Course Hero has millions of student submitted documents similar to the one

below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

26
Capacitance Chapter and Dielectrics
CHAPTE R OUTLI N E
26.1 Deﬁnition of Capacitance 26.2 Calculating Capacitance 26.3 Combinations of Capacitors 26.4 Energy Stored in a Charged Capacitor 26.5 Capacitors with Dielectrics 26.6 Electric Dipole in an Electric Field 26.7 An Atomic Description of Dielectrics
L All of these devices are capacitors, which store electric charge and energy. A capacitor is
one type of circuit element that we can combine with others to make electric circuits. (Paul Silverman/ Fundamental Photographs)
795 795
795
In this chapter, we will introduce the ﬁrst of three simple circuit elements that can be
–Q
+Q
connected with wires to form an electric circuit. Electric circuits are the basis for the vast majority of the devices that we use in current society. We shall discuss capacitors— devices that store electric charge. This discussion will be followed by the study of resistors in Chapter 27 and inductors in Chapter 32. In later chapters, we will study more sophisticated circuit elements such as diodes and transistors. Capacitors are commonly used in a variety of electric circuits. For instance, they are used to tune the frequency of radio receivers, as ﬁlters in power supplies, to eliminate sparking in automobile ignition systems, and as energy-storing devices in electronic ﬂash units. A capacitor consists of two conductors separated by an insulator. The capacitance of a given capacitor depends on its geometry and on the material—called a dielectric— that separates the conductors.
Figure 26.1 A capacitor consists of two conductors. When the capacitor is charged, the conductors carry charges of equal magnitude and opposite sign.
26.1
Deﬁnition of Capacitance
L
PITFALL PR EVE NTI O N
26.1 Capacitance Is a Capacity
To understand capacitance, think of similar notions that use a similar word. The capacity of a milk carton is the volume of milk that it can store. The heat capacity of an object is the amount of energy an object can store per unit of temperature difference. The capacitance of a capacitor is the amount of charge the capacitor can store per unit of potential difference.
Deﬁnition of capacitance
Consider two conductors carrying charges of equal magnitude and opposite sign, as shown in Figure 26.1. Such a combination of two conductors is called a capacitor. The conductors are called plates. A potential difference V exists between the conductors due to the presence of the charges. What determines how much charge is on the plates of a capacitor for a given voltage? Experiments show that the quantity of charge Q on a capacitor1 is linearly proportional to the potential difference between the conductors; that is, Q V. The proportionality constant depends on the shape and separation of the conductors.2 We can write this relationship as Q C V if we deﬁne capacitance as follows:
The capacitance C of a capacitor is deﬁned as the ratio of the magnitude of the charge on either conductor to the magnitude of the potential difference between the conductors: C Q V (26.1)
1
Although the total charge on the capacitor is zero (because there is as much excess positive charge on one conductor as there is excess negative charge on the other), it is common practice to refer to the magnitude of the charge on either conductor as “the charge on the capacitor.’’
2
The proportionality between V and Q can be proved from Coulomb’s law or by experiment.
796
S ECTI O N 2 6.2 • Calculating Capacitance
797
Note that by deﬁnition capacitance is always a positive quantity. Furthermore, the charge Q and the potential difference V are always expressed in Equation 26.1 as positive quantities. Because the potential difference increases linearly with the stored charge, the ratio Q / V is constant for a given capacitor. Therefore, capacitance is a measure of a capacitor’s ability to store charge. Because positive and negative charges are separated in the system of two conductors in a capacitor, there is electric potential energy stored in the system. From Equation 26.1, we see that capacitance has SI units of coulombs per volt. The SI unit of capacitance is the farad (F), which was named in honor of Michael Faraday: 1F 1 C/V
L
PITFALL PR EVE NTI O N
26.2 Potential Difference is V, not V
We use the symbol V for the potential difference across a circuit element or a device because this is consistent with our deﬁnition of potential difference and with the meaning of the delta sign. It is a common, but confusing, practice to use the symbol V without the delta sign for a potential difference. Keep this in mind if you consult other texts.
The farad is a very large unit of capacitance. In practice, typical devices have capacitances ranging from microfarads (10 6 F) to picofarads (10 12 F). We shall use the symbol F to represent microfarads. To avoid the use of Greek letters, in practice, physical capacitors often are labeled “mF’’ for microfarads and “mmF’’ for micromicrofarads or, equivalently, “pF’’ for picofarads. Let us consider a capacitor formed from a pair of parallel plates, as shown in Figure 26.2. Each plate is connected to one terminal of a batter y, which acts as a source of potential difference. If the capacitor is initially uncharged, the batter y establishes an electric ﬁeld in the connecting wires when the connections are made. Let us focus on the plate connected to the negative terminal of the batter y. The electric ﬁeld applies a force on electrons in the wire just outside this plate; this force causes the electrons to move onto the plate. This movement continues until the plate, the wire, and the terminal are all at the same electric potential. Once this equilibrium point is attained, a potential difference no longer exists between the terminal and the plate, and as a result no electric ﬁeld is present in the wire, and the movement of electrons stops. The plate now carries a negative charge. A similar process occurs at the other capacitor plate, with electrons moving from the plate to the wire, leaving the plate positively charged. In this ﬁnal conﬁguration, the potential difference across the capacitor plates is the same as that between the terminals of the batter y. Suppose that we have a capacitor rated at 4 pF. This rating means that the capacitor can store 4 pC of charge for each volt of potential difference between the two conductors. If a 9-V battery is connected across this capacitor, one of the conductors ends up with a net charge of 36 pC and the other ends up with a net charge of 36 pC.
–Q +Q
Area = A d
+
–
Quick Quiz 26.1
A capacitor stores charge Q at a potential difference V. If the voltage applied by a battery to the capacitor is doubled to 2 V, (a) the capacitance falls to half its initial value and the charge remains the same (b) the capacitance and the charge both fall to half their initial values (c) the capacitance and the charge both double (d) the capacitance remains the same and the charge doubles.
Figure 26.2 A parallel-plate capacitor consists of two parallel conducting plates, each of area A, separated by a distance d. When the capacitor is charged by connecting the plates to the terminals of a battery, the plates carry equal amounts of charge. One plate carries positive charge, and the other carries negative charge.
26.2
Calculating Capacitance
L
PITFALL PR EVE NTI O N
We can derive an expression for the capacitance of a pair of oppositely charged conductors in the following manner: assume a charge of magnitude Q , and calculate the potential difference using the techniques described in the preceding chapter. We then use the expression C Q / V to evaluate the capacitance. As we might expect, we can per form this calculation relatively easily if the geometry of the capacitor is simple.
26.3 Too Many C’s
Do not confuse italic C for capacitance with non-italic C for the unit coulomb.
798
C H A P T E R 2 6 • Capacitance and Dielectrics
While the most common situation is that of two conductors, a single conductor also has a capacitance. For example, imagine a spherical charged conductor. The electric ﬁeld lines around this conductor are exactly the same as if there were a conducting shell of inﬁnite radius, concentric with the sphere and carrying a charge of the same magnitude but opposite sign. Thus, we can identify the imaginary shell as the second conductor of a two-conductor capacitor. We now calculate the capacitance for this situation. The electric potential of the sphere of radius R is simply ke Q /R , and setting V 0 for the inﬁnitely large shell, we have
Capacitance of an isolated charged sphere
C
Q V
Q k e Q /R
R ke
4
0R
(26.2)
This expression shows that the capacitance of an isolated charged sphere is proportional to its radius and is independent of both the charge on the sphere and the potential difference. The capacitance of a pair of conductors depends on the geometry of the conductors. Let us illustrate this with three familiar geometries, namely, parallel plates, concentric cylinders, and concentric spheres. In these examples, we assume that the charged conductors are separated by a vacuum. The effect of a dielectric material placed between the conductors is treated in Section 26.5.
Parallel-Plate Capacitors
Two parallel metallic plates of equal area A are separated by a distance d, as shown in Figure 26.2. One plate carries a charge Q , and the other carries a charge Q . Let us consider how the geometr y of these conductors influences the capacity of the combination to store charge. Recall that charges of the same sign repel one another. As a capacitor is being charged by a batter y, electrons flow into the negative plate and out of the positive plate. If the capacitor plates are large, the accumulated charges are able to distribute themselves over a substantial area, and the amount of charge that can be stored on a plate for a given potential difference increases as the plate area is increased. Thus, we expect the capacitance to be proportional to the plate area A. Now let us consider the region that separates the plates. If the batter y has a constant potential difference between its terminals, then the electric ﬁeld between the plates must increase as d is decreased. Let us imagine that we move the plates closer together and consider the situation before any charges have had a chance to move in response to this change. Because no charges have moved, the electric ﬁeld between the plates has the same value but extends over a shorter distance. Thus, the magnitude of the potential difference between the plates V Ed (Eq. 25.6) is now smaller. The difference between this new capacitor voltage and the terminal voltage of the batter y now exists as a potential difference across the wires connecting the batter y to the capacitor. This potential difference results in the electric ﬁeld in the wires that drives more charge onto the plates, increasing the potential difference between the plates. When the potential difference between the plates again matches that of the batter y, the potential difference across the wires falls back to zero, and the flow of charge stops. Thus, moving the plates closer together causes the charge on the capacitor to increase. If d is increased, the charge decreases. As a result, we expect the capacitance of the pair of plates to be inversely proportional to d. We can verify these physical arguments with the following derivation. The Q /A. If the plates are very close sur face charge density on either plate is together (in comparison with their length and width), we can assume that the electric ﬁeld is uniform between the plates and is zero elsewhere. According to the What If ? feature in Example 24.8, the value of the electric ﬁeld between
S ECTI O N 2 6.2 • Calculating Capacitance
799
the plates is E Q 0A
0
Because the ﬁeld between the plates is uniform, the magnitude of the potential difference between the plates equals Ed (see Eq. 25.6); therefore, V Ed Qd 0A
Substituting this result into Equation 26.1, we ﬁnd that the capacitance is C Q V
0A
Q Qd/ 0A (26.3)
Capacitance of parallel plates
C
d
That is, the capacitance of a parallel-plate capacitor is proportional to the area of its plates and inversely proportional to the plate separation, just as we expected from our conceptual argument. A careful inspection of the electric ﬁeld lines for a parallel-plate capacitor reveals that the ﬁeld is uniform in the central region between the plates, as shown in Figure 26.3a. However, the ﬁeld is nonuniform at the edges of the plates. Figure 26.3b is a photograph of the electric ﬁeld pattern of a parallel-plate capacitor. Note the nonuniform nature of the electric ﬁeld at the ends of the plates. Such end effects can be neglected if the plate separation is small compared with the length of the plates. Figure 26.4 shows a battery connected to a single parallel-plate capacitor with a switch in the circuit. Let us identify the circuit as a system. When the switch is closed, the battery establishes an electric ﬁeld in the wires and charges ﬂow between the wires and the capacitor. As this occurs, there is a transformation of energy within the system. Before the switch is closed, energy is stored as chemical energy in the battery. This energy is transformed during the chemical reaction that occurs within the battery when it is operating in an electric circuit. When the switch is closed, some of the chemical energy in the battery is converted to electric potential energy related to the separation of positive and negative charges on the plates. As a result, we can describe a capacitor as a device that stores energy as well as charge. We will explore this energy storage in more detail in Section 26.4.
+Q
–Q
Courtesy of Harold M. Waage, Princeton University
(a)
(b)
Figure 26.3 (a) The electric ﬁeld between the plates of a parallel-plate capacitor is uniform near the center but nonuniform near the edges. (b) Electric ﬁeld pattern of two oppositely charged conducting parallel plates. Small pieces of thread on an oil sur face align with the electric ﬁeld.
800
C H A P T E R 2 6 • Capacitance and Dielectrics
Separation of charges represents potential Electrons move energy from the plate to the wire, – leaving the plate + positively + – charged + – + + + Electric field in wire + – ∆V – –
Electrons move from the wire to the plate
– E Electric field between plates
+
Electric field in wire Chemical energy in batter y is reduced
– ∆V
(a)
(b)
At the Active Figures link at http://www.pse6.com, you can adjust the battery voltage and see the resulting charge on the plates and the electric ﬁeld between the plates.
Active Figure 26.4 (a) A circuit consisting of a capacitor, a battery, and a switch. (b) When the switch is closed, the battery establishes an electric ﬁeld in the wire that causes electrons to move from the left plate into the wire and into the right plate from the wire. As a result, a separation of charge exists on the plates, which represents an increase in electric potential energy of the system of the circuit. This energy in the system has been transformed from chemical energy in the battery.
Quick Quiz 26.2
Many computer keyboard buttons are constructed of capacitors, as shown in Figure 26.5. When a key is pushed down, the soft insulator between the movable plate and the ﬁxed plate is compressed. When the key is pressed, the capacitance (a) increases, (b) decreases, or (c) changes in a way that we cannot determine because the complicated electric circuit connected to the keyboard button may cause a change in V.
B
Key Movable plate Soft insulator Fixed plate
Figure 26.5 (Quick Quiz 26.2) One type of computer keyboard button.
Example 26.1
Parallel-Plate Capacitor C
0A
A parallel-plate capacitor with air between the plates has an area A 2.00 10 4 m2 and a plate separation d 1.00 mm. Find its capacitance. Solution From Equation 26.3, we ﬁnd that
(8.85 10
12
10 F
12
d 1.77
C 2/N m2)(2.00 1.00 10 3 m
10
4
m2)
1.77 pF
S ECTI O N 2 6.2 • Calculating Capacitance
801
Cylindrical and Spherical Capacitors
From the deﬁnition of capacitance, we can, in principle, ﬁnd the capacitance of any geometric arrangement of conductors. The following examples demonstrate the use of this deﬁnition to calculate the capacitance of the other familiar geometries that we mentioned: cylinders and spheres.
Example 26.2
The Cylindrical Capacitor not contribute to the electric ﬁeld inside it. Using this result and noting from Figure 26.6b that E is along r, we ﬁnd that Vb Va
b a
A solid cylindrical conductor of radius a and charge Q is coaxial with a cylindrical shell of negligible thickness, radius b a, and charge Q (Fig. 26.6a). Find the capacitance of this cylindrical capacitor if its length is . Solution It is difﬁcult to apply physical arguments to this conﬁguration, although we can reasonably expect the capacitance to be proportional to the cylinder length for the same reason that parallel-plate capacitance is proportional to plate area: stored charges have more room in which to be distributed. If we assume that is much greater than a and b, we can neglect end effects. In this case, the electric ﬁeld is perpendicular to the long axis of the cylinders and is conﬁned to the region between them (Fig. 26.6b). We must ﬁrst calculate the potential difference between the two cylinders, which is given in general by Vb Va
b a
E r dr
2k e
b a
dr r
2k e ln
b a
Substituting this result into Equation 26.1 and using the fact that Q / , we obtain C Q V Q (2k e Q / )ln(b/a) (26.4)
2k e ln(b/a)
E ds
where E is the electric ﬁeld in the region between the cylinders. In Chapter 24, we showed using Gauss’s law that the magnitude of the electric ﬁeld of a cylindrical charge distribution having linear charge density is E 2ke /r (Eq. 24.7). The same result applies here because, according to Gauss’s law, the charge on the outer cylinder does
where V is the magnitude of the potential difference between the cylinders, given by V Va Vb 2ke ln(b/a), a positive quantity. As predicted, the capacitance is proportional to the length of the cylinders. As we might expect, the capacitance also depends on the radii of the two cylindrical conductors. From Equation 26.4, we see that the capacitance per unit length of a combination of concentric cylindrical conductors is C 1 2k e ln(b/a) (26.5)
b a
b a r Q –Q
An example of this type of geometric arrangement is a coaxial cable, which consists of two concentric cylindrical conductors separated by an insulator. You are likely to have a coaxial cable attached to your television set or VCR if you are a subscriber to cable television. The cable carries electrical signals in the inner and outer conductors. Such a geometry is especially useful for shielding the signals from any possible external inﬂuences. What If? Suppose b
2.00a for the cylindric al c apacitor. We would like to increase the c apacit ance, and we c an do so by choosing to increase by 1 0% or by increasing a by 1 0%. Which choice is more effective at increasing the c apacit ance?
Gaussian sur face (a) (b)
Answer According to Equation 26.4, C is proportional to , so increasing by 10% results in a 10% increase in C. For the result of the change in a, let us ﬁrst evaluate C for b 2.00a : C
Figure 26.6 (Example 26.2) (a) A cylindrical capacitor consists of a solid cylindrical conductor of radius a and length surrounded by a coaxial cylindrical shell of radius b. (b) End view. The electric ﬁeld lines are radial. The dashed line represents the end of the cylindrical gaussian sur face of radius r and length .
2k e ln(b/a) 0.721 ke
2k e ln(2.00)
2k e (0.693)
802
C H A P T E R 2 6 • Capacitance and Dielectrics
Now, for a 10% increase in a, the new value is a C 2k e ln(b/a ) 2k e ln(2.00a/1.10a) 2k e (0.598)
1.10a, so
2k e ln(2.00/1.10)
0.836
ke
The ratio of the new and old capacitances is C C Example 26.3 0.836 /k e 0.721 /k e 1.16
corresponding to a 16% increase in capacitance. Thus, it is more effective to increase a than to increase . Note two more extensions of this problem. First, the advantage goes to increasing a only for a range of relationships between a and b. It is a valuable exercise to show that if b 2.85a, increasing by 10% is more effective than increasing a (Problem 77). Second, if we increase b, we reduce the capacitance, so we would need to decrease b to increase the capacitance. Increasing a and decreasing b both have the effect of bringing the plates closer together, which increases the capacitance.
The Spherical Capacitor Vb Va
b a
A spherical capacitor consists of a spherical conducting shell of radius b and charge Q concentric with a smaller conducting sphere of radius a and charge Q (Fig. 26.7). Find the capacitance of this device. Solution As we showed in Chapter 24, the ﬁeld outside a spherically symmetric charge distribution is radial and given by the expression ke Q /r 2. In this case, this result applies to the ﬁeld between the spheres (a r b). From Gauss’s law we see that only the inner sphere contributes to this ﬁeld. Thus, the potential difference between the spheres is
–Q
E r dr 1 b 1 a
keQ
b a
dr r2
keQ
1 r
b a
keQ
The magnitude of the potential difference is ∆V Vb Va k eQ (b ab a)
Substituting this value for V into Equation 26.1, we obtain C Q ∆V ab k e(b a) (26.6)
+Q a
What If? What if the radius b of the outer sphere
b
approaches inﬁnity? What does the capacitance become?
Answer In Equation 26.6, we let b : :
Figure 26.7 (Example 26.3) A spherical capacitor consists of an inner sphere of radius a surrounded by a concentric spherical shell of radius b. The electric ﬁeld between the spheres is directed radially outward when the inner sphere is positively charged.
C
b:
lim
ab k e(b a)
ab k e(b)
a ke
4
0a
Note that this is the same expression as Equation 26.2, the capacitance of an isolated spherical conductor.
Capacitor symbol
26.3
Combinations of Capacitors
Batter y symbol
–
+
Switch symbol
Figure 26.8 Circuit symbols for capacitors, batteries, and switches. Note that capacitors are in blue and batteries and switches are in red.
Two or more capacitors often are combined in electric circuits. We can calculate the equivalent capacitance of certain combinations using methods described in this section. Throughout this section, we assume that the capacitors to be combined are initially uncharged. In studying electric circuits, we use a simpliﬁed pictorial representation called a circuit diagram. Such a diagram uses circuit symbols to represent various circuit elements. The circuit symbols are connected by straight lines that represent the wires between the circuit elements. The circuit symbols for capacitors and batteries, as well as the color codes used for them in this text, are given in Figure 26.8. The symbol for the capacitor reflects the geometr y of the most common model for a capacitor—a pair of parallel plates. The positive terminal of the batter y is at the higher potential and is represented in the circuit symbol by the longer line.
S ECTI O N 2 6.3 • Combinations of Capacitors
803
Parallel Combination
Two capacitors connected as shown in Figure 26.9a are known as a parallel combination of capacitors. Figure 26.9b shows a circuit diagram for this combination of capacitors. The left plates of the capacitors are connected by a conducting wire to the positive terminal of the battery and are therefore both at the same electric potential as the positive terminal. Likewise, the right plates are connected to the negative terminal and are therefore both at the same potential as the negative terminal. Thus, the individual potential differences across capacitors connected in parallel are the same and are equal to the potential difference applied across the combination. In a circuit such as that shown in Figure 26.9, the voltage applied across the combination is the terminal voltage of the battery. Situations can occur in which the parallel combination is in a circuit with other circuit elements; in such situations, we must determine the potential difference across the combination by analyzing the entire circuit. When the capacitors are ﬁrst connected in the circuit shown in Figure 26.9, electrons are transferred between the wires and the plates; this transfer leaves the left plates positively charged and the right plates negatively charged. The ﬂow of charge ceases when the voltage across the capacitors is equal to that across the battery terminals. The capacitors reach their maximum charge when the ﬂow of charge ceases. Let us call the maximum charges on the two capacitors Q 1 and Q 2. The total charge Q stored by the two capacitors is Q Q1 Q2
(26.7)
That is, the total charge on capacitors connected in parallel is the sum of the charges on the individual capacitors. Because the voltages across the capacitors are the same, the charges that they carry are Q1 C 1 ∆V Q2 C 2 ∆V
Suppose that we wish to replace these two capacitors by one equivalent capacitor having a capacitance C eq, as in Figure 26.9c. The effect this equivalent capacitor has on the circuit must be exactly the same as the effect of the combination of the two
C1 + – ∆V1 = ∆V2 = ∆V C1 Q1 C2 + – Q2 C2 C eq = C 1 + C 2
+
– ∆V
+ ∆V
–
+ ∆V
–
(a)
(b)
(c)
Active Figure 26.9 (a) A parallel combination of two capacitors in an electric circuit in which the potential difference across the battery terminals is V. (b) The circuit diagram for the parallel combination. (c) The equivalent capacitance is C eq C 1 C 2.
At the Active Figures link at http://www.pse6.com, you can adjust the battery voltage and the individual capacitances to see the resulting charges and voltages on the capacitors. You can combine up to four capacitors in parallel.
804
C H A P T E R 2 6 • Capacitance and Dielectrics
individual capacitors. That is, the equivalent capacitor must store Q units of charge when connected to the battery. We can see from Figure 26.9c that the voltage across the equivalent capacitor also is V because the equivalent capacitor is connected directly across the battery terminals. Thus, for the equivalent capacitor, Q C eq ∆V C 1 ∆V C2 V
Substituting these three relationships for charge into Equation 26.7, we have C eq ∆V C eq C1 C2 (parallel combination)
If we extend this treatment to three or more capacitors connected in parallel, we ﬁnd the equivalent capacitance to be
Capacitors in parallel
C eq
C1
C2
C3
(parallel combination)
(26.8)
Thus, the equivalent capacitance of a parallel combination of capacitors is the algebraic sum of the individual capacitances and is greater than any of the individual capacitances. This makes sense because we are essentially combining the areas of all the capacitor plates when we connect them with conducting wire, and capacitance of parallel plates is proportional to area (Eq. 26.3).
Series Combination
Two capacitors connected as shown in Figure 26.10a and the equivalent circuit diagram in Figure 26.10b are known as a series combination of capacitors. The left plate of capacitor 1 and the right plate of capacitor 2 are connected to the terminals of a battery. The other two plates are connected to each other and to nothing else; hence, they form an isolated conductor that is initially uncharged and must continue to have zero net charge. To analyze this combination, let us begin by considering the uncharged capacitors and follow what happens just after a battery is connected to the circuit. When the battery is connected, electrons are transferred out of the left plate of C 1 and into the right plate of C 2. As this negative charge accumulates on the right plate of C 2, an equivalent amount of negative charge is forced off the left plate of C 2, and this left plate therefore has an excess positive charge. The negative charge leaving
∆V 1
C1
∆V 2
C2 C eq
1 =1+1
C1
C2
+Q
–Q
+Q
–Q
C1
Q1 = Q2 = Q C2 ∆V2
+
–
∆V
∆V1
At the Active Figures link at http://www.pse6.com, you can adjust the battery voltage and the individual capacitances to see the resulting charges and voltages on the capacitors. You can combine up to four capacitors in series.
+– ∆V
+
∆V (c)
–
(a)
(b)
Active Figure 26.10 (a) A series combination of two capacitors. The charges on the two capacitors are the same. (b) The circuit diagram for the series combination. (c) The equivalent capacitance can be calculated from the relationship 1 C eq 1 C1 1 . C2
S ECTI O N 2 6.3 • Combinations of Capacitors
805
the left plate of C 2 causes negative charges to accumulate on the right plate of C 1. As a result, all the right plates end up with a charge Q , and all the left plates end up with a charge Q . Thus, the charges on capacitors connected in series are the same. From Figure 26.10a, we see that the voltage V across the battery terminals is split between the two capacitors: ∆V ∆V1 ∆V2
(26.9)
where V1 and V2 are the potential differences across capacitors C 1 and C 2, respectively. In general, the total potential difference across any number of capacitors connected in series is the sum of the potential differences across the individual capacitors. Suppose that the equivalent single capacitor in Figure 26.10c has the same effect on the circuit as the series combination when it is connected to the battery. After it is fully charged, the equivalent capacitor must have a charge of Q on its right plate and a charge of Q on its left plate. Applying the deﬁnition of capacitance to the circuit in Figure 26.10c, we have ∆V Q C eq
Because we can apply the expression Q C V to each capacitor shown in Figure 26.10b, the potential differences across them are ∆V1 Q C1 ∆V2 Q C2
Substituting these expressions into Equation 26.9, we have Q C eq Canceling Q , we arrive at the relationship 1 C eq 1 C1 1 C2 (series combination) Q C1 Q C2
When this analysis is applied to three or more capacitors connected in series, the relationship for the equivalent capacitance is 1 C eq 1 C1 1 C2 1 C3 (series combination)
(26.10)
Capacitors in series
This shows that the inverse of the equivalent capacitance is the algebraic sum of the inverses of the individual capacitances and the equivalent capacitance of a series combination is always less than any individual capacitance in the combination.
Quick Quiz 26.3
Two capacitors are identical. They can be connected in series or in parallel. If you want the smallest equivalent capacitance for the combination, do you connect them in (a) series, in (b) parallel, or (c) do the combinations have the same capacitance? Consider the two capacitors in Quick Quiz 26.3 again. Each capacitor is charged to a voltage of 10 V. If you want the largest combined potential difference across the combination, do you connect them in (a) series, in (b) parallel, or (c) do the combinations have the same potential difference?
Quick Quiz 26.4
806
C H A P T E R 2 6 • Capacitance and Dielectrics
P R O B L E M - S O LV I N G H I N T S
Capacitors
• •
Be careful with units. When you calculate capacitance in farads, make sure that distances are expressed in meters. When checking consistency of units, remember that the unit for electric ﬁelds can be either N/C or V/m. When two or more capacitors are connected in parallel, the potential difference across each is the same. The charge on each capacitor is proportional to its capacitance; hence, the capacitances can be added directly to give the equivalent capacitance of the parallel combination. The equivalent capacitance is always larger than the individual capacitances. When two or more capacitors are connected in series, they carr y the same charge, and the sum of the potential differences equals the total potential difference applied to the combination. The sum of the reciprocals of the capacitances equals the reciprocal of the equivalent capacitance, which is always less than the capacitance of the smallest individual capacitor.
•
Example 26.4
Equivalent Capacitance 1 C eq C eq 1 C1 2.0 F 1 C2 1 4.0 F 1 4.0 F
Interactive 1 2.0 F
Find the equivalent capacitance between a and b for the combination of capacitors shown in Figure 26.11a. All capacitances are in microfarads. Solution Using Equations 26.8 and 26.10, we reduce the combination step by step as indicated in the ﬁgure. The 1.0- F and 3.0- F capacitors are in parallel and combine according to the expression C eq C 1 C 2 4.0 F. The 2.0- F and 6.0- F capacitors also are in parallel and have an equivalent capacitance of 8.0 F. Thus, the upper branch in Figure 26.11b consists of two 4.0- F capacitors in series, which combine as follows:
The lower branch in Figure 26.11b consists of two 8.0- F capacitors in series, which combine to yield an equivalent capacitance of 4.0 F. Finally, the 2.0- F and 4.0- F capacitors in Figure 26.11c are in parallel and thus have an equivalent capacitance of 6.0 F.
1.0 4.0 3.0 a 6.0 8.0 (a) b a 8.0 (b) b a 4.0 (c) (d) b a 6.0 b 4.0 4.0 2.0
2.0
8.0
Figure 26.11 (Example 26.4) To ﬁnd the equivalent capacitance of the capacitors in part (a), we reduce the various combinations in steps as indicated in parts (b), (c), and (d), using the series and parallel rules described in the text. Practice reducing a combination of capacitors to a single equivalent capacitance at the Interactive Worked Example link at http://www.pse6.com.
S ECTI O N 2 6.4 • Energy Stored in a Charged Capacitor
807
26.4
Energy Stored in a Charged Capacitor
Almost everyone who works with electronic equipment has at some time veriﬁed that a capacitor can store energy. If the plates of a charged capacitor are connected by a conductor, such as a wire, charge moves between each plate and its connecting wire until the capacitor is uncharged. The discharge can often be observed as a visible spark. If you should accidentally touch the opposite plates of a charged capacitor, your ﬁngers act as a pathway for discharge, and the result is an electric shock. The degree of shock you receive depends on the capacitance and on the voltage applied to the capacitor. Such a shock could be fatal if high voltages are present, such as in the power supply of a television set. Because the charges can be stored in a capacitor even when the set is turned off, unplugging the television does not make it safe to open the case and touch the components inside. To calculate the energy stored in the capacitor, we shall assume a charging process that is different from the actual process described in Section 26.1 but which gives the same ﬁnal result. We can make this assumption because the energy in the ﬁnal conﬁguration does not depend on the actual charge-transfer process. We imagine that the charge is transferred mechanically through the space between the plates. We reach in and grab a small amount of positive charge on the plate connected to the negative terminal and apply a force that causes this positive charge to move over to the plate connected to the positive terminal. Thus, we do work on the charge as we transfer it from one plate to the other. At ﬁrst, no work is required to transfer a small amount of charge dq from one plate to the other.3 However, once this charge has been transferred, a small potential difference exists between the plates. Therefore, work must be done to move additional charge through this potential difference. As more and more charge is transferred from one plate to the other, the potential difference increases in proportion, and more work is required. Suppose that q is the charge on the capacitor at some instant during the charging process. At the same instant, the potential difference across the capacitor is V q /C. From Section 25.2, we know that the work necessary to transfer an increment of charge dq from the plate carrying charge q to the plate carrying charge q (which is at the higher electric potential) is dW ∆V dq q dq C
∆V
Q dq
q
This is illustrated in Figure 26.12. The total work required to charge the capacitor from q 0 to some ﬁnal charge q Q is W
Q 0
Figure 26.12 A plot of potential difference versus charge for a capacitor is a straight line having a slope 1/C. The work required to move charge dq through the potential difference V existing at the time across the capacitor plates is given approximately by the area of the shaded rectangle. The total work required to charge the capacitor to a ﬁnal charge Q is the triangular area under the straight line, W 1Q V . (Don’t 2 forget that 1 V J/C; hence, the unit for the triangular area is the joule.)
q dq C
1 C
Q 0
q dq
Q2 2C
The work done in charging the capacitor appears as electric potential energy U stored in the capacitor. Using Equation 26.1, we can express the potential energy stored in a charged capacitor in the following forms: U Q2 2C
1 Q 2
∆V
1 C (∆V )2 2
(26.11)
Energy stored in a charged capacitor
This result applies to any capacitor, regardless of its geometry. We see that for a given capacitance, the stored energy increases as the charge increases and as the potential difference increases. In practice, there is a limit to the maximum energy (or charge) that can be stored because, at a sufﬁciently great value of V, discharge ultimately occurs between the plates. For this reason, capacitors are usually labeled with a maximum operating voltage.
3
We shall use lowercase q for the time-varying charge on the capacitor while it is charging, to distinguish it from uppercase Q , which is the total charge on the capacitor after it is completely charged.
808
C H A P T E R 2 6 • Capacitance and Dielectrics
We can consider the energy stored in a capacitor as being stored in the electric ﬁeld created between the plates as the capacitor is charged. This description is reasonable because the electric ﬁeld is proportional to the charge on the capacitor. For a parallel-plate capacitor, the potential difference is related to the electric ﬁeld through the relationship V Ed. Furthermore, its capacitance is C 0A/d (Eq. 26.3). Substituting these expressions into Equation 26.11, we obtain U
1 2 0A (E 2d 2) d 1 ( Ad )E 2 20
(26.12)
Because the volume occupied by the electric ﬁeld is Ad, the energy per unit volume uE U/Ad, known as the energy density, is
Energy density in an electric ﬁeld
uE
1 E2 20
(26.13)
L
PITFALL PR EVE NTI O N
26.4 Not a New Kind of Energy
The energy given by Equation 26.13 is not a new kind of energy. It is familiar electric potential energy associated with a system of separated source charges. Equation 26.13 provides a new interpretation, or a new way of modeling the energy, as energy associated with the electric ﬁeld, regardless of the source of the ﬁeld.
Although Equation 26.13 was derived for a parallel-plate capacitor, the expression is generally valid, regardless of the source of the electric ﬁeld. That is, the energy density in any electric ﬁeld is proportional to the square of the magnitude of the electric ﬁeld at a given point.
Quick Quiz 26.5
You have three capacitors and a battery. In which of the following combinations of the three capacitors will the maximum possible energy be stored when the combination is attached to the battery? (a) series (b) parallel (c) Both combinations will store the same amount of energy.
Quick Quiz 26.6
You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching each other. When you pull the plates apart to a larger separation, do the following quantities increase, decrease, or stay the same? (a) C ; (b) Q ; (c) E between the plates; (d) V ; (e) energy stored in the capacitor.
Quick Quiz 26.7
Repeat Quick Quiz 26.6, but this time answer the questions for the situation in which the battery remains connected to the capacitor while you pull the plates apart.
Example 26.5
Rewiring Two Charged Capacitors
Q 1i C1 – Q 1f
Interactive
C1 –
Two capacitors C 1 and C 2 (where C 1 C 2) are charged to the same initial potential difference V i . The charged capacitors are removed from the batter y, and their plates are connected with opposite polarity as in Figure 26.13a. The switches S1 and S2 are then closed, as in Figure 26.13b. (A) Find the ﬁnal potential difference after the switches are closed. Vf between a and b
+
+
a S1 Q 2i – + C2 S2
b
a S1 Q 2f + – C2 S2
b
Solution Figure 26.13 helps us conceptualize the initial and ﬁnal conﬁgurations of the system. In Figure 26.13b, it might appear as if the capacitors are connected in parallel, but there is no battery in this circuit that is applying a voltage across the combination. Thus, we cannot categorize this as a problem in which capacitors are connected in parallel. We can categorize this as a problem involving an isolated system for electric charge—the left-hand plates of the capac-
(a)
(b)
Figure 26.13 (Example 26.5) (a) Two capacitors are charged to the same initial potential difference and connected together with plates of opposite sign to be in contact when the switches are closed. (b) When the switches are closed, the charges redistribute.
itors form an isolated system because they are not connected to the right-hand plates by conductors. To analyze
S ECTI O N 2 6.4 • Energy Stored in a Charged Capacitor
809
the problem, note that the charges on the left-hand plates before the switches are closed are Q 1i C1 ∆Vi and Q 2i C 2 ∆Vi
(B) Find the total energy stored in the capacitors before and after the switches are closed and the ratio of the ﬁnal energy to the initial energy. Solution Before the switches are closed, the total energy stored in the capacitors is Ui
1 C (∆Vi)2 21 1 C (∆Vi)2 22 1 (C 21
The negative sign for Q 2i is necessary because the charge on the left plate of capacitor C 2 is negative. The total charge Q in the system is (1) Q Q 1i Q 2i (C1 C 2 )∆Vi
C 2)(∆Vi)2
After the switches are closed, the total charge Q in the system remains the same but the charges on the individual capacitors change to new values Q 1f and Q 2f . Because the system is isolated, (2) Q Q 1f Q 2f
After the switches are closed, the total energy stored in the capacitors is Uf
1 C (∆Vf )2 21 1 C (∆Vf )2 22 1 (C 1 2
C 2)( Vf )2
The charges redistribute until the potential difference is the same across both capacitors, Vf . To satisfy this requirement, the charges on the capacitors after the switches are closed are Q 1f C1 ∆Vf and Q 2f C 2 ∆Vf
Using the results of part (A), we can express this as
1 2
Uf
Dividing the ﬁrst equation by the second, we have (3) Q 1f C1 Q C 2 2f
(C 1 C 2)2(∆Vi)2 (C 1 C 2)
Therefore, the ratio of the ﬁnal energy stored to the initial energy stored is (8) Uf Ui C1 C1
1 (C 1 2
Combining Equations (2) and (3), we obtain Q Q 1f Q 2f (4) C1 Q C 2 2f Q 2f Q Q 2f C2 C1 C2 Q 2f 1 C1 C2
C 2)2(∆Vi)2/(C1
1 (C 21
C 2)
C 2)(∆Vi)2
2
C2 C2
Using Equations (3) and (4) to ﬁnd Q 1f in terms of Q , we have (5) Q 1f C1 Q C 2 2f Q C1 C1 C2 C1 Q C2 C2 C1 C2
To ﬁnalize this problem, note that this ratio is less than unity, indicating that the ﬁnal energy is less than the initial energy. At ﬁrst, you might think that the law of energy conservation has been violated, but this is not the case. The “missing’’ energy is transferred out of the system of the capacitors by the mechanism of electromagnetic waves, as we shall see in Chapter 34. What If? What if the two capacitors have the same capacitance? What would we expect to happen when the switches are closed?
Finally, using Equation 26.1 to ﬁnd the voltage across each capacitor, we ﬁnd that (6) (7) ∆V1f ∆V2f Q 1f C1 Q 2f C2 Q [C 1/(C 1 C1 Q [C 2/(C 1 C2 C 2)] C1 C 2)] C1 Q C2 Q C2
As noted earlier, V1f V 2f Vf . To express Vf in terms of the given quantities C 1, C 2, and Vi , we substitute the value of Q from Equation (1) into either Equation (6) or (7) to obtain C1 C1 C2 C2
∆Vf
∆Vi
Answer The equal-magnitude charges on the two capacitors should simply cancel each other and the capacitors will be uncharged afterward. Let us test our results to see if this is the case mathematically. In Equation (1), because the charges are of equal magnitude and opposite sign, we see that Q 0. Thus, Equations (4) and (5) show us that Q 1f Q 2f 0, consistent with our prediction. Furthermore, Equations (6) and (7) show us that V1f V2f 0, which is consistent with uncharged capacitors. Finally, if C 1 C 2, Equation (8) shows us that Uf 0, which is also consistent with uncharged capacitors.
At the Interactive Worked Example link at http://www.pse6.com, explore this situation for various initial values of the voltage and the capacitances.
810
C H A P T E R 2 6 • Capacitance and Dielectrics
Figure 26.14 In a hospital or at an emergency scene, you might see a patient being revived with a deﬁbrillator. The deﬁbrillator’s paddles are applied to the patient’s chest, and an electric shock is sent through the chest cavity. The aim of this technique is to restore the heart’s normal rhythm pattern.
One device in which capacitors have an important role is the deﬁbrillator (Fig. 26.14). Up to 360 J is stored in the electric ﬁeld of a large capacitor in a deﬁbrillator when it is fully charged. The deﬁbrillator can deliver all this energy to a patient in about 2 ms. (This is roughly equivalent to 3 000 times the power delivered to a 60-W lightbulb!) Under the proper conditions, the deﬁbrillator can be used to stop cardiac ﬁbrillation (random contractions) in heart attack victims. When ﬁbrillation occurs, the heart produces a rapid, irregular pattern of beats. A fast discharge of energy through the heart can return the organ to its normal beat pattern. Emergency medical teams use portable deﬁbrillators that contain batteries capable of charging a capacitor to a high voltage. (The circuitry actually permits the capacitor to be charged to a much higher voltage than that of the battery.) The stored energy is released through the heart by conducting electrodes, called paddles, that are placed on both sides of the victim’s chest. The paramedics must wait between applications of the energy due to the time necessary for the capacitors to become fully charged. In this case and others (e.g., camera ﬂash units and lasers used for fusion experiments), capacitors serve as energy reservoirs which can be slowly charged and then discharged quickly to provide large amounts of energy in a short pulse. A camera’s ﬂash unit also uses a capacitor, although the total amount of energy stored is much less than that stored in a deﬁbrillator. After the ﬂash unit’s capacitor is charged, tripping the camera’s shutter causes the stored energy to be sent through a special lightbulb that brieﬂy illuminates the subject being photographed.
26.5
Adam Hart-Davis/SPL/Custom Medical Stock
Capacitors with Dielectrics
A dielectric is a nonconducting material, such as rubber, glass, or waxed paper. When a dielectric is inserted between the plates of a capacitor, the capacitance increases. If the dielectric completely ﬁlls the space between the plates, the capacitance increases by a dimensionless factor , which is called the dielectric constant of the material. The dielectric constant varies from one material to another. In this section, we analyze this change in capacitance in terms of electrical parameters such as electric charge, electric ﬁeld, and potential difference; in Section 26.7, we shall discuss the microscopic origin of these changes.
S ECTI O N 2 6.5 • Capacitors with Dielectrics
811
Dielectric C0 + Q0 – C + Q0 –
∆V0
∆V
(a)
(b)
Figure 26.15 A charged capacitor (a) before and (b) after insertion of a dielectric between the plates. The charge on the plates remains unchanged, but the potential difference decreases from V0 to V V0/ . Thus, the capacitance increases from C 0 to C 0.
L
PITFALL PR EVE NTI O N
We can per form the following experiment to illustrate the effect of a dielectric in a capacitor. Consider a parallel-plate capacitor that without a dielectric has a charge Q 0 and a capacitance C 0. The potential difference across the capacitor is V 0 Q 0/C 0. Figure 26.15a illustrates this situation. The potential difference is measured by a voltmeter, which we shall study in greater detail in Chapter 28. Note that no battery is shown in the ﬁgure; also, we must assume that no charge can ﬂow through an ideal voltmeter. Hence, there is no path by which charge can ﬂow and alter the charge on the capacitor. If a dielectric is now inserted between the plates, as in Figure 26.15b, the voltmeter indicates that the voltage between the plates decreases to a value V. The voltages with and without the dielectric are related by the factor as follows: ∆V ∆V0
26.5 Is the Capacitor Connected to a Battery?
In problems in which you are modifying a capacitor (by insertion of a dielectric, for example), you must note whether modiﬁcations to the capacitor are being made while the capacitor is connected to a battery or after it is disconnected. If the capacitor remains connected to the battery, the voltage across the capacitor necessarily remains the same. If you disconnect the capacitor from the battery before making any modiﬁcations to the capacitor, the capacitor is an isolated system and its charge remains the same.
Because V V0, we see that 1. Because the charge Q 0 on the capacitor does not change, we conclude that the capacitance must change to the value C Q0 ∆V C Q0 ∆V0/ C0 Q0 ∆V0
(26.14)
That is, the capacitance increases by the factor when the dielectric completely ﬁlls the region between the plates.4 For a parallel-plate capacitor, where C 0 0 A/d (Eq. 26.3), we can express the capacitance when the capacitor is ﬁlled with a dielectric as C
0A
Capacitance of a capacitor ﬁlled with a material of dielectric constant
d
(26.15)
From Equations 26.3 and 26.15, it would appear that we could make the capacitance very large by decreasing d, the distance between the plates. In practice, the lowest value of d is limited by the electric discharge that could occur through the dielectric medium separating the plates. For any given separation d, the maximum voltage that can be applied to a capacitor without causing a discharge depends on the
If the dielectric is introduced while the potential difference is held constant by a battery, the charge increases to a value Q Q 0. The additional charge comes from the wires attached to the capacitor, and the capacitance again increases by the factor .
4
812
C H A P T E R 2 6 • Capacitance and Dielectrics
Figure 26.16 Dielectric breakdown in air. Sparks are produced when the high voltage between the wires causes the electric ﬁeld to exceed the dielectric strength of air.
dielectric strength (maximum electric ﬁeld) of the dielectric. If the magnitude of the electric ﬁeld in the dielectric exceeds the dielectric strength, then the insulating properties break down and the dielectric begins to conduct. Figure 26.16 shows the effect of exceeding the dielectric strength of air. Sparks appear between the two wires, due to ionization of atoms and recombination with electrons in the air, similar to the process that produced corona discharge in Section 25.6. Physical capacitors have a speciﬁcation called by a variety of names, including working voltage, breakdown voltage, and rated voltage. This parameter represents the largest voltage that can be applied to the capacitor without exceeding the dielectric strength of the dielectric material in the capacitor. Consequently, when selecting a capacitor for a given application, you must consider the capacitance of the device along with the expected voltage across the capacitor in the circuit, making sure that the expected voltage will be smaller than the rated voltage of the capacitor. You can see the rated voltage on several of the capacitors in the opening photograph for this chapter. Insulating materials have values of greater than unity and dielectric strengths greater than that of air, as Table 26.1 indicates. Thus, we see that a dielectric provides the following advantages: • • • Increase in capacitance Increase in maximum operating voltage Possible mechanical support between the plates, which allows the plates to be close together without touching, thereby decreasing d and increasing C.
© Loren Winters/Visuals Unlimited
Types of Capacitors
Commercial capacitors are often made from metallic foil interlaced with thin sheets of either parafﬁn-impregnated paper or Mylar as the dielectric material. These alternate layers of metallic foil and dielectric are rolled into a cylinder to form a small package (Fig. 26.17a). High-voltage capacitors commonly consist of a number of interwoven
Table 26.1
Approximate Dielectric Constants and Dielectric Strengths of Various Materials at Room Temperature
Material Air (dry) Bakelite Fused quartz Mylar Neoprene rubber Nylon Paper Parafﬁn-impregnated paper Polystyrene Polyvinyl chloride Porcelain Pyrex glass Silicone oil Strontium titanate Teﬂon Vacuum Water
a
Dielectric Constant 1.000 59 4.9 3.78 3.2 6.7 3.4 3.7 3.5 2.56 3.4 6 5.6 2.5 233 2.1 1.000 00 80
Dielectric Strengtha (106 V/m) 3 24 8 7 12 14 16 11 24 40 12 14 15 8 60 — —
The dielectric strength equals the maximum electric ﬁeld that can exist in a dielectric without electrical breakdown. Note that these values depend strongly on the presence of impurities and ﬂaws in the materials.
S ECTI O N 2 6.5 • Capacitors with Dielectrics
813
Metal foil Plates Electrolyte Case
Contacts Paper (a) Oil (b) Metallic foil + oxide layer (c)
Figure 26.17 Three commercial capacitor designs. (a) A tubular capacitor, whose plates are separated by paper and then rolled into a cylinder. (b) A high-voltage capacitor consisting of many parallel plates separated by insulating oil. (c) An electrolytic capacitor.
metallic plates immersed in silicone oil (Fig. 26.17b). Small capacitors are often constructed from ceramic materials. Often, an electrolytic capacitor is used to store large amounts of charge at relatively low voltages. This device, shown in Figure 26.17c, consists of a metallic foil in contact with an electrolyte—a solution that conducts electricity by virtue of the motion of ions contained in the solution. When a voltage is applied between the foil and the electrolyte, a thin layer of metal oxide (an insulator) is formed on the foil, and this layer serves as the dielectric. Very large values of capacitance can be obtained in an electrolytic capacitor because the dielectric layer is very thin, and thus the plate separation is very small. Electrolytic capacitors are not reversible as are many other capacitors—they have a polarity, which is indicated by positive and negative signs marked on the device. When electrolytic capacitors are used in circuits, the polarity must be aligned properly. If the polarity of the applied voltage is opposite that which is intended, the oxide layer is removed and the capacitor conducts electricity instead of storing charge. Variable capacitors (typically 10 to 500 pF) usually consist of two interwoven sets of metallic plates, one ﬁxed and the other movable, and contain air as the dielectric (Fig. 26.18). These types of capacitors are often used in radio tuning circuits.
George Semple
Figure 26.18 A variable capacitor. When one set of metal plates is rotated so as to lie between a ﬁxed set of plates, the capacitance of the device changes.
Quick Quiz 26.8
If you have ever tried to hang a picture or a mirror, you know it can be difﬁcult to locate a wooden stud in which to anchor your nail or screw. A carpenter’s stud-ﬁnder is basically a capacitor with its plates arranged side by side instead of facing one another, as shown in Figure 26.19. When the device is moved over a stud, does the capacitance increase or decrease?
Stud
Capacitor plates
Stud-finder Wall board (a) (b)
Figure 26.19 (Quick Quiz 26.8) A stud-ﬁnder. (a) The materials between the plates of the capacitor are the wallboard and air. (b) When the capacitor moves across a stud in the wall, the materials between the plates are the wallboard and the wood. The change in the dielectric constant causes a signal light to illuminate.
814
C H A P T E R 2 6 • Capacitance and Dielectrics
Quick Quiz 26.9
A fully charged parallel-plate capacitor remains connected to a battery while you slide a dielectric between the plates. Do the following quantities increase, decrease, or stay the same? (a) C ; (b) Q ; (c) E between the plates; (d) V.
Example 26.6
A Paper-Filled Capacitor Solution From Table 26.1 we see that the dielectric strength of paper is 16 106 V/m. Because the thickness of the paper is 1.0 mm, the maximum voltage that can be applied before breakdown is ∆Vmax E max d 16 10
12 C 2/N
A parallel-plate capacitor has plates of dimensions 2.0 cm by 3.0 cm separated by a 1.0-mm thickness of paper. (A) Find its capacitance. Solution Because C 3.7 20
0A
3.7 for paper (see Table 26.1), we have
(16 V
106 V/m)(1.0
10
3
m)
103
d (8.85 10
12 F
Hence, the maximum charge is 1.0 20 pF m2)(6.0 10 3 m 10
4
m2) Q max C Vmax 0.32 C (20 10
12 F )(16
103 V )
(B) What is the maximum charge that can be placed on the capacitor? Example 26.7 Energy Stored Before and After Solution From Equation 26.11, we see that the energy stored in the absence of the dielectric is
A parallel-plate capacitor is charged with a battery to a charge Q 0, as shown in Figure 26.20a. The battery is then removed, and a slab of material that has a dielectric constant is inserted between the plates, as shown in Figure 26.20b. Find the energy stored in the capacitor before and after the dielectric is inserted.
Q0 –
U0
Q 02 2C 0
C0 +
After the battery is removed and the dielectric inserted, the charge on the capacitor remains the same. Hence, the energy stored in the presence of the dielectric is
U
Q 02 2C
∆V 0 (a) Dielectric Q0 +
But the capacitance in the presence of the dielectric is C C 0, so U becomes
U
Q 02 2 C0
U0
–
(b)
Figure 26.20 (Example 26.7) (a) A battery charges up a parallel-plate capacitor. (b) The battery is removed and a slab of dielectric material is inserted between the plates.
Because 1, the ﬁnal energy is less than the initial energy. We can account for the “missing’’ energy by noting that the dielectric, when inserted, is pulled into the device (see Section 26.7). An external agent must do negative work to keep the dielectric from accelerating. This work is simply the difference U U 0. (Alternatively, the positive work done by the system on the external agent is U 0 U.)
S ECTI O N 2 6.6 • Electric Dipole in an Electric Field
815
+q +
26.6
Electric Dipole in an Electric Field
2a – –q p
We have discussed the effect on the capacitance of placing a dielectric between the plates of a capacitor. In Section 26.7, we shall describe the microscopic origin of this effect. Before we can do so, however, we need to expand upon the discussion of the electric dipole that we began in Section 23.4 (see Example 23.6). The electric dipole consists of two charges of equal magnitude and opposite sign separated by a distance 2a, as shown in Figure 26.21. The electric moment dipole of this conﬁguration is deﬁned as the vector p directed from q toward q along the line joining the charges and having magnitude 2aq : p 2aq
(26.16)
Figure 26.21 An electric dipole consists of two charges of equal magnitude and opposite sign separated by a distance of 2a. The electric dipole moment p is directed from q toward q.
Now suppose that an electric dipole is placed in a uniform electric ﬁeld E, as shown in Figure 26.22. We identify E as the ﬁeld external to the dipole, distinguishing it from the ﬁeld due to the dipole, which we discussed in Section 23.4. The ﬁeld E is established by some other charge distribution, and we place the dipole into this ﬁeld. Let us imagine that the dipole moment makes an angle with the ﬁeld. The electric forces acting on the two charges are equal in magnitude (F q E ) and opposite in direction as shown in Figure 26.22. Thus, the net force on the dipole is zero. However, the two forces produce a net torque on the dipole; as a result, the dipole rotates in the direction that brings the dipole moment vector into greater alignment with the ﬁeld. The torque due to the force on the positive charge about an axis through O in Figure 26.22 has magnitude Fa sin , where a sin is the moment arm of F about O. This force tends to produce a clockwise rotation. The torque about O on the negative charge is also of magnitude Fa sin ; here again, the force tends to produce a clockwise rotation. Thus, the magnitude of the net torque about O is 2Fa sin Because F qE and p 2aq, we can express as 2aqE sin p sin
(26.17)
+q
+
F
θ
O –F E – –q
Figure 26.22 An electric dipole in a uniform external electric ﬁeld. The dipole moment p is at an angle to the ﬁeld, causing the dipole to experience a torque.
It is convenient to express the torque in vector form as the cross product of the vectors p and E: p E
(26.18)
Torque on an electric dipole in an external electric ﬁeld
We can determine the potential energy of the system—an electric dipole in an external electric ﬁeld—as a function of the orientation of the dipole with respect to the ﬁeld. To do this, we recognize that work must be done by an external agent to rotate the dipole through an angle so as to cause the dipole moment vector to become less aligned with the ﬁeld. The work done is then stored as potential energy in the system. The work dW d (Eq. 10.22). Because required to rotate the dipole through an angle d is dW pE sin and because the work results in an increase in the potential energy U, we ﬁnd that for a rotation from i to f the change in potential energy of the system is Uf Ui
i f
d
i f i
f
pE sin d pE(cos
i
pE
i
f
sin
d
pE [ cos ]
cos
f)
The term that contains cos i is a constant that depends on the initial orientation of the dipole. It is convenient for us to choose a reference angle of i 90°, so that cos i cos 90° 0. Furthermore, let us choose Ui 0 at i 90° as our reference of potential energy. Hence, we can express a general value of U Uf as U pE cos
(26.19)
We can write this expression for the potential energy of a dipole in an electric ﬁeld as the dot product of the vectors p and E:
816
C H A P T E R 2 6 • Capacitance and Dielectrics
Potential energy of the system of an electric dipole in an external electric ﬁeld
U
pE
(26.20)
O − − H + 105° H +
Figure 26.23 The water molecule, H2O, has a permanent polarization resulting from its nonlinear geometry. The center of the positive charge distribution is at the point .
+
− (a)
+
E + (b) −− +
Figure 26.24 (a) A linear symmetric molecule has no permanent polarization. (b) An external electric ﬁeld induces a polarization in the molecule.
To develop a conceptual understanding of Equation 26.19, compare this expression with the expression for the potential energy of the system of an object in the gravitational ﬁeld of the Earth, U mgh (see Chapter 8). The gravitational expression includes a parameter associated with the object we place in the ﬁeld—its mass m. Likewise, Equation 26.19 includes a parameter of the object in the electric ﬁeld—its dipole moment p. The gravitational expression includes the magnitude of the gravitational ﬁeld g. Similarly, Equation 26.19 includes the magnitude of the electric ﬁeld E. So far, these two contributions to the potential energy expressions appear analogous. However, the ﬁnal contribution is somewhat different in the two cases. In the gravitational expression, the potential energy depends on how high we lift the object, measured by h. In Equation 26.19, the potential energy depends on the angle through which we rotate the dipole. In both cases, we are making a change in the conﬁguration of the system. In the gravitational case, the change involves moving an object in a translational sense, whereas in the electrical case, the change involves moving an object in a rotational sense. In both cases, however, once the change is made, the system tends to return to the original conﬁguration when the object is released: the object of mass m falls back to the ground, and the dipole begins to rotate back toward the conﬁguration in which it is aligned with the ﬁeld. Thus, apart from the type of motion, the expressions for potential energy in these two cases are similar. Molecules are said to be polarized when a separation exists between the average position of the negative charges and the average position of the positive charges in the molecule. In some molecules, such as water, this condition is always present—such molecules are called polar molecules. Molecules that do not possess a permanent polarization are called nonpolar molecules. We can understand the permanent polarization of water by inspecting the geometry of the water molecule. In the water molecule, the oxygen atom is bonded to the hydrogen atoms such that an angle of 105° is formed between the two bonds (Fig. 26.23). The center of the negative charge distribution is near the oxygen atom, and the center of the positive charge distribution lies at a point midway along the line joining the hydrogen atoms (the point labeled in Fig. 26.23). We can model the water molecule and other polar molecules as dipoles because the average positions of the positive and negative charges act as point charges. As a result, we can apply our discussion of dipoles to the behavior of polar molecules. Microwave ovens take advantage of the polar nature of the water molecule. When in operation, microwave ovens generate a rapidly changing electric ﬁeld that causes the polar molecules to swing back and forth, absorbing energy from the ﬁeld in the process. Because the jostling molecules collide with each other, the energy they absorb from the ﬁeld is converted to internal energy, which corresponds to an increase in temperature of the food. Another household scenario in which the dipole structure of water is exploited is washing with soap and water. Grease and oil are made up of nonpolar molecules, which are generally not attracted to water. Plain water is not very useful for removing this type of grime. Soap contains long molecules called surfactants. In a long molecule, the polarity characteristics of one end of the molecule can be different from those at the other end. In a surfactant molecule, one end acts like a nonpolar molecule and the other acts like a polar molecule. The nonpolar end can attach to a grease or oil molecule, and the polar end can attach to a water molecule. Thus, the soap serves as a chain, linking the dirt and water molecules together. When the water is rinsed away, the grease and oil go with it. A symmetric molecule (Fig. 26.24a) has no permanent polarization, but polarization can be induced by placing the molecule in an electric ﬁeld. A ﬁeld directed to the left, as shown in Figure 26.24b, would cause the center of the positive charge distribution to shift to the left from its initial position and the center of the negative charge distribution to shift to the right. This induced polarization is the effect that predominates in most materials used as dielectrics in capacitors.
S ECTI O N 2 6.7 • An Atomic Description of Dielectrics
817
Example 26.8
The H2O Molecule we obtain W U90 pE 1.6 U0 (6.3 10
24
The water (H 2 O) molecule has an electric dipole moment of 6.3 10 30 C m. A sample contains 10 21 water molecules, with the dipole moments all oriented in the direction of an electric ﬁeld of magnitude 2.5 105 N/C. How much work is required to rotate the dipoles from this orientation ( 0°) to one in which all the moments are perpendicular to the ﬁeld ( 90°)? Solution The work required to rotate one molecule 90° is equal to the difference in potential energy between the 90° orientation and the 0° orientation. Using Equation 26.19,
( pE cos 90 ) 10 J
30
( pE cos 0 ) 105 N/C)
C m)(2.5
Because there are work required is Wtotal
1021
molecules in the sample, the total 10
24
(1021)(1.6
J)
1.6
10
3
J
26.7
An Atomic Description of Dielectrics
In Section 26.5 we found that the potential difference V0 between the plates of a capacitor is reduced to V0/ when a dielectric is introduced. The potential difference is reduced because the magnitude of the electric ﬁeld decreases between the plates. In particular, if E0 is the electric ﬁeld without the dielectric, the ﬁeld in the presence of a dielectric is E E0
(26.21)
Let us ﬁrst consider a dielectric made up of polar molecules placed in the electric ﬁeld between the plates of a capacitor. The dipoles (that is, the polar molecules making up the dielectric) are randomly oriented in the absence of an electric ﬁeld, as shown in Figure 26.25a. When an external ﬁeld E0 due to charges on the capacitor plates is applied, a torque is exerted on the dipoles, causing them to partially align with the ﬁeld, as shown in Figure 26.25b. We can now describe the dielectric as being polarized. The degree of alignment of the molecules with the electric ﬁeld depends on temperature and on the magnitude of the ﬁeld. In general, the alignment increases with decreasing temperature and with increasing electric ﬁeld. If the molecules of the dielectric are nonpolar, then the electric ﬁeld due to the plates produces some charge separation and an induced dipole moment. These induced dipole moments tend to align with the external ﬁeld, and the dielectric is polarized. Thus, we can polarize a dielectric with an external ﬁeld regardless of whether the molecules are polar or nonpolar.
+
+– – +
–
+
(a)
Figure 26.25 (a) Polar molecules are randomly oriented in the absence of an external electric ﬁeld. (b) When an external electric ﬁeld is applied, the molecules partially align with the ﬁeld. (c) The charged edges of the dielectric can be modeled as an additional pair of parallel plates establishing an electric ﬁeld E ind in the direction opposite to that of E 0.
–+
–
–+
–+
–+
–
–
–+
+
+
–+
+ + + + + +
–+ –+ –+
–+
–+ –+
–+
–+
–+ –+ –+
–+
E0
– – – – – –
+– + + + + + –σ ind (c) E ind E0
+
– – – – – –
(b)
σind
818
C H A P T E R 2 6 • Capacitance and Dielectrics
With these ideas in mind, consider a slab of dielectric material placed between the plates of a capacitor so that it is in a uniform electric ﬁeld E 0, as shown in Figure 26.25b. The electric ﬁeld due to the plates is directed to the right and polarizes the dielectric. The net effect on the dielectric is the formation of an induced positive sur face charge density ind on the right face and an equal-magnitude negative sur face charge density ind on the left face, as shown in Figure 26.25c. Because we can model these sur face charge distributions as being due to parallel plates, the induced sur face charges on the dielectric give rise to an induced electric ﬁeld Eind in the direction opposite the external ﬁeld E 0. Therefore, the net electric ﬁeld E in the dielectric has a magnitude E
σ + + + + + + + + + + + + +
– σ ind σ ind – σ σ + – – – + – – – + – – – + – – – + – – – + – – – + – –
E0
E ind
(26.22)
In the parallel-plate capacitor shown in Figure 26.26, the external ﬁeld E 0 is related to the charge density on the plates through the relationship E 0 / 0. The induced electric ﬁeld in the dielectric is related to the induced charge density ind through the relationship E ind E 0/ / 0, substitution into Equation ind/ 0. Because E 26.22 gives
ind 0 ind 0 0
1
(26.23)
Figure 26.26 Induced charge on a dielectric placed between the plates of a charged capacitor. Note that the induced charge density on the dielectric is less than the charge density on the plates.
Because 1, this expression shows that the charge density ind induced on the dielectric is less than the charge density on the plates. For instance, if 3 we see that the induced charge density is two-thirds the charge density on the plates. If no 1 and ind 0 as expected. However, if the dielectric is dielectric is present, then replaced by an electrical conductor, for which E 0, then Equation 26.22 indicates . That is, the sur face charge induced on that E 0 E ind; this corresponds to ind the conductor is equal in magnitude but opposite in sign to that on the plates, resulting in a net electric ﬁeld of zero in the conductor (see Fig. 24.16). We can use the existence of the induced sur face charge distributions on the dielectric to explain the result of Example 26.7. As we saw there, the energy of a capacitor not connected to a battery is lowered when a dielectric is inserted between the plates; this means that negative work is done on the dielectric by the external agent inserting the dielectric into the capacitor. This, in turn, implies that a force must be acting on the dielectric that draws it into the capacitor. This force originates from the nonuniform nature of the electric ﬁeld of the capacitor near its edges, as indicated in Figure 26.27.
+Q
– +
– +
– +
– +
– –– + ++
–Q
Figure 26.27 The nonuniform electric ﬁeld near the edges of a parallel-plate capacitor causes a dielectric to be pulled into the capacitor. Note that the ﬁeld acts on the induced sur face charges on the dielectric, which are nonuniformly distributed.
S ECTI O N 2 6.7 • An Atomic Description of Dielectrics
819
The horizontal component of this fringe ﬁeld acts on the induced charges on the sur face of the dielectric, producing a net horizontal force directed into the space between the capacitor plates.
Example 26.9
Effect of a Metallic Slab (B) Show that the capacitance of the original capacitor is unaffected by the insertion of the metallic slab if the slab is inﬁnitesimally thin. Solution In the result for part (A), we let a : 0: C
a:0
A parallel-plate capacitor has a plate separation d and plate area A. An uncharged metallic slab of thickness a is inserted midway between the plates. (A) Find the capacitance of the device. Solution We can solve this problem by noting that any charge that appears on one plate of the capacitor must induce a charge of equal magnitude and opposite sign on the near side of the slab, as shown in Figure 26.28a. Consequently, the net charge on the slab remains zero, and the electric ﬁeld inside the slab is zero. Hence, the capacitor is equivalent to two capacitors in series, each having a plate separation (d a)/2, as shown in Figure 26.28b. Using Eq. 26.3 and the rule for adding two capacitors in series (Eq. 26.10), we obtain 1 C 1 C1
0A
lim
0A
d
a
0A d
which is the original capacitance. What If? What if the met allic slab in part (A) is not midway between the plates? How does this affect the c apacit ance? Answer Let us imagine that the slab in Figure 26.27a is moved upward so that the distance between the upper edge of the slab and the upper plate is b. Then, the distance between the lower edge of the slab and the lower plate is d b a. As in part (A), we ﬁnd the total capacitance of the series combination: 1 C 1 C1 b
0A
1 C2
1 0A (d a)/2
1 0A (d a)/2
C
d
a
1 C2 d
1 ( 0 A/b) b
0A
1
0 A/(d
b
a)
a
d
a
0A
Note that C approaches inﬁnity as a approaches d. Why? C d
0A
a
(d – a)/2 + da – + – +++ (d – a)/2 ––– +++ (d – a)/2 ––– +σ – –σ +σ – –σ (d – a)/2
This is the same result as in part (A). It is independent of the value of b, so it does not matter where the slab is located. In Figure 26.28b, when the central structure is moved up or down, the decrease in plate separation of one capacitor is compensated by the increase in plate separation for the other.
(a)
(b)
Figure 26.28 (Example 26.9) (a) A parallel-plate capacitor of plate separation d partially ﬁlled with a metallic slab of thickness a. (b) The equivalent circuit of the device in part (a) consists of two capacitors in series, each having a plate separation (d a)/2.
Example 26.10
A Partially Filled Capacitor Solution In Example 26.9, we found that we could insert a metallic slab between the plates of a capacitor and consider the combination as two capacitors in series. The resulting capacitance was independent of the location of the slab. Furthermore, if the thickness of the slab approaches zero,
A parallel-plate capacitor with a plate separation d has a capacitance C 0 in the absence of a dielectric. What is the capacitance when a slab of dielectric material of dielectric constant and thickness 1 d is inserted between the plates 3 (Fig. 26.29a)?
820
C H A P T E R 2 6 • Capacitance and Dielectrics
1 –d 3 2 –d 3 d
κ
(a)
then the capacitance of the system approaches the capacitance when the slab is absent. From this, we conclude that we can insert an inﬁnitesimally thin metallic slab anywhere between the plates of a capacitor without affecting the capacitance. Thus, let us imagine sliding an inﬁnitesimally thin metallic slab along the bottom face of the dielectric shown in Figure 26.29a. We can then consider this system to be the series combination of the two capacitors shown in Figure 26.29b: one having a plate separation d/3 and ﬁlled with a dielectric, and the other having a plate separation 2d/3 and air between its plates. From Equations 26.15 and 26.3, the two capacitances are C1
0A
d/3
and
C2
0A
2d/3
1 –d 3
κ
C1
Using Equation 26.10 for two capacitors combined in series, we have 1 C 1 C C 1 C1 d 3 2
0A
1 C2 1 3 1 2
d/3 0A d 3
2d/3 0A 1 2
0A
2 –d 3
C2
0A
d
0
Because the capacitance without the dielectric is C 0 we see that
(b)
A/d,
Figure 26.29 (Example 26.10) (a) A parallel-plate capacitor of plate separation d partially ﬁlled with a dielectric of thickness d/3. (b) The equivalent circuit of the capacitor consists of two capacitors connected in series.
C
3 2 1
C0
S U M MARY
Take a practice test for this chapter by clicking on the Practice Test link at http://www.pse6.com.
A capacitor consists of two conductors carrying charges of equal magnitude and opposite sign. The capacitance C of any capacitor is the ratio of the charge Q on either conductor to the potential difference V between them: C Q ∆V
(26.1)
The capacitance depends only on the geometry of the conductors and not on an external source of charge or potential difference. The SI unit of capacitance is coulombs per volt, or the farad (F), and 1 F 1 C/V. Capacitance expressions for various geometries are summarized in Table 26.2. If two or more capacitors are connected in parallel, then the potential difference is the same across all of them. The equivalent capacitance of a parallel combination of capacitors is C eq C1 C2 C3
(26.8)
If two or more capacitors are connected in series, the charge is the same on all of them, and the equivalent capacitance of the series combination is given by
Questions
821
Table 26.2
Capacitance and Geometr y
Geometr y Isolated sphere of radius R (second spherical conductor assumed to have inﬁnite radius) Parallel-plate capacitor of plate area A and plate separation d Cylindrical capacitor of length and inner and outer radii a and b, respectively Spherical capacitor with inner and outer radii a and b, respectively Capacitance C 4
0R
Equation 26.2
C C C
0
A d
26.3 26.4 26.6
2k e ln(b/a) ab k e (b a)
1 C eq
1 C1
1 C2
1 C3
(26.10)
These two equations enable you to simplify many electric circuits by replacing multiple capacitors with a single equivalent capacitance. Energy is stored in a capacitor because the charging process is equivalent to the transfer of charges from one conductor at a lower electric potential to another conductor at a higher potential. The energy stored in a capacitor with charge Q is U Q2 2C
1 Q 2
∆V
1 C(∆V )2 2
(26.11)
When a dielectric material is inserted between the plates of a capacitor, the capacitance increases by a dimensionless factor , called the dielectric constant: C C0
(26.14)
where C 0 is the capacitance in the absence of the dielectric. The increase in capacitance is due to a decrease in the magnitude of the electric ﬁeld in the presence of the dielectric. The decrease in the magnitude of E arises from an internal electric ﬁeld produced by aligned dipoles in the dielectric. The electric dipole moment p of an electric dipole has a magnitude p 2aq
(26.16)
The direction of the electric dipole moment vector is from the negative charge toward the positive charge. The torque acting on an electric dipole in a uniform electric ﬁeld E is p E
(26.18)
The potential energy of the system of an electric dipole in a uniform external electric ﬁeld E is U pE
(26.20)
QU ESTIONS
1. The plates of a capacitor are connected to a battery. What happens to the charge on the plates if the connecting wires are removed from the battery? What happens to the charge if the wires are removed from the battery and connected to each other? 2. A farad is a very large unit of capacitance. Calculate the length of one side of a square, air-ﬁlled capacitor that has a capacitance of 1 F and a plate separation of 1 m. 3. A pair of capacitors are connected in parallel while an identical pair are connected in series. Which pair would be
822
C H A P T E R 2 6 • Capacitance and Dielectrics
more dangerous to handle after being connected to the same battery? Explain. 4. If you are given three different capacitors C 1, C 2, C 3, how many different combinations of capacitance can you produce? 5. What advantage might there be in using two identical capacitors in parallel connected in series with another identical parallel pair, rather than using a single capacitor? 6. Is it always possible to reduce a combination of capacitors to one equivalent capacitor with the rules we have developed? Explain. 7. The sum of the charges on both plates of a capacitor is zero. What does a capacitor store? 8. Because the charges on the plates of a parallel-plate capacitor are opposite in sign, they attract each other. Hence, it would take positive work to increase the plate separation. What type of energy in the system changes due to the external work done in this process? 9. Why is it dangerous to touch the terminals of a highvoltage capacitor even after the applied potential difference has been turned off? What can be done to make the capacitor safe to handle after the voltage source has been removed? 10. Explain why the work needed to move a charge Q through a potential difference V is W Q V whereas the energy stored in a charged capacitor is U 1 Q V. 2 Where does the 1 factor come from? 2 11. If the potential difference across a capacitor is doubled, by what factor does the energy stored change? 12. It is possible to obtain large potential differences by ﬁrst charging a group of capacitors connected in parallel and then activating a switch arrangement that in effect dis-
connects the capacitors from the charging source and from each other and reconnects them in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten capacitors each of 500 F and a charging source of 800 V? 13. Assume you want to increase the maximum operating voltage of a parallel-plate capacitor. Describe how you can do this for a ﬁxed plate separation. 14. An air-ﬁlled capacitor is charged, then disconnected from the power supply, and ﬁnally connected to a voltmeter. Explain how and why the potential difference changes when a dielectric is inserted between the plates of the capacitor. 15. Using the polar molecule description of a dielectric, explain how a dielectric affects the electric ﬁeld inside a capacitor. 16. Explain why a dielectric increases the maximum operating voltage of a capacitor although the physical size of the capacitor does not change. 17. What is the difference between dielectric strength and the dielectric constant? 18. Explain why a water molecule is permanently polarized. What type of molecule has no permanent polarization? 19. If a dielectric-filled capacitor is heated, how will its capacitance change? (Ignore thermal expansion and assume that the dipole orientations are temperaturedependent.) 20. If you were asked to design a capacitor where small size and large capacitance were required, what factors would be important in your design?
PROBLEMS
1, 2, 3 = straightforward, intermediate, challenging = paired numerical and symbolic problems = full solution available in the Student Solutions Manual and Study Guide = computer useful in solving problem = coached solution with hints available at http://www.pse6.com
Section 26.1
Deﬁnition of Capacitance
1. (a) How much charge is on each plate of a 4.00- F capacitor when it is connected to a 12.0-V battery? (b) If this same capacitor is connected to a 1.50-V battery, what charge is stored? 2. Two conductors having net charges of 10.0 C and 10.0 C have a potential difference of 10.0 V between them. (a) Determine the capacitance of the system. (b) What is the potential difference between the two conductors if the charges on each are increased to 100 C and 100 C?
21.0 cm from its center. (a) What is its sur face charge density? (b) What is its capacitance? 4. (a) If a drop of liquid has capacitance 1.00 pF, what is its radius? (b) If another drop has radius 2.00 mm, what is its capacitance? (c) What is the charge on the smaller drop if its potential is 100 V? 5. Two conducting spheres with diameters of 0.400 m and 1.00 m are separated by a distance that is large compared with the diameters. The spheres are connected by a thin wire and are charged to 7.00 C. (a) How is this total charge shared between the spheres? (Ignore any charge on the wire.) (b) What is the potential of the system of spheres when the reference potential is taken to be V 0 at r ? 6. Regarding the Earth and a cloud layer 800 m above the Earth as the “plates’’ of a capacitor, calculate the
Section 26.2
Calculating Capacitance
3. An isolated charged conducting sphere of radius 12.0 cm creates an electric ﬁeld of 4.90 10 4 N/C at a distance
Problems
823
capacitance. Assume the cloud layer has an area of 1.00 km2 and that the air between the cloud and the ground is pure and dr y. Assume charge builds up on the cloud and on the ground until a uniform electric ﬁeld of 3.00 106 N/C throughout the space between them makes the air break down and conduct electricity as a lightning bolt. What is the maximum charge the cloud can hold? 7. An air-ﬁlled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.80 mm. A 20.0-V potential difference is applied to these plates. Calculate (a) the electric ﬁeld between the plates, (b) the sur face charge density, (c) the capacitance, and (d) the charge on each plate.
other. The region between the spheres is a vacuum. Determine the volume of this region. 13. An air-ﬁlled spherical capacitor is constructed with inner and outer shell radii of 7.00 and 14.0 cm, respectively. (a) Calculate the capacitance of the device. (b) What potential difference between the spheres results in a charge of 4.00 C on the capacitor? 14. A small object of mass m carries a charge q and is suspended by a thread between the vertical plates of a parallel-plate capacitor. The plate separation is d. If the thread makes an angle with the vertical, what is the potential difference between the plates? 15. Find the capacitance of the Earth. (Suggestion: The outer conductor of the “spherical capacitor’’ may be considered as a conducting sphere at inﬁnity where V approaches zero.) Section 26.3 Combinations of Capacitors
8. A 1-megabit computer memory chip contains many 60.0-f F capacitors. Each capacitor has a plate area of 21.0 10 12 m2. Determine the plate separation of such a capacitor (assume a parallel-plate conﬁguration). The order of magnitude of the diameter of an atom is 10 10 m 0.1 nm. Express the plate separation in nanometers. 9. When a potential difference of 150 V is applied to the plates of a parallel-plate capacitor, the plates carry a sur face charge density of 30.0 nC/cm2. What is the spacing between the plates? 10. A variable air capacitor used in a radio tuning circuit is made of N semicircular plates each of radius R and positioned a distance d from its neighbors, to which it is electrically connected. As shown in Figure P26.10, a second identical set of plates is enmeshed with its plates halfway between those of the ﬁrst set. The second set can rotate as a unit. Determine the capacitance as a function of the angle of rotation , where 0 corresponds to the maximum capacitance.
16. Two capacitors, C 1 5.00 F and C 2 12.0 F, are connected in parallel, and the resulting combination is connected to a 9.00-V battery. (a) What is the equivalent capacitance of the combination? What are (b) the potential difference across each capacitor and (c) the charge stored on each capacitor? 17. What If ? The two capacitors of Problem 16 are now connected in series and to a 9.00-V battery. Find (a) the equivalent capacitance of the combination, (b) the potential difference across each capacitor, and (c) the charge on each capacitor. 18. Evaluate the equivalent capacitance of the conﬁguration shown in Figure P26.18. All the capacitors are identical, and each has capacitance C.
C C
d
C C C
C
θ
R
Figure P26.18
Figure P26.10
19. Two capacitors when connected in parallel give an equivalent capacitance of 9.00 pF and give an equivalent capacitance of 2.00 pF when connected in series. What is the capacitance of each capacitor ? 20. Two capacitors when connected in parallel give an equivalent capacitance of C p and an equivalent capacitance of C s when connected in series. What is the capacitance of each capacitor? 21. Four capacitors are connected as shown in Figure P26.21. (a) Find the equivalent capacitance between points a and b. (b) Calculate the charge on each capacitor if Vab 15.0 V.
11.
A 50.0-m length of coaxial cable has an inner conductor that has a diameter of 2.58 mm and carries a charge of 8.10 C. The surrounding conductor has an inner diameter of 7.27 mm and a charge of 8.10 C. (a) What is the capacitance of this cable? (b) What is the potential difference between the two conductors? Assume the region between the conductors is air.
12. A 20.0- F spherical capacitor is composed of two concentric metal spheres, one having a radius twice as large as the
824
C H A P T E R 2 6 • Capacitance and Dielectrics
15.0 µF 3.00 µF 20.0 µF a b
line with capacitance 29.8 F between A and B. What additional capacitor should be installed in series or in parallel in that circuit, to meet the speciﬁcation? 25. A group of identical capacitors is connected ﬁrst in series and then in parallel. The combined capacitance in parallel is 100 times larger than for the series connection. How many capacitors are in the group? 26. Consider three capacitors C 1, C 2 , C 3 , and a battery. If C 1 is connected to the battery, the charge on C 1 is 30.8 C. Now C 1 is disconnected, discharged, and connected in series with C 2. When the series combination of C 2 and C 1 is connected across the battery, the charge on C 1 is 23.1 C. The circuit is disconnected and the capacitors discharged. Capacitor C 3 , capacitor C 1, and the battery are connected in series, resulting in a charge on C 1 of 25.2 C. If, after being disconnected and discharged, C 1, C 2 , and C 3 are connected in series with one another and with the battery, what is the charge on C 1? 27. Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in Figure P26.27. Take C 1 5.00 F, C 2 10.0 F, and C 3 2.00 F.
6.00 µ F Figure P26.21
22. Three capacitors are connected to a battery as shown in Figure P26.22. Their capacitances are C 1 3C , C 2 C , and C 3 5C . (a) What is the equivalent capacitance of this set of capacitors? (b) State the ranking of the capacitors according to the charge they store, from largest to smallest. (c) Rank the capacitors according to the potential differences across them, from largest to smallest. (d) What If ? If C 3 is increased, what happens to the charge stored by each of the capacitors?
C1
C1
a
C1
C2
C3
C2 C3 C2
Figure P26.22
C2 b C2
23. Consider the circuit shown in Figure P26.23, where C1 6.00 F, C 2 3.00 F, and V 20.0 V. Capacitor C 1 is ﬁrst charged by the closing of switch S 1. Switch S 1 is then opened, and the charged capacitor is connected to the uncharged capacitor by the closing of S 2. Calculate the initial charge acquired by C 1 and the ﬁnal charge on each capacitor.
Figure P26.27 Problems 27 and 28.
28. For the network described in the previous problem, if the potential difference between points a and b is 60.0 V, what charge is stored on C 3? 29. Find the equivalent capacitance between points a and b in the combination of capacitors shown in Figure P26.29.
∆V
C1
C2
S1
S2
a
4.0 µF 7.0 µF 5.0 µF 6.0 µ F b
Figure P26.23
24. According to its design speciﬁcation, the timer circuit delaying the closing of an elevator door is to have a capacitance of 32.0 F between two points A and B. (a) When one circuit is being constructed, the inexpensive but durable capacitor installed between these two points is found to have capacitance 34.8 F. To meet the speciﬁcation, one additional capacitor can be placed between the two points. Should it be in series or in parallel with the 34.8- F capacitor? What should be its capacitance? (b) What If ? The next circuit comes down the assembly
Figure P26.29
30. Some physical systems possessing capacitance continuously distributed over space can be modeled as an inﬁnite array of discrete circuit elements. Examples are a microwave waveguide and the axon of a nerve cell. To practice analy-
Problems
825
sis of an inﬁnite array, determine the equivalent capacitance C between terminals X and Y of the inﬁnite set of capacitors represented in Figure P26.30. Each capacitor has capacitance C 0. Suggestion: Imagine that the ladder is cut at the line AB, and note that the equivalent capacitance of the inﬁnite section to the right of AB is also C.
A C0 X C0 Y C0 B Figure P26.30 C0
person. A particular device can be destroyed by a discharge releasing an energy of 250 J. To what voltage on the body does this correspond? 36. A uniform electric ﬁeld E 3 000 V/m exists within a certain region. What volume of space contains an energy equal to 1.00 10 7 J? Express your answer in cubic meters and in liters. 37. A parallel-plate capacitor has a charge Q and plates of area A. What force acts on one plate to attract it toward the other plate? Because the electric ﬁeld between the plates is E Q /A 0, you might think that the force is F Q E Q 2/A 0. This is wrong, because the ﬁeld E includes contributions from both plates, and the ﬁeld created by the positive plate cannot exert any force on the positive plate. Show that the force exerted on each plate is actually F Q 2/2 0 A. (Suggestion: Let C 0 A/x for an arbitrary plate separation x ; then require that the work done in separating the two charged plates be W F dx.) The force exerted by one charged plate on another is sometimes used in a machine shop to hold a workpiece stationary.
Section 26.4
Energy Stored in a Charged Capacitor
31. (a) A 3.00- F capacitor is connected to a 12.0-V battery. How much energy is stored in the capacitor? (b) If the capacitor had been connected to a 6.00-V battery, how much energy would have been stored? 32. The immediate cause of many deaths is ventricular ﬁbrillation, uncoordinated quivering of the heart as opposed to proper beating. An electric shock to the chest can cause momentary paralysis of the heart muscle, after which the heart will sometimes start organized beating again. A deﬁbrillator (Fig. 26.14) is a device that applies a strong electric shock to the chest over a time interval of a few milliseconds. The device contains a capacitor of several microfarads, charged to several thousand volts. Electrodes called paddles, about 8 cm across and coated with conducting paste, are held against the chest on both sides of the heart. Their handles are insulated to prevent injury to the operator, who calls, “Clear!’’ and pushes a button on one paddle to discharge the capacitor through the patient’s chest. Assume that an energy of 300 J is to be delivered from a 30.0- F capacitor. To what potential difference must it be charged? 33. Two capacitors, C 1 25.0 F and C 2 5.00 F, are connected in parallel and charged with a 100-V power supply. (a) Draw a circuit diagram and calculate the total energy stored in the two capacitors. (b) What If? What potential difference would be required across the same two capacitors connected in series in order that the combination stores the same amount of energy as in (a)? Draw a circuit diagram of this circuit. 34. A parallel-plate capacitor is charged and then disconnected from a battery. By what fraction does the stored energy change (increase or decrease) when the plate separation is doubled? 35. As a person moves about in a dry environment, electric charge accumulates on his body. Once it is at high voltage, either positive or negative, the body can discharge via sometimes noticeable sparks and shocks. Consider a human body well separated from ground, with the typical capacitance 150 pF. (a) What charge on the body will produce a potential of 10.0 kV? (b) Sensitive electronic devices can be destroyed by electrostatic discharge from a
38. The circuit in Figure P26.38 consists of two identical parallel metal plates connected by identical metal springs to a 100-V battery. With the switch open, the plates are uncharged, are separated by a distance d 8.00 mm, and have a capacitance C 2.00 F. When the switch is closed, the distance between the plates decreases by a factor of 0.500. (a) How much charge collects on each plate and (b) what is the spring constant for each spring? (Suggestion : Use the result of Problem 37.)
d k k
S
+ ∆V
–
Figure P26.38
39. Review problem. A certain storm cloud has a potential of 1.00 108 V relative to a tree. If, during a lightning storm, 50.0 C of charge is transferred through this potential difference and 1.00% of the energy is absorbed by the tree, how much sap in the tree can be boiled away? Model the sap as water initially at 30.0°C. Water has a speciﬁc heat of 4 186 J/kg°C, a boiling point of 100°C, and a latent heat of vaporization of 2.26 106 J/kg. 40. Two identical parallel-plate capacitors, each with capacitance C , are charged to potential difference V and connected in parallel. Then the plate separation in one of the capacitors is doubled. (a) Find the total energy of the system of two capacitors before the plate separation is doubled. (b) Find the potential difference across each capacitor after the plate separation is doubled. (c) Find the
826
C H A P T E R 2 6 • Capacitance and Dielectrics
total energy of the system after the plate separation is doubled. (d) Reconcile the difference in the answers to parts (a) and (c) with the law of conservation of energy. 41. Show that the energy associated with a conducting sphere of radius R and charge Q surrounded by a vacuum is U k e Q 2/2R . 42. Consider two conducting spheres with radii R 1 and R 2. They are separated by a distance much greater than either radius. A total charge Q is shared between the spheres, subject to the condition that the electric potential energy of the system has the smallest possible value. The total charge Q is equal to q 1 q 2 , where q 1 represents the charge on the ﬁrst sphere and q 2 the charge on the second. Because the spheres are very far apart, you can assume that the charge of each is uniformly distributed over its sur face. You may use the result of Problem 41. (a) Determine the values of q 1 and q 2 in terms of Q , R 1, and R 2. (b) Show that the potential difference between the spheres is zero. (We saw in Chapter 25 that two conductors joined by a conducting wire will be at the same potential in a static situation. This problem illustrates the general principle that static charge on a conductor will distribute itself so that the electric potential energy of the system is a minimum.) Section 26.5 Capacitors with Dielectrics
48. A wafer of titanium dioxide ( 173) of area 1.00 cm2 has a thickness of 0.100 mm. Aluminum is evaporated on the parallel faces to form a parallel-plate capacitor. (a) Calculate the capacitance. (b) When the capacitor is charged with a 12.0-V battery, what is the magnitude of charge delivered to each plate? (c) For the situation in part (b), what are the free and induced sur face charge densities? (d) What is the magnitude of the electric ﬁeld? 49. Each capacitor in the combination shown in Figure P26.49 has a breakdown voltage of 15.0 V. What is the breakdown voltage of the combination?
20.0 µF 10.0 µ F 20.0 µF 20.0 µF 20.0 µF
Figure P26.49
Section 26.6
Electric Dipole in an Electric Field
43. Determine (a) the capacitance and (b) the maximum potential difference that can be applied to a Teﬂon-ﬁlled parallel-plate capacitor having a plate area of 1.75 cm2 and plate separation of 0.040 0 mm. 44. (a) How much charge can be placed on a capacitor with air between the plates before it breaks down, if the area of each of the plates is 5.00 cm2 ? (b) What If ? Find the maximum charge if polystyrene is used between the plates instead of air. 45. A commercial capacitor is to be constructed as shown in Figure 26.17a. This particular capacitor is made from two strips of aluminum separated by a strip of parafﬁn-coated paper. Each strip of foil and paper is 7.00 cm wide. The foil is 0.004 00 mm thick, and the paper is 0.025 0 mm thick and has a dielectric constant of 3.70. What length should the strips have, if a capacitance of 9.50 10 8 F is desired before the capacitor is rolled up? (Adding a second strip of paper and rolling the capacitor effectively doubles its capacitance, by allowing charge storage on both sides of each strip of foil.) 46. The supermarket sells rolls of aluminum foil, of plastic wrap, and of waxed paper. Describe a capacitor made from supermarket materials. Compute order-of-magnitude estimates for its capacitance and its breakdown voltage. 47. A parallel-plate capacitor in air has a plate separation of 1.50 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 250 V and disconnected from the source. The capacitor is then immersed in distilled water. Determine (a) the charge on the plates before and after immersion, (b) the capacitance and potential difference after immersion, and (c) the change in energy of the capacitor. Assume the liquid is an insulator.
50. A small rigid object carries positive and negative 3.50-nC charges. It is oriented so that the positive charge has coordinates ( 1.20 mm, 1.10 mm) and the negative charge is at the point (1.40 mm, 1.30 mm). (a) Find the electric dipole moment of the object. The object is placed in an electric ﬁeld E (7 800ˆ 4 900ˆ) N/C. (b) Find the i j torque acting on the object. (c) Find the potential energy of the object–ﬁeld system when the object is in this orientation. (d) If the orientation of the object can change, ﬁnd the difference between the maximum and minimum potential energies of the system. 51. A small object with electric dipole moment p is placed in a nonuniform electric ﬁeld E E(x)ˆ. That is, the ﬁeld is in i the x direction and its magnitude depends on the coordinate x. Let represent the angle between the dipole moment and the x direction. (a) Prove that the dipole feels a net force F p dE dx cos
in the direction toward which the ﬁeld increases. (b) Consider a spherical balloon centered at the origin, with radius 15.0 cm and carrying charge 2.00 C. Evaluate dE/dx at the point (16 cm, 0, 0). Assume a water droplet at this point has an induced dipole moment of 6.30ˆ nC m. i Find the force on it. Section 26.7 An Atomic Description of Dielectrics 52. A detector of radiation called a Geiger tube consists of a closed, hollow, conducting cylinder with a ﬁne wire along its axis. Suppose that the internal diameter of the cylinder is 2.50 cm and that the wire along the axis has a diameter of 0.200 mm. The dielectric strength of the gas between the central wire and the cylinder is 1.20 106 V/m. Calculate the maximum potential difference that can be applied between the wire and the cylinder before breakdown occurs in the gas.
Problems
827
53. The general form of Gauss’s law describes how a charge creates an electric ﬁeld in a material, as well as in vacuum. It is E dA q
capacitance of the three-plate system P1P2P3? (b) What is the charge on P2? (c) If P4 is now connected to the positive terminal of the battery, what is the capacitance of the fourplate system P1P2P3P4? (d) What is the charge on P4? 56. One conductor of an overhead electric transmission line is a long aluminum wire 2.40 cm in radius. Suppose that at a particular moment it carries charge per length 1.40 C/m and is at potential 345 kV. Find the potential 12.0 m below the wire. Ignore the other conductors of the transmission line and assume the electric ﬁeld is everywhere purely radial. 57. Two large parallel metal plates are oriented horizontally and separated by a distance 3d. A grounded conducting wire joins them, and initially each plate carries no charge. Now a third identical plate carrying charge Q is inserted between the two plates, parallel to them and located a distance d from the upper plate, as in Figure P26.57. (a) What induced charge appears on each of the two original plates? (b) What potential difference appears between the middle plate and each of the other plates? Each plate has area A.
where 0 is the permittivity of the material. (a) A sheet with charge Q uniformly distributed over its area A is surrounded by a dielectric. Show that the sheet creates a uniform electric ﬁeld at nearby points, with magnitude E Q /2 A . (b) Two large sheets of area A, carrying opposite charges of equal magnitude Q , are a small distance d apart. Show that they create uniform electric ﬁeld in the space between them, with magnitude E Q /A . (c) Assume that the negative plate is at zero potential. Show that the positive plate is at potential Q d/A . (d) Show that the capacitance of the pair of plates is A /d A 0/d. Additional Problems 54. For the system of capacitors shown in Figure P26.54, ﬁnd (a) the equivalent capacitance of the system, (b) the potential across each capacitor, (c) the charge on each capacitor, and (d) the total energy stored by the group.
3.00 µ F 6.00 µF
d
2d
2.00 µF 4.00 µ F
Figure P26.57
90.0 V Figure P26.54
58. A 2.00-nF parallel-plate capacitor is charged to an initial potential difference Vi 100 V and then isolated. The dielectric material between the plates is mica, with a dielectric constant of 5.00. (a) How much work is required to withdraw the mica sheet? (b) What is the potential difference of the capacitor after the mica is withdrawn? 59. A parallel-plate capacitor is constructed using a dielectric material whose dielectric constant is 3.00 and whose dielectric strength is 2.00 108 V/m. The desired capacitance is 0.250 F, and the capacitor must withstand a maximum potential difference of 4 000 V. Find the minimum area of the capacitor plates.
55. Four parallel metal plates P1, P2, P3, and P4, each of area 7.50 cm2, are separated successively by a distance d 1.19 mm, as shown in Figure P26.55. P1 is connected to the negative terminal of a battery, and P2 to the positive terminal. The battery maintains a potential difference of 12.0 V. (a) If P3 is connected to the negative terminal, what is the
P1
P2
P3
P4
60. A 10.0- F capacitor has plates with vacuum between them. Each plate carries a charge of magnitude 1 000 C. A particle with charge 3.00 C and mass 2.00 10 16 kg is ﬁred from the positive plate toward the negative plate with an initial speed of 2.00 106 m/s. Does it reach the negative plate? If so, ﬁnd its impact speed. If not, what fraction of the way across the capacitor does it travel? 61. A parallel-plate capacitor is constructed by ﬁlling the space between two square plates with blocks of three dielectric materials, as in Figure P26.61. You may assume that d. (a) Find an expression for the capacitance of the device in terms of the plate area A and d, 1, 2, and 3. (b) Calculate the capacitance using the values A 1.00 cm2, d 2.00 mm, 1 4.90, 2 5.60, and 2.10. 3
12.0 V
d
d
d
Figure P26.55
828
C H A P T E R 2 6 • Capacitance and Dielectrics
d
κ1
κ2 κ3
d/2
sent the potential difference. (c) Find the direction and magnitude of the force exerted on the dielectric, assuming a constant potential difference V. Ignore friction. (d) Obtain a numerical value for the force assuming that 5.00 cm, V 2 000 V, d 2.00 mm, and the dielectric is glass ( 4.50). (Suggestion: The system can be considered as two capacitors connected in parallel.) 65. A capacitor is constructed from two square plates of sides and separation d, as suggested in Figure P26.64. You may assume that d is much less than . The plates carry charges Q 0 and Q 0. A block of metal has a width , a length , and a thickness slightly less than d. It is inserted a distance x into the capacitor. The charges on the plates are not disturbed as the block slides in. In a static situation, a metal prevents an electric ﬁeld from penetrating inside it. The metal can be thought of as a per fect dielectric, with : . (a) Calculate the stored energy as a function of x. (b) Find the direction and magnitude of the force that acts on the metallic block. (c) The area of the advancing front face of the block is essentially equal to d. Considering the force on the block as acting on this face, ﬁnd the stress (force per area) on it. (d) For comparison, express the energy density in the electric ﬁeld between the capacitor plates in terms of Q 0 , , d , and 0. 66. When considering the energy supply for an automobile, the energy per unit mass of the energy source is an important parameter. Using the following data, compare the energy per unit mass ( J/kg) for gasoline, lead–acid batteries, and capacitors. (The ampere A will be introduced in the next chapter as the SI unit of electric current. 1 A 1 C/s.) Gasoline: 126 000 Btu/gal; density 670 kg/m3. Lead–acid batter y: 12.0 V; 100 A h; mass 16.0 kg. Capacitor: potential difference at full charge 12.0 V; capacitance 0.100 F; mass 0.100 kg. 67. An isolated capacitor of unknown capacitance has been charged to a potential difference of 100 V. When the charged capacitor is then connected in parallel to an uncharged 10.0- F capacitor, the potential difference across the combination is 30.0 V. Calculate the unknown capacitance. 68. To repair a power supply for a stereo ampliﬁer, an electronics technician needs a 100- F capacitor capable of withstanding a potential difference of 90 V between the plates. The only available supply is a box of ﬁve 100- F capacitors, each having a maximum voltage capability of 50 V. Can the technician substitute a combination of these capacitors that has the proper electrical characteristics? If so, what will be the maximum voltage across any of the capacitors used? (Suggestion: The technician may not have to use all the capacitors in the box.) 69. A parallel-plate capacitor of plate separation d is charged to a potential difference V0. A dielectric slab of thickness d and dielectric constant is introduced between the plates while the battery remains connected to the plates. (a) Show that the ratio of energy stored after the dielectric is introduced to the energy stored in the empty capacitor is U/ U 0 . Give a physical explanation for this increase in stored energy. (b) What happens to the charge on the capacitor? (Note that this situation is not the same as in
/2 Figure P26.61
62. A 10.0- F capacitor is charged to 15.0 V. It is next connected in series with an uncharged 5.00- F capacitor. The series combination is ﬁnally connected across a 50.0-V battery, as diagrammed in Figure P26.62. Find the new potential differences across the 5- F and 10- F capacitors.
5.00 µ F
10.0 µ F +– ∆Vi = 15.0 V
50.0 V
Figure P26.62
63. (a) Two spheres have radii a and b and their centers are a distance d apart. Show that the capacitance of this system is C 4 1 a 1 b
0
2 d
provided that d is large compared with a and b. (Suggestion: Because the spheres are far apart, assume that the potential of each equals the sum of the potentials due to each sphere, and when calculating those potentials assume that V keQ /r applies.) (b) Show that as d approaches inﬁnity the above result reduces to that of two spherical capacitors in series. 64. A capacitor is constructed from two square plates of sides and separation d. A material of dielectric constant is inserted a distance x into the capacitor, as shown in Figure P26.64. Assume that d is much smaller than x. (a) Find the equivalent capacitance of the device. (b) Calculate the energy stored in the capacitor, letting V repre-
κ
x
d
Figure P26.64 Problems 64 and 65.
Answers to Quick Quizzes
829
Example 26.7, in which the battery was removed from the circuit before the dielectric was introduced.) 70. A vertical parallel-plate capacitor is half ﬁlled with a dielectric for which the dielectric constant is 2.00 (Fig. P26.70a). When this capacitor is positioned horizontally, what fraction of it should be ﬁlled with the same dielectric (Fig. P26.70b) in order for the two capacitors to have equal capacitance?
75. Determine the equivalent capacitance of the combination shown in Figure P26.75. (Suggestion: Consider the symmetry involved.)
C
2C
3C
C
2C
Figure P26.75
(a)
(b)
Figure P26.70
76. Consider two long, parallel, and oppositely charged wires of radius d with their centers separated by a distance D. Assuming the charge is distributed uniformly on the sur face of each wire, show that the capacitance per unit length of this pair of wires is C ln[(D
0
71. Capacitors C 1 6.00 F and C 2 2.00 F are charged as a parallel combination across a 250-V battery. The capacitors are disconnected from the battery and from each other. They are then connected positive plate to negative plate and negative plate to positive plate. Calculate the resulting charge on each capacitor. 72. Calculate the equivalent capacitance between the points a and b in Figure P26.72. Note that this is not a simple series or parallel combination. (Suggestion: Assume a potential difference V between points a and b. Write expressions for V ab in terms of the charges and capacitances for the various possible pathways from a to b, and require conservation of charge for those capacitor plates that are connected to each other.)
d)/d]
77. Example 26.2 explored a cylindrical capacitor of length and radii a and b of the two conductors. In the What If ? section, it was claimed that increasing by 10% is more effective in terms of increasing the capacitance than increasing a by 10% if b 2.85a. Verify this claim mathematically. Answers to Quick Quizzes 26.1 (d). The capacitance is a property of the physical system and does not vary with applied voltage. According to Equation 26.1, if the voltage is doubled, the charge is doubled. 26.2 (a). When the key is pressed, the plate separation is decreased and the capacitance increases. Capacitance depends only on how a capacitor is constructed and not on the external circuit.
a
4.00 µF
2.00 µ F
8.00 µF 2.00 µF
4.00 µF
26.3 (a). When connecting capacitors in series, the inverses of the capacitances add, resulting in a smaller overall equivalent capacitance. 26.4 (a). When capacitors are connected in series, the voltages add, for a total of 20 V in this case. If they are combined in parallel, the voltage across the combination is still 10 V. 26.5 (b). For a given voltage, the energy stored in a capacitor is proportional to C : U C( V )2/2. Thus, you want to maximize the equivalent capacitance. You do this by connecting the three capacitors in parallel, so that the capacitances add. 26.6 (a) C decreases (Eq. 26.3). (b) Q stays the same because there is no place for the charge to ﬂow. (c) E remains constant (see Eq. 24.8 and the paragraph following it). (d) V increases because V Q /C , Q is constant (part b), and C decreases (part a). (e) The energy stored in the capacitor is proportional to both Q and V (Eq.
b
Figure P26.72
73. The inner conductor of a coaxial cable has a radius of 0.800 mm, and the outer conductor’s inside radius is 3.00 mm. The space between the conductors is ﬁlled with polyethylene, which has a dielectric constant of 2.30 and a dielectric strength of 18.0 106 V/m. What is the maximum potential difference that this cable can withstand? 74. You are optimizing coaxial cable design for a major manufacturer. Show that for a given outer conductor radius b, maximum potential difference capability is attained when the radius of the inner conductor is a b/e where e is the base of natural logarithms.
830
C H A P T E R 2 6 • Capacitance and Dielectrics
26.11) and thus increases. The additional energy comes from the work you do in pulling the two plates apart. 26.7 (a) C decreases (Eq. 26.3). (b) Q decreases. The battery supplies a constant potential difference V ; thus, charge must ﬂow out of the capacitor if C Q / V is to decrease. (c) E decreases because the charge density on the plates decreases. (d) V remains constant because of the presence of the battery. (e) The energy stored in the capacitor decreases (Eq. 26.11). 26.8 Increase. The dielectric constant of wood (and of all other insulating materials, for that matter) is greater than 1; therefore, the capacitance increases (Eq. 26.14). This
increase is sensed by the stud-ﬁnder’s special circuitry, which causes an indicator on the device to light up. 26.9 (a) C increases (Eq. 26.14). (b) Q increases. Because the battery maintains a constant V, Q must increase if C increases. (c) E between the plates remains constant because V Ed and neither V nor d changes. The electric ﬁeld due to the charges on the plates increases because more charge has ﬂowed onto the plates. The induced sur face charges on the dielectric create a ﬁeld that opposes the increase in the ﬁeld caused by the greater number of charges on the plates (see Section 26.7). (d) The battery maintains a constant V.

**Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.**

Below is a small sample set of documents:

Nanjing University - PHYS - 102

Chapter 27Current and ResistanceCHAPTE R OUTLI N E27.1 Electric Current 27.2 Resistance 27.3 A Model for Electrical Conduction 27.4 Resistance and Temperature 27.5 Superconductors 27.6 Electrical PowerL These power lines transfer energy from the power

Nanjing University - PHYS - 102

858Chapter 28C HAPTE R 2 8 Direct Current CircuitsDirect Current CircuitsCHAPTE R OUTLI N E28.1 Electromotive Force 28.2 Resistors in Series and Parallel 28.3 Kirchhoffs Rules 28.4 RC Circuits 28.5 Electrical Meters 28.6 Household Wiring and Electric

Nanjing University - PHYS - 102

Chapter 29Magnetic FieldsCHAPTE R OUTLI N E29.1 Magnetic Fields and Forces 29.2 Magnetic Force Acting on a Current-Carrying Conductor 29.3 Torque on a Current Loop in a Uniform Magnetic Field 29.4 Motion of a Charged Particle in a Uniform Magnetic Fiel

Nanjing University - PHYS - 102

Chapter 30Sources of the Magnetic FieldCHAPTE R OUTLI N E30.1 The BiotSavart Law 30.2 The Magnetic Force Between Two Parallel Conductors 30.3 Ampres Law 30.4 The Magnetic Field of a Solenoid 30.5 Magnetic Flux 30.6 Gausss Law in Magnetism 30.7 Displace

Nanjing University - PHYS - 102

Chapter 31Faradays LawCHAPTE R OUTLI N E31.1 Faradays Law of Induction 31.2 Motional emf 31.3 Lenzs Law 31.4 Induced emf and Electric Fields 31.5 Generators and Motors 31.6 Eddy Currents 31.7 Maxwells EquationsL In a commercial electric power plant, l

Nanjing University - PHYS - 102

Chapter 32InductanceCHAPTE R OUTLI N E32.1 Self-Inductance 32.2 RL Circuits 32.3 Energy in a Magnetic Field 32.4 Mutual Inductance 32.5 Oscillations in an LC Circuit 32.6 The RLC CircuitL An airport metal detector contains a large coil of wire around

Nanjing University - PHYS - 102

Chapter 33Alternating Current CircuitsCHAPTE R OUTLI N E33.1 AC Sources 33.2 Resistors in an AC Circuit 33.3 Inductors in an AC Circuit 33.4 Capacitors in an AC Circuit 33.5 The RLC Series Circuit 33.6 Power in an AC Circuit 33.7 Resonance in a Series

Nanjing University - PHYS - 102

Chapter 34Electromagnetic WavesCHAPTE R OUTLI N E34.1 Maxwells Equations and Hertzs Discoveries 34.2 Plane Electromagnetic Waves 34.3 Energy Carried by Electromagnetic Waves 34.4 Momentum and Radiation Pressure 34.5 Production of Electromagnetic Waves

Nanjing University - PHYS - 102

Light and Opticsight is basic to almost all life on the Earth. Plants convert the energy transferred by sunlight to chemical energy through photosynthesis. In addition, light is the principal means by which we are able to transmit and receive information

Nanjing University - PHYS - 102

Chapter 36Image FormationCHAPTE R OUTLI N E36.1 Images Formed by Flat Mirrors 36.2 Images Formed by Spherical Mirrors 36.3 Images Formed by Refraction 36.4 Thin Lenses 36.5 Lens Aberrations 36.6 The Camera 36.7 The Eye 36.8 The Simple Magnier 36.9 The

Nanjing University - PHYS - 102

Chapter 37Interference of Light WavesCHAPTE R OUTLI N E37.1 Conditions for Interference 37.2 Youngs Double-Slit Experiment 37.3 Intensity Distribution of the Double-Slit Interference Pattern 37.4 Phasor Addition of Waves 37.5 Change of Phase Due to Ree

Nanjing University - PHYS - 102

Chapter 38Diffraction Patterns and PolarizationCHAPTE R OUTLI N E38.1 Introduction to Diffraction Patterns 38.2 Diffraction Patterns from Narrow Slits 38.3 Resolution of Single-Slit and Circular Apertures 38.4 The Diffraction Grating 38.5 Diffraction o

Nanjing University - PHYS - 102

Modern Physicst the end of the nineteenth century, many scientists believed that they had learned most of what there was to know about physics. Newtons laws of motion and his theory of universal gravitation, Maxwells theoretical work in unifying electric

Nanjing University - PHYS - 102

Answers to Odd-Numbered ProblemsCHAPTE R 11. 0.141 nm 3. 2.15 5. 4 104 kg/m3 r 13)/3 10 24 g (b) 55.9 u 10 22 g 1026 atoms 9.28 10235. (a) 3.75 m/s 7. (a) 11. 1.34 g(b) 0 3.8 m/s (c) 4.0 s 2.5 m/s (c) 0 (d) 5.0 m/s2.4 m/s (b) 104 m/s2 (b)9. (a) 5.0

Nanjing University - PHYS - 102

Appendix A TablesTable A.1Conversion FactorsLength m 1 meter 1 centimeter 1 kilometer 1 inch 1 foot 1 mile Mass kg 1 kilogram 1 gram 1 slug 1 atomic mass unitNote : 1 metric toncm 102 1 105 2.540 30.48 1.609km 10 3 10 5 1 2.540 3.048 1.609in . 39.3

Nanjing University - PHYS - 102

Appendix B Mathematics ReviewThese appendices in mathematics are intended as a brief review of operations and methods. Early in this course, you should be totally familiar with basic algebraic techniques, analytic geometry, and trigonometry. The appendic

Nanjing University - PHYS - 102

Appendix C Periodic Table of the ElementsGroup IGroup IITransition elementsH1.007 91s1Li6.9412s 13Be9.01222s 24SymbolCa40.0784s 220Atomic number Electron configurationNa22.9903s 111Mg24.3053s 212Atomic mass K39.0984s 119

Nanjing University - PHYS - 102

Appendix D SI UnitsTable D.1SI UnitsSI Base Unit Base Quantity Length Mass Time Electric current Temperature Amount of substance Luminous intensity Name Meter Kilogram Second Ampere Kelvin Mole Candela Symbol m kg s A K mol cdTable D.2Some Derived SI

Nanjing University - PHYS - 102

Appendix E Nobel PrizesAll Nobel Prizes in physics are listed (and marked with a P), as well as relevant Nobel Prizes in Chemistry (C). The key dates for some of the scientic work are supplied; they often antedate the prize considerably. 1901 1902 1903 1

UT Dallas - CHEM - 2325

CHEM 2325 NAME (PLEASE PRINT): WebCT User Name: ANSWER SHEET SAMPLE TEST 1Detach the first page as your answer sheet and turn in for grading. You may keep the rest for yourself. 2. (5 pts)1. (5 pts)3. (5 pts)4. (5 pts)5. (5 pts)6. (5 pts)7. (5 pts

UT Dallas - CHEM - 2325

CHEM 2325 NAME (PLEASE PRINT): WebCT User Name: ANSWER SHEET SAMPLE TEST 2Detach the first page as your answer sheet and turn in for grading. You may keep the rest for yourself. 2. (5 pts)1. (5 pts)3. (5 pts)4. (5 pts)5. (5 pts)6. (5 pts)7. (5 pts

UT Dallas - CHEM - 2325

CHEM 2325SAMPLE TEST 31-12. Draw a structure of ONE major product for each of the following reactions in a blank box on the answer sheet. Indicate all stereochemistry if applicable. 1. (5 pts)O O NaOCH2CH3 CH3CH2OH2. (5 pts)O1)NH22) NaBH43. (5 pt

The University of Oklahoma - ECON - 101

Approved by CET 6/11/2007Spring 2011 Academic CalendarApril 1, 2010 Thursday October 1, 2010 Friday October 25, 2010 Monday Oct. 25-Dec. 3, 2010 November 1, 2010 Monday November 1, 2010 Monday December 3, 2010 Friday December 16, 2010 January 1, 2011 De

Santa Rosa - GEOG - 3

Five Themes of Geography in the context of SexualityClick to edit Master subtitle stylePrepared by: Mohammad O. Rasid 2/25/11Five themes of geography exemplify the geographical imagination, and illustrate how geographers have researched sexuality.2/25

Santa Rosa - GEOG - 3

H ow is English Related to L anguages Spoken Elsewhere inI ndo-European Language Sir William Jones (1786) theory about the similarities between Latin, Greek and SanskritEmerged from Proto-Indo-European language which is believed to be the mother tongue

Santa Rosa - GEOG - 3

Group 1Geography of SexualityBasic ConceptsGEOGRAPHY OF SEXUALITY Sexuality and space is a field of study within human geography. The phrase encompasses all relationships and interactions between human sexuality, space and place.GEOGRAPHY OF SEXUALIT

Santa Rosa - GEOG - 3

Spatialconstraints onhomosexualityClick to edit Master subtitle style2/25/11GayandLesbianConsumerismthey expresses their identities Click to edit Master subtitle style through the purchase of certain clothes, CDs, beverages, and accessories or through

Santa Rosa - GEOG - 3

W hy Do People Living in Different Locations Speak English Differently?ClicktoeditMastersubtitlestyleGeographyof 2/25/11 LanguageOutlineGeographyandDialect DevelopmentofDialectsinEnglish BritishReceivedPronunciation(BRP) DialectsinEnglandCurrentdia

York University - PHIL - 1100

Alice Furnari Freedom and Reason in Kant24 /2/97Morality, Kant says, cannot be regarded as a set of rules which prescribe the means necessary to the achievement of a given end; its rules must be obeyed without consideration of the consequences that will

York University - PHIL - 1100

We ought then regard the present state of the universe as the effect of its previous state and the cause of the one which is to follow. An intelligence knowing at a given instant of time all the forces operating in nature, as well as the position at that

York University - PHIL - 1100

Free Will Versus Determinism for years. is the The controversy between free will and determinism has been argued about What is the difference between the two? Looking in a dictionary, free willpower, attributed to human beings, of making free choices tha

York University - PHIL - 1100

Human Nature Is Inherently Bad There are many theories as to Human nature. One of Which exists, under the thoughts of a prominent philosopher, and founder of Psychoanalysis Sigmund Freud. His deductive argument, entails his conclusion that man is bad, or

York University - PHIL - 1100

The Life & Philosophy of Friedrich Nietzsche Philosophy Class Essay Born: 1844. Rocken, Germany Died: 1900. Weimar, Germany Major Works: The Gay Science (1882), Thus Spoke Zarathustra (1883-1885), Beyond Good & Evil (1886), On the Genealogy of Morals (188

York University - PHIL - 1100

While acquiring knowledge on the topics of Functionalism and Physicalism, I ran across many disagreement between the two. Interestingly, those disagreements gave me an impression of different sides arguing with their own support from their own theories. A

York University - PHIL - 1100

2/4/97 G Galileo and Newton Galileo believed the physical world to be bounded. He says that all material things have "this or that shape" and are small or large in relation to other things. He also says that material objects are either in motion or at res

York University - PHIL - 1100

Hollywood has been showing it to us for years. Frankenstein, The Six Million Dollar Man, Jurassic Park, etc.; the list goes on. All these movies show man's instinct to create. This fiction of playing God in recent years is becoming a reality. In 1952, deo

York University - PHIL - 1100

Title: "Power Comes From the Barrel of a Gun" - took the opposing view "Would you respect me, If I didnt have this gun? Cause without it, I dont get it, And thats why I carry one." -Phil Collins Power. A word from which many meanings derive. To each indiv

York University - PHIL - 1100

GOODNESS K KantThe philosopher I used is Immanuel Kant. He was very practical in his thinking of goodness. A quote of his was I ought, therefore I can. His view was good anything is under good will . He believed good will was the primary goodness, good i

York University - PHIL - 1100

Hammurabi In his position as King of Babylonia, Hammurabi managed to organize the world's first code of laws and establish Babylon as the dominant and successful Amorite city of its time. "Records written on clay tablets show that Hammurabi was a very cap

York University - PHIL - 1100

The standard definition of happiness is that it is a condition of supreme well-being and good spirits. There can be many definitions of the word happiness. It can be applied to many examples. Many people are often in search of happiness. One meaning can b

York University - PHIL - 1100

HAPPINESS Happiness: In one word, this concept exemplifies the American dream. People go to any means by which to obtain the many varied materials and issues that induce pleasures in each individual, and intrinsically, this emotion remains the ultimate go

York University - PHIL - 1100

Hemingway and Nada In "The light of the world" written by Ernest Hemingway Steve Ketchel, a boxer symbolizes a Jesus figure for a woman called Alice. Alice, a 350 pound, unpleasant prostitute struggles with her current life. Her central being focuses at t

York University - PHIL - 1100

The Quarrel About Historical ExplanationThe discussion of the philosophical question of historical explanation is in reality a disagreement concerning the nature of the philosophic method. There are primarily two sides taken in this argument, those who a

York University - PHIL - 1100

Thomas Hobbes Paper - What is the difference between obligations in foro interno and in foro externo, and when do we h have such obligations? According to Thomas Hobbes, there are certain laws of nature which exist in the absence of an organized governmen

York University - PHIL - 1100

Hollywoods Attack on Religion The section that I have chosen to analyze from the book Hollywood vs. America is The Attack on Religion. In this part of the book, Michael Medved discusses the shift in attitude Hollywood has made toward religion, from accept

York University - PHIL - 1100

Homosexuality has been on debate for numerous years. It is mentioned in the Bible which is thousands of years old. But recently two philosophers have spoken how they feel about Homosexuality. Michael Levin and Richard Mohr's views on the subject are in co

York University - PHIL - 1100

15 Feb 97 dayz: T Today's Lesson: H Hooked on Ebonics: Leroy is a 20 year old 9th grader. This be Leroy's homework assignment. He must use each vocabulary word in a sentence. Foreclose: If I pay alimony this month, I'll have no money foreclose. Rectum: I

York University - PHIL - 1100

How Technology Effects Modern America i don't know if you have this one yet. U U.S. Wage Trends The microeconomic picture of the U.S. has changed immensely since 1973, and the trends are proving to be consistently downward for the nations high school grad

York University - PHIL - 1100

How to get married and stay married to the perfect mate! This book will talk about the ways and theories of how to stay married one you are married. It will cover stuff like communication, similarity, physical attractiveness, similarity, balance and equit

York University - PHIL - 1100

Human Perception An intimate look into the most intriguing aspect of modern psychology. It determines what we see, what we do, what we feel. It controls our emotions, our thoughts, and our conscience. What is this remarkable element of the human mind? It

York University - PHIL - 1100

Kevin Clark Philosophy p.2 H Humanism The word "humanism" has a number of meanings, and because there are so many different meanings it can be quite confusing if you don't know what kind of humanism s someone is talking about. L Literary Humanism is a dev

York University - PHIL - 1100

HUME AND DESCARTES ON THE THEORY OF IDEAS David Hume and Rene Descartes are philosophers with opposing views about the origination of ideas. Descartes believed there were three types of ideas which are, innate, adventitious and those from imagination. He

York University - PHIL - 1100

Hume, DavidIn An Inquiry Concerning Human Understanding, David Hume demonstrates how there is no way to rationally make any claims about future occurrences. According to Hume knowledge of matters of fact come from previous experience. From building on th

York University - PHIL - 1100

Travis Slaby phil-101 Alex Clarkson 2 2-17-97H Hume's Mind Game The human mind is a very intricate machine. There have been many people that have attempted, and failed, to explain how the human mind operates. After reading Hume, I was in agreement with a

York University - PHIL - 1100

I Am Me, and You Are You Existentialists view mankind as individuals whose unique past experiences establish personal characteristics that set all of us apart. This idea can be best expressed in an intuitive statement by a celebrated individualist, Tarzan

York University - PHIL - 1100

. In a world with such a vast amount of people their exists virtually every different belief, thought, and ideology. This means that for every argument and every disagreement that their exists two sides of relative equal strength. It is through these disa

York University - PHIL - 1100

Impermanence, Selflessness, and DissatisfactionBuddhism is neither a religion nor a philosophy, but rather a way of life. This does not imply that Buddhism is nothing more than an ethical code: it is a way of moral, spiritual and intellectual training le

York University - PHIL - 1100

INTRODUCTION The study of the way people think and behave is called psychology. The field of psychology has a number of sub-disciplines devoted to the study of the different levels and contexts of human thought and behavior. Social psychology, for example

York University - PHIL - 1100

Take-Home Assignment #1 T TU 11:05 I Investing in the FutureTobin Lichti Psych 101 Jon DrummondWelfare and school reform are two of the most widely discussed issues in politics today. Many people are calling for reduction or elimination welfare programs

York University - PHIL - 1100

The Existence of God The existence of a God has for generations been the topic of fierce debate. This most usually occurring between members of the religious society and, everybody else. As a matter of fact the religious world itself has not always been a

York University - PHIL - 1100

Topic: Philosophy of Religion T Thesis: Is there a God or is He(?) an illusion? Bibliography: Benedict Spinoza - rationalist Freidrick Schleiermacher - religion S Sigmund Freud - atheist An illusion is one's own interpretation and perception of someone or