of capacitors shown in Fig.23.25, assume that C1= 1 µF,
C2= 2 µF, C3= 3 µF, and V = 6 V (a) Find the equivalent capacitance of
the combination (b) Find the charge on each capacitor (c) Find the potential difference across each capacitor
Fig 23.25 See Exercise (23)
3
C
1
C
2
C ΔV
(24) Three capacitors, C1= 6 µF, C2= 4 µF, and C3= 12 µF, are connected in four
different ways, as shown in Fig.23.26 In all configurations, the potential dif-ference is 22 V How many coulombs of charge pass from the battery to each
combination?
(25) When the three capacitors C1= 2 µF, C2= 1 µF, and C3= 4 µF are connected
to a source of a potential differenceV, as shown Fig.23.27, the charge Q2on
C2is found to be 10µC (a) Find the values of the charges on the two capacitors
C1and C3 (b) Determine the value of V.
Trang 2804 23 Capacitors and Capacitance
1
C
2
1
2
C
3
C
3
C
1
C
2
C
Δ V
Fig 23.26 See Exercise (24)
Fig 23.27 See Exercise (25)
1
C
2
V
Δ
(26) For the circuit shown in Fig.23.28, C1= 3 µF, C2= 6 µF, C3= 6 µF, C4=
12µF, and V = 12 V (a) Find the equivalent capacitance of the combination.
(b) Find the potential difference across each capacitor
Fig 23.28 See Exercise (26)
2
C
1
4
C V
Δ
(27) For each of the combinations shown in Fig.23.29, find a formula that represents the equivalent capacitance between the terminals A and B
(28) Assume that in Exercise 27, C = 12 µF and VBA= 12 V For each
combina-tion, find the magnitude of the total charge that the source between A and B will distribute on the capacitors
Trang 323.6 Exercises 805
C
C
C
C C
A
A
A
B
C
C
C
C
C
C
Fig 23.29 See Exercise (27)
(29) Two capacitors, C1= 25 µF and C2= 40 µF, are charged by being connected to
batteries that have a potential differenceV = 50 V, see part (a) of Fig.23.30 They are then disconnected from their batteries and connected to each other, with each positive plate connected to the other’s negative plate; see part (b) of Fig.23.30 (a) Find the equivalent capacitance between A and B (b) What is
the charge Q on the equivalent capacitor? (c) What is the potential difference
VBAbetween A and B? (d) Find the final charge on each capacitor
Δ V
Q2 C2
Δ V
C2
Ceq
Q
Q1 f
Q2 f
Fig 23.30 See Exercise (29)
(30) A parallel-plate capacitor has an area A and separation d A slab of copper of thickness a is inserted midway between the plates, see part (a) of Fig.23.31 Show that the capacitor is equivalent to two capacitors in series, each having a plate separation(d − a)/2, as shown in part (b) of the figure, and show that the
capacitance after inserting the slab is given by:
C= ◦A
d − a
Trang 4806 23 Capacitors and Capacitance
Fig 23.31 See Exercise (30)
(d-a) / 2
(d-a) / 2
(31) Show that the capacitance of the capacitor in Fig.23.10b can be obtained by finding the equivalent capacitance of two capacitors in series, one capacitor with
a dielectric of thickness a and the second an air-filled capacitor of thickness
d − a.
(32) A parallel-plate capacitor of plate area A and separation d is filled in two different
ways with two dielectricsκ1andκ2as shown in parts (a) and (b) of Fig.23.32 Show that the capacitances of the two capacitors of parts (a) and (b) are:
C=◦A
d
κ1 + κ2
2 and C= 2◦A
d
κ1 κ2 κ1 + κ2
respectively,
A /2 A /2
2
κ
d/2
d/2
d
Fig 23.32 See Exercise (32)
Section 23.6 Energy Stored in a charged Capacitor
(33) How much energy is stored in one cubic meter of air due to an electric field of magnitude 100 V/m?
(34) The two capacitors shown in Fig.23.33are uncharged when the switch S is
open Assume that C1= 4 µF, C2= 6 µF, and V = 10 V The two capacitors
become fully charged when the switch S is closed (a) Find the energy stored
Trang 523.6 Exercises 807
in these two capacitors (b) Does the stored potential energy in the equivalent capacitor equal the total stored energy in the two capacitors?
Fig 23.33 See Exercise (34)
1
S
V
Δ
(35) Redo Example 23.9 when C1= C2= 5 µF, and Vi= 10 V Does the initial
and final stored potential energy remains the same?
(36) A capacitor is charged to a potential difference V How much should you
increaseV so that the stored potential energy is increased by 20%?
(37) Calculate the electric field, the energy density, and the stored potential energy
in the parallel-plate capacitor of Exercise 7
(38) A parallel-plate capacitor has a capacitance of 4µF when a mica sheet with dielectric constantκ = 5 fills the space between the plates The capacitor is
charged by a battery that has a potential difference 50 V, and is later
discon-nected How much work must be done to slowly pull the dielectric from the capacitor?
(39) For the circuit shown in Fig.23.34, C1= 2 µF, C2= 3 µF, C3= 6 µF, C4=
1µF, and C5= 2 µF (a) Find the potential difference between A and B needed
to give C3a charge of 20µC (b) Under these considerations, what is the electric
potential energy stored in the combination?
C1
B
C2
C3
C4
C5
Trang 6808 23 Capacitors and Capacitance (40) Confirm the relationships shown in Fig.23.35, whereV◦is shortened by V◦
andV is shortened by V.
Q
+ °
⇒
⇒
Q
− °
Q
+ ° −Q°
Dielectric
Voltmeter Voltmeter
C°
V°
E°
u°
Fixed charge
Before inserting the dielectric
Fixed charge After inserting the dielectric Relationships
C=κC°
(Fixed)
Q° V
V=κ°
E E
κ
= °
/
u u= °κ
Fixed voltage
Before inserting the dielectric
Q
°(Fixed)
Fixed voltage After inserting the dielectric Relationships
B
Q
+ ° C° −Q°
E°
C
Q
B
E
(Fixed)
C°
Q
°
V°
E°
u°
(Fixed)
C Q
V° E u
C κC°
Q κQ°
(Fixed)
V°
E=E°
u κu°
κ
κ
Dielectric
C
V E u
(Fixed)
Q°
Fig 23.35 See Exercise (40)
Trang 7Electric Circuits 24
In this chapter we analyze simple electric circuits that contain devices such as bat-teries, resistors, and capacitors in various combinations We begin by introducing
steady-state electric circuits and the concept of a constant rate of flow of electric
charges, known as direct current (dc) We also introduce Kirchhoff’s two rules,
which are used to simplify and analyze more complicated circuits Finally, we
con-sider circuits containing resistors and capacitors, in which currents can vary with
time
Electric Current
When there is a net flow of charge across any area, we say there is an electric
current (or simply current) across that area To maintain a continuous current, we
must maintain a net force on the mobile charge in some way The net force may result, for example, from an electrostatic field We assume that an electric field →
E is
maintained within a conductor such that the charged particle q is acted on by a force
→
F = q E→ We refer to this force as the particle’s driving force.
To define the current, we consider positive charges moving perpendicularly onto
a surface area A as shown in Fig.24.1
Spotlight
The current I across an area A is defined as the net charge flowing
perpendic-ularly to that area per unit time
H A Radi and J O Rasmussen, Principles of Physics, 809 Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_24,
© Springer-Verlag Berlin Heidelberg 2013
Trang 8810 24 Electric Circuits Thus, if a net chargeQ flows across an area A in a time t, the average current
Iavacross the area is:
Iav= Q
A I +
+ +
+
Fig 24.1 Charged particles in motion perpendicular onto an area A The current I represents the time
rate of flow of charges and has by convention the direction of the motion of positive charges
When the rate of flow varies with time, we define the instantaneous current (or the
current) I as:
I= dQ
The SI unit of the current is ampere (abbreviated by A) That is:
1 A = 1 C
Thus, 1 A is equivalent to 1 C of charge passing through the surface area in 1 s
Small currents are more conveniently expressed in milliamperes (1 mA= 10−3A)
or microamperes (1µA = 10−6A).
Currents can be due to positive charges, or negative charges, or both In conductors, the current is due to the motion of only negatively charged free electrons (called
conduction electrons) By convention, the direction of the current is the direction
of the flow of positive charges Therefore, the direction of the current is opposite to
the direction of the flow of electrons, see Fig.24.2b A moving charge, positive or
negative, is usually referred to as a mobile charge carrier.
+
+
-+ +
-I2
I1
I = +
Fig 24.2 Direction of current due to (a) positive charges, (b) negative charges, and (c) both positive and negative charges
Trang 924.1 Electric Current and Electric Current Density 811
Electric Current Density
The current across an area can be expressed in terms of the motion of the charge carri-ers To achieve this we consider a portion of a cylindrical rod that has a cross-sectional
area A, length x, and carries a constant current I, see Fig.24.3 For convenience we
consider positive charge carriers each having a charge q , and the number of carriers per unit volume in the rod is n Therefore, in this portion, the number of carriers is
n A x and the total charge Q is:
Fig 24.3 A portion of a
straight rod of uniform
cross-sectional area A,
carrying a constant current I.
The mobile charge carriers are
assumed to be positive and
x
I A
d
+
d
+
d
+
Suppose that all the carriers move with an average speedvd(called the drift speed).
Therefore, during a time interval t, all carriers must achieve a displacement
x = vd t in the x direction Now, let us choose t such that the carriers in the
cylindrical portion move through a displacement whose magnitude is equal to the length of the cylinder, see Fig.24.3 During such a time interval, all the charge
carri-ers in this cylindrical portion must pass through the circular area A at the right end.
Accordingly, we write the last relation as:
Therefore, the current I = Q/t in the rod will be given by:
The charge carriers in a solid conductor are all free electrons If the conductor
is isolated, these electrons move with speeds of the order of 106m/s, and because
of their collisions with the scatterers (atoms or molecules in the conductor), they
move randomly in all directions This results in a zero drift velocity and hence no net
Trang 10812 24 Electric Circuits
charge transport, which means zero current When an electric field→
Eis established across the conductor, this field exerts an electric force→
F = −e E→on each electron, producing a current Of course, the electrons do not move in a straight line along the conductor, but their resultant motion is complicated and zigzagged, see Fig.24.4 Regardless of the collisions of these electrons, they move slowly along the conductor
in a direction opposite to→
Ewith a drift velocity→vd, see Fig.24.4
Fig 24.4 A schematic
representation of the random
zigzag motion and the drift of
a free electron with an average
speedvd in a conductor, due to
the effect of an external
electric field →
E
d
I
E
Scatterer
Conductor
J
-The current density J is defined as the current per unit area, i.e.:
J= I
Using the relation I = n q vdA, we get:
where the SI unit of the current density is A/m2 Equation24.8is valid only if J is uniform and the direction of I is perpendicular to the cross-sectional area A Generally,
the current density is a vector quantity that has the direction of q→vd, for both signs
of q; that is:
→
The amount of current that passes through an element of area dA, can be written
as →
J • d→
A , where d→
A is the vector area of the element The current that passes
throughout the entire area A is thus:
I = →J • d→
If the current density is uniform across the area and parallel to d→
A, then this equation leads to Eq.24.7
Trang 1124.1 Electric Current and Electric Current Density 813
Example 24.1
Estimate the drift speed of the conduction electrons in a copper wire that is 2 mm
in diameter and carries a current of 1 A Comment on your result The density of copper is 8.92 × 103kg/m3.
[Hint: Assume that each copper atom contributes one free conduction electron to the current.]
Solution: To get the drift speedvd , we need to find the free-electron density n.
To get n, we need to know the volume occupied by one kmol of copper From
the periodic table of elements, see Appendix C, the molar mass of copper is
M (Cu) = 63.546 kg/kmol Recall that the mass of one kmol of63.5Cu contains
Avogadro’s number of atoms (NA= 6.022 × 1026atoms/kmol) Thus:
Volume of 1 kmol=Mass of 1 kmol
ρ
Number of copper atoms/m3= Avogadro’s number
Volume of 1 kmol ⇒ n = NA
V = NAρ M
Therefore:
n = NAρ
M = (6.022 × 1026atoms/kmol)(8.92 × 103kg/m3)
63.546 kg/kmol
= 8.45 × 1028atoms/m3
Since the density of free-electrons is equal to the density of copper atoms, then
we use Eq.24.6to find the drift speed as follows:
vd= I
(8.45 × 1028electrons/m3)(1.6 × 10−19C)(π × (10−3m)2)
= 2.35 × 10−5m/s= 8.46 cm/h (Very small speed)
You might ask why, even thoughvdis so small, that regular light bulbs light up very quickly when one turns on its circuit switch? The answer is that the electric field travels along the connecting wires of the circuit at almost the speed of light,
so electrons everywhere in the wires all begin to drift at once with a small drift
speed
Trang 12814 24 Electric Circuits
Example 24.2
One end of the copper wire in example 1 is welded to one end of an aluminum wire with a 4 mm diameter The composite wire carries a steady current equal to
that of Example 24.1 (i.e I= 1 A) (a) What is the current density in each wire? (b) What is the value of the drift speedvdin the aluminum? [Aluminum has one free electron per atom and density 2.7 × 103kg/m3]
Solution: (a) Except near the junction, the current density in a copper wire of
radius rCu= 1 mm and aluminum wire of radius rAl= 2 mm are:
JCu= I
ACu = I
πr2 Cu
=π × (101 A−3
m)2 = 3.18 × 105A/m2
JAl= I
AAl = I
πr2 Al
= π × (2 × 101 A −3
m)2 = 7.96 × 104A/m2 (b) From the periodic table of elements, see Appendix C, the molar mass of
aluminum is M (Al) = 26.98 kg/kmol As in Example 24.1, we find:
n= NAρ
M = (6.022 × 1026atoms/kmol)(2.7 × 103kg/m3)
26.98 kg/kmol
= 6.03 × 1028
atoms/m3
vd= I
(6.03 × 1028electrons/m3)(1.6 × 10−19C)(π × (2 × 10−3m)2)
= 8.25 × 10−6m/s= 2.97 cm/h
As a result of maintaining a potential difference V across a conductor, an
elec-tric field→
Eand a current density→
J are established in the conductor For materials with electrical properties that are the same in all directions (isotropic materials), the electric field is found to be proportional to the current density That is:
→
where the constantρ1is called the resistivity of the conductor Materials that obey this relation are said to obey Ohm’s law:
1 Not to be confused withρ referring to mass density or charge density.
Trang 1324.2 Ohm’s Law and Electric Resistance 815
Spotlight
For many materials and most metals, the ratio of the magnitude of the electric field to the magnitude of the current density is a constant and does not depend
on the electric field producing the current
Since it is difficult to measure→
Eand→
Jdirectly, we need to put Ohm’s law into a more practical form This can be obtained by considering a portion of a straight
con-ductor that has a uniform cross-sectional area A and length L, as shown in Fig.24.5
In addition, a potential differenceV = V b − Vabetween the ends of the conductor
(denoted by a and b) will create a straight electric field and current, as also shown in
Fig.24.5 Since charge carriers in conductors are electrons, they will drift from face
a to face b, against the field→
E
Fig 24.5 A potential
differenceV = V b − V a
across a conductor of
cross-sectional area A and
length L sets up a field→
E and
current I
b
L
I A
a b
J
Recall that for uniform electric fields we have:
Using this relation to eliminate E from the scalar form of Eq.24.11, we get:
V
Also, using J = I/A, the potential difference V can be written as:
V =
ρ L A
The quantity in brackets is called the electrical resistance (or simply resistance) of
the conductor and is denoted by the symbol R; that is:
R = ρ L