36
795 795 795 Capacitance and Dielectrics CHAPTER OUTLINE 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 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) Chapter 26

Chapter 26 Capacitance and Dielectrics - … · Capacitance and Dielectrics ... whic h store electric charge and energy . ... The capacitance C of a capacitor is deÞned as the ratio

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Page 1: Chapter 26 Capacitance and Dielectrics - … · Capacitance and Dielectrics ... whic h store electric charge and energy . ... The capacitance C of a capacitor is deÞned as the ratio

795795 795

Capacitance and Dielectrics

CHAPTE R OUTL I N E

26.1 Definition of Capacitance

26.2 Calculating Capacitance

26.3 Combinations of Capacitors

26.4 Energy Stored in a ChargedCapacitor

26.5 Capacitors with Dielectrics

26.6 Electric Dipole in an ElectricField

26.7 An Atomic Description ofDielectrics

! All of these devices are capacitors, which store electric charge and energy. A capacitor isone type of circuit element that we can combine with others to make electric circuits.(Paul Silverman/Fundamental Photographs)

Chapter 26

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796

In this chapter, we will introduce the first of three simple circuit elements that can beconnected with wires to form an electric circuit. Electric circuits are the basis for thevast 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 resis-tors in Chapter 27 and inductors in Chapter 32. In later chapters, we will study moresophisticated circuit elements such as diodes and transistors.

Capacitors are commonly used in a variety of electric circuits. For instance, they areused to tune the frequency of radio receivers, as filters in power supplies, to eliminatesparking in automobile ignition systems, and as energy-storing devices in electronicflash units.

A capacitor consists of two conductors separated by an insulator. The capacitanceof a given capacitor depends on its geometry and on the material—called a dielectric—that separates the conductors.

26.1 Definition of Capacitance

Consider two conductors carrying charges of equal magnitude and opposite sign, asshown 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 con-ductors due to the presence of the charges.

What determines how much charge is on the plates of a capacitor for a givenvoltage? Experiments show that the quantity of charge Q on a capacitor1 is linearlyproportional to the potential difference between the conductors; that is, Q " !V.The proportionality constant depends on the shape and separation of the con-ductors.2 We can write this relationship as Q # C !V if we define capacitance asfollows:

1 Although the total charge on the capacitor is zero (because there is as much excess positive chargeon one conductor as there is excess negative charge on the other), it is common practice to refer to themagnitude 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.

The capacitance C of a capacitor is defined as the ratio of the magnitude of thecharge on either conductor to the magnitude of the potential difference betweenthe conductors:

(26.1)C ! Q

!V

–Q

+Q

Figure 26.1 A capacitor consists oftwo conductors. When the capaci-tor is charged, the conductors carrycharges of equal magnitude andopposite sign.

! PITFALL PREVENTION 26.1 Capacitance Is a

CapacityTo understand capacitance, thinkof similar notions that use a simi-lar word. The capacity of a milkcarton is the volume of milk thatit can store. The heat capacity ofan object is the amount of energyan object can store per unitof temperature difference. Thecapacitance of a capacitor is theamount of charge the capacitorcan store per unit of potentialdifference.

Definition of capacitance

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Note that by definition capacitance is always a positive quantity. Furthermore, the chargeQ and the potential difference !V are always expressed in Equation 26.1 as positivequantities. Because the potential difference increases linearly with the stored charge,the ratio Q /!V is constant for a given capacitor. Therefore, capacitance is a measureof a capacitor’s ability to store charge. Because positive and negative charges are sepa-rated in the system of two conductors in a capacitor, there is electric potential energystored in the system.

From Equation 26.1, we see that capacitance has SI units of coulombs per volt. TheSI unit of capacitance is the farad (F), which was named in honor of Michael Faraday:

The farad is a very large unit of capacitance. In practice, typical devices have capaci-tances ranging from microfarads (10$6 F) to picofarads (10$12 F). We shall use thesymbol %F to represent microfarads. To avoid the use of Greek letters, in practice,physical capacitors often are labeled “mF’’ for microfarads and “mmF’’ for micromicro-farads or, equivalently, “pF’’ for picofarads.

Let us consider a capacitor formed from a pair of parallel plates, as shown inFigure 26.2. Each plate is connected to one terminal of a battery, which acts asa source of potential difference. If the capacitor is initially uncharged, the batteryestablishes 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 battery. Theelectric field applies a force on electrons in the wire just outside this plate; thisforce causes the electrons to move onto the plate. This movement continues untilthe plate, the wire, and the terminal are all at the same electric potential. Once thisequilibrium point is attained, a potential difference no longer exists between theterminal and the plate, and as a result no electric field is present in the wire, andthe movement of electrons stops. The plate now carries a negative charge. A similarprocess occurs at the other capacitor plate, with electrons moving from the plate tothe wire, leaving the plate positively charged. In this final configuration, the poten-tial difference across the capacitor plates is the same as that between the terminalsof the battery.

Suppose that we have a capacitor rated at 4 pF. This rating means that the capaci-tor can store 4 pC of charge for each volt of potential difference between the twoconductors. If a 9-V battery is connected across this capacitor, one of the conductorsends up with a net charge of $ 36 pC and the other ends up with a net charge of& 36 pC.

1 F # 1 C/V

S ECT I O N 26 . 2 • Calculating Capacitance 797

Quick Quiz 26.1 A capacitor stores charge Q at a potential difference !V. Ifthe voltage applied by a battery to the capacitor is doubled to 2 !V, (a) the capacitancefalls to half its initial value and the charge remains the same (b) the capacitance andthe charge both fall to half their initial values (c) the capacitance and the charge bothdouble (d) the capacitance remains the same and the charge doubles.

d

–Q+Q

Area = A

+ –

Figure 26.2 A parallel-plate capac-itor consists of two parallel con-ducting plates, each of area A,separated by a distance d. Whenthe capacitor is charged by con-necting the plates to the terminalsof a battery, the plates carry equalamounts of charge. One platecarries positive charge, and theother carries negative charge.

! PITFALL PREVENTION 26.2 Potential Difference

is !V, not VWe use the symbol !V for the potential difference across a circuit element or a devicebecause this is consistent with ourdefinition of potential differenceand with the meaning of the deltasign. It is a common, but confus-ing, practice to use the symbol Vwithout the delta sign for a poten-tial difference. Keep this in mindif you consult other texts.

! PITFALL PREVENTION 26.3 Too Many C’sDo not confuse italic C for capac-itance with non-italic C for theunit coulomb.

26.2 Calculating Capacitance

We can derive an expression for the capacitance of a pair of oppositely charged con-ductors in the following manner: assume a charge of magnitude Q , and calculate thepotential difference using the techniques described in the preceding chapter. Wethen use the expression C # Q /!V to evaluate the capacitance. As we might expect,we can perform this calculation relatively easily if the geometry of the capacitor issimple.

Page 4: Chapter 26 Capacitance and Dielectrics - … · Capacitance and Dielectrics ... whic h store electric charge and energy . ... The capacitance C of a capacitor is deÞned as the ratio

While the most common situation is that of two conductors, a single conductor alsohas a capacitance. For example, imagine a spherical charged conductor. The electricfield lines around this conductor are exactly the same as if there were a conductingshell of infinite radius, concentric with the sphere and carrying a charge of the samemagnitude but opposite sign. Thus, we can identify the imaginary shell as the secondconductor of a two-conductor capacitor. We now calculate the capacitance for this situ-ation. The electric potential of the sphere of radius R is simply keQ /R , and settingV # 0 for the infinitely large shell, we have

(26.2)

This expression shows that the capacitance of an isolated charged sphere is propor-tional to its radius and is independent of both the charge on the sphere and the poten-tial difference.

The capacitance of a pair of conductors depends on the geometry of the conduc-tors. Let us illustrate this with three familiar geometries, namely, parallel plates, con-centric cylinders, and concentric spheres. In these examples, we assume that thecharged conductors are separated by a vacuum. The effect of a dielectric materialplaced 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 shownin Figure 26.2. One plate carries a charge Q , and the other carries a charge $ Q .Let us consider how the geometry of these conductors influences the capacity of thecombination to store charge. Recall that charges of the same sign repel oneanother. As a capacitor is being charged by a battery, electrons flow into the nega-tive plate and out of the positive plate. If the capacitor plates are large, the accumu-lated charges are able to distribute themselves over a substantial area, and theamount of charge that can be stored on a plate for a given potential differenceincreases as the plate area is increased. Thus, we expect the capacitance to be pro-portional to the plate area A.

Now let us consider the region that separates the plates. If the battery has aconstant potential difference between its terminals, then the electric field betweenthe plates must increase as d is decreased. Let us imagine that we move the platescloser together and consider the situation before any charges have had a chanceto move in response to this change. Because no charges have moved, the electricfield 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 theterminal voltage of the battery now exists as a potential difference across the wiresconnecting the battery to the capacitor. This potential difference results in theelectric field in the wires that drives more charge onto the plates, increasing thepotential difference between the plates. When the potential difference betweenthe plates again matches that of the battery, the potential difference across the wiresfalls back to zero, and the flow of charge stops. Thus, moving the plates closertogether causes the charge on the capacitor to increase. If d is increased, the chargedecreases. As a result, we expect the capacitance of the pair of plates to be inverselyproportional to d.

We can verify these physical arguments with the following derivation. Thesurface charge density on either plate is ' # Q /A. If the plates are very closetogether (in comparison with their length and width), we can assume that theelectric field is uniform between the plates and is zero elsewhere. Accordingto the What If? feature in Example 24.8, the value of the electric field between

C #Q

!V#

Qke Q /R

#Rke

# 4()0R

798 C H A P T E R 2 6 • Capacitance and Dielectrics

Capacitance of an isolatedcharged sphere

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the plates is

Because the field between the plates is uniform, the magnitude of the potential differ-ence between the plates equals Ed (see Eq. 25.6); therefore,

Substituting this result into Equation 26.1, we find that the capacitance is

(26.3)

That is, the capacitance of a parallel-plate capacitor is proportional to the areaof its plates and inversely proportional to the plate separation, just as we ex-pected from our conceptual argument.

A careful inspection of the electric field lines for a parallel-plate capacitorreveals that the field is uniform in the central region between the plates, as shownin Figure 26.3a. However, the field is nonuniform at the edges of the plates. Figure26.3b is a photograph of the electric field pattern of a parallel-plate capacitor. Notethe nonuniform nature of the electric field at the ends of the plates. Such endeffects can be neglected if the plate separation is small compared with the length ofthe plates.

Figure 26.4 shows a battery connected to a single parallel-plate capacitor with aswitch 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 wiresand 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. Thisenergy is transformed during the chemical reaction that occurs within the batterywhen it is operating in an electric circuit. When the switch is closed, some of thechemical energy in the battery is converted to electric potential energy related to theseparation of positive and negative charges on the plates. As a result, we can describe acapacitor as a device that stores energy as well as charge. We will explore this energystorage in more detail in Section 26.4.

C #)0Ad

C #Q

!V#

QQd/)0A

!V # Ed #Qd)0A

E #'

)0#

Q)0A

S ECT I O N 26 . 2 • Calculating Capacitance 799

+Q

–Q

(a)

Cour

tesy

of H

arol

d M

. Waa

ge, P

rince

ton

Unive

rsity

Figure 26.3 (a) The electric field between the plates of a parallel-plate capacitor isuniform near the center but nonuniform near the edges. (b) Electric field pattern oftwo oppositely charged conducting parallel plates. Small pieces of thread on an oilsurface align with the electric field.

Capacitance of parallel plates

(b)

Page 6: Chapter 26 Capacitance and Dielectrics - … · Capacitance and Dielectrics ... whic h store electric charge and energy . ... The capacitance C of a capacitor is deÞned as the ratio

800 C H A P T E R 2 6 • Capacitance and Dielectrics

Example 26.1 Parallel-Plate Capacitor

A parallel-plate capacitor with air between the plates has anarea A # 2.00 * 10$4 m2 and a plate separation d # 1.00 mm.Find its capacitance.

Solution From Equation 26.3, we find that1.77 pF # 1.77 * 10$12 F #

C #)0Ad

#(8.85 * 10$12 C2/N+m2)(2.00 * 10$4 m2)

1.00 * 10$3 m

+ –

(b)

+ –

+ –

+ –

+ –

+ –

+ –

∆V

Chemicalenergy inbattery isreduced

Electricfield inwire

Electricfield inwire

Electricfield betweenplates

E

Electronsmovefrom thewire tothe plate

Separationof chargesrepresentspotentialenergyElectrons move

from the plateto the wire,leaving the platepositivelycharged

+ –

(a)

∆V

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 wirethat causes electrons to move from the left plate into the wire and into the right platefrom the wire. As a result, a separation of charge exists on the plates, which representsan increase in electric potential energy of the system of the circuit. This energy in thesystem has been transformed from chemical energy in the battery.

At the Active Figures linkat http://www.pse6.com, youcan adjust the battery voltageand see the resulting charge onthe plates and the electric fieldbetween the plates.

Quick Quiz 26.2 Many computer keyboard buttons are constructed ofcapacitors, as shown in Figure 26.5. When a key is pushed down, the soft insulatorbetween 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 cannotdetermine because the complicated electric circuit connected to the keyboard buttonmay cause a change in !V.

Figure 26.5 (Quick Quiz 26.2) One type of com-puter keyboard button.

Key

Movableplate

Softinsulator

Fixedplate

B

Page 7: Chapter 26 Capacitance and Dielectrics - … · Capacitance and Dielectrics ... whic h store electric charge and energy . ... The capacitance C of a capacitor is deÞned as the ratio

S ECT I O N 26 . 2 • Calculating Capacitance 801

Example 26.2 The Cylindrical Capacitor

A solid cylindrical conductor of radius a and charge Q iscoaxial with a cylindrical shell of negligible thickness, radiusb , a, and charge $ Q (Fig. 26.6a). Find the capacitance ofthis cylindrical capacitor if its length is !.

Solution It is difficult to apply physical arguments to thisconfiguration, although we can reasonably expect the capac-itance to be proportional to the cylinder length ! for thesame reason that parallel-plate capacitance is proportionalto plate area: stored charges have more room in which to bedistributed. If we assume that ! is much greater than a andb, we can neglect end effects. In this case, the electric fieldis perpendicular to the long axis of the cylinders and isconfined to the region between them (Fig. 26.6b). We mustfirst calculate the potential difference between the two cylin-ders, which is given in general by

where E is the electric field in the region between thecylinders. In Chapter 24, we showed using Gauss’s law thatthe magnitude of the electric field of a cylindrical chargedistribution having linear charge density - is E # 2ke-/r(Eq. 24.7). The same result applies here because, accord-ing to Gauss’s law, the charge on the outer cylinder does

Vb $ Va # $"b

a E+d s

Figure 26.6 (Example 26.2) (a) A cylindrical capacitor consistsof a solid cylindrical conductor of radius a and length !surrounded by a coaxial cylindrical shell of radius b. (b) Endview. The electric field lines are radial. The dashed linerepresents the end of the cylindrical gaussian surface of radiusr and length !.

ba

!

(a) (b)

Gaussiansurface

–Qa

Q

b

r

not contribute to the electric field inside it. Using this re-sult and noting from Figure 26.6b that E is along r, we findthat

Substituting this result into Equation 26.1 and using the factthat - # Q /!, we obtain

(26.4)

where !V is the magnitude of the potential differencebetween the cylinders, given by !V # #Va $ Vb # #2ke -ln(b/a), a positive quantity. As predicted, the capaci-tance is proportional to the length of the cylinders. As wemight expect, the capacitance also depends on the radii ofthe two cylindrical conductors. From Equation 26.4, wesee that the capacitance per unit length of a combinationof concentric cylindrical conductors is

(26.5)

An example of this type of geometric arrangement is a coax-ial cable, which consists of two concentric cylindrical conduc-tors separated by an insulator. You are likely to have acoaxial cable attached to your television set or VCR if youare a subscriber to cable television. The cable carries electri-cal signals in the inner and outer conductors. Such a geom-etry is especially useful for shielding the signals from anypossible external influences.

What If? Suppose b " 2.00a for the cylindrical capacitor.We would like to increase the capacitance, and we can doso by choosing to increase ! by 10% or by increasing aby 10%. Which choice is more effective at increasing thecapacitance?

Answer According to Equation 26.4, C is proportional to !,so increasing ! by 10% results in a 10% increase in C. Forthe result of the change in a, let us first evaluate C forb # 2.00a:

# 0.721 !

ke

C #!

2ke ln(b/a)#

!

2ke ln(2.00)#

!

2ke (0.693)

C!

#1

2ke ln(b/a)

!

2ke ln(b/a)C #

Q!V

#Q

(2ke Q /!)ln(b/a)#

Vb $ Va # $"b

a Er dr # $2ke - "b

a

drr

# $2ke - ln $ ba %

Cylindrical and Spherical Capacitors

From the definition of capacitance, we can, in principle, find the capacitance of anygeometric arrangement of conductors. The following examples demonstrate the use ofthis definition to calculate the capacitance of the other familiar geometries that wementioned: cylinders and spheres.

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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. # 1.10a, so

The ratio of the new and old capacitances is

C .

C#

0.836 !/ke

0.721 !/ke# 1.16

#!

2ke ln(2.00/1.10)#

!

2ke (0.598)# 0.836

!

ke

C . #!

2ke ln(b/a.)#

!

2ke ln(2.00a/1.10a)

corresponding to a 16% increase in capacitance. Thus, it ismore effective to increase a than to increase !.

Note two more extensions of this problem. First, theadvantage goes to increasing a only for a range of relation-ships between a and b. It is a valuable exercise to show that ifb , 2.85a, increasing ! by 10% is more effective thanincreasing a (Problem 77). Second, if we increase b, wereduce the capacitance, so we would need to decrease b toincrease the capacitance. Increasing a and decreasing b bothhave the effect of bringing the plates closer together, whichincreases the capacitance.

Example 26.3 The Spherical Capacitor

A spherical capacitor consists of a spherical conducting shellof radius b and charge $ Q concentric with a smaller con-ducting sphere of radius a and charge Q (Fig. 26.7). Findthe capacitance of this device.

Solution As we showed in Chapter 24, the field outside aspherically symmetric charge distribution is radial and givenby the expression keQ /r 2. In this case, this result applies tothe field between the spheres (a / r / b). From Gauss’s lawwe see that only the inner sphere contributes to this field.Thus, the potential difference between the spheres is

The magnitude of the potential difference is

Substituting this value for !V into Equation 26.1, we obtain

(26.6)

What If? What if the radius b of the outer sphereapproaches infinity? What does the capacitance become?

Answer In Equation 26.6, we let b : 0:

Note that this is the same expression as Equation 26.2, thecapacitance of an isolated spherical conductor.

C # limb : 0

ab

ke(b $ a)#

abke(b)

#ake

# 4()0a

ab

ke(b $ a)C #

Q∆V

#

∆V # & Vb $ Va & # keQ (b $ a)

ab

# ke Q $ 1b

$1a %

Vb $ Va # $"b

a E r dr # $ke Q "b

a drr

2 # ke Q ' 1r (

b

a

Figure 26.7 (Example 26.3) A spherical capacitor consists of aninner sphere of radius a surrounded by a concentric sphericalshell of radius b. The electric field between the spheres isdirected radially outward when the inner sphere is positivelycharged.

a

b

– Q

+Q

26.3 Combinations of Capacitors

Two or more capacitors often are combined in electric circuits. We can calculate theequivalent capacitance of certain combinations using methods described in thissection. Throughout this section, we assume that the capacitors to be combined areinitially uncharged.

In studying electric circuits, we use a simplified pictorial representation calleda circuit diagram. Such a diagram uses circuit symbols to represent variouscircuit elements. The circuit symbols are connected by straight lines that representthe wires between the circuit elements. The circuit symbols for capacitors andbatteries, as well as the color codes used for them in this text, are given in Figure26.8. The symbol for the capacitor reflects the geometry of the most commonmodel for a capacitor—a pair of parallel plates. The positive terminal of the batteryis at the higher potential and is represented in the circuit symbol by the longer line.

Capacitorsymbol

Batterysymbol +–

Switchsymbol

Figure 26.8 Circuit symbols forcapacitors, batteries, and switches.Note that capacitors are in blue andbatteries and switches are in red.

Page 9: Chapter 26 Capacitance and Dielectrics - … · Capacitance and Dielectrics ... whic h store electric charge and energy . ... The capacitance C of a capacitor is deÞned as the ratio

Parallel Combination

Two capacitors connected as shown in Figure 26.9a are known as a parallel combinationof 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 positiveterminal of the battery and are therefore both at the same electric potential as thepositive terminal. Likewise, the right plates are connected to the negative terminal andare therefore both at the same potential as the negative terminal. Thus, the individualpotential differences across capacitors connected in parallel are the same andare 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 combi-nation is the terminal voltage of the battery. Situations can occur in which the parallelcombination is in a circuit with other circuit elements; in such situations, we must deter-mine 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, elec-trons are transferred between the wires and the plates; this transfer leaves the leftplates positively charged and the right plates negatively charged. The flow of chargeceases when the voltage across the capacitors is equal to that across the battery termi-nals. The capacitors reach their maximum charge when the flow of charge ceases. Letus call the maximum charges on the two capacitors Q 1 and Q 2. The total charge Qstored by the two capacitors is

(26.7)

That is, the total charge on capacitors connected in parallel is the sum of thecharges on the individual capacitors. Because the voltages across the capacitors arethe same, the charges that they carry are

Suppose that we wish to replace these two capacitors by one equivalent capacitorhaving a capacitance Ceq, as in Figure 26.9c. The effect this equivalent capacitor hason the circuit must be exactly the same as the effect of the combination of the two

Q 1 # C 1 ∆V Q 2 # C 2 ∆V

Q # Q 1 & Q 2

S ECT I O N 26 . 3 • Combinations of Capacitors 803

(a)

+ –

C2

+ –

C1

+ –

(b)

∆V

+ –

Q2

C2

Q1

C1

∆V1 = ∆V2 = ∆V

∆V

+ –

Ceq = C1 + C2

(c)

∆V

Active Figure 26.9 (a) A parallel combination of two capacitors in an electric circuitin which the potential difference across the battery terminals is !V. (b) The circuit dia-gram for the parallel combination. (c) The equivalent capacitance is Ceq # C1 & C2.

At the Active Figures linkat http://www.pse6.com, youcan adjust the battery voltageand the individual capacitancesto see the resulting chargesand voltages on the capacitors.You can combine up to fourcapacitors in parallel.

Page 10: Chapter 26 Capacitance and Dielectrics - … · Capacitance and Dielectrics ... whic h store electric charge and energy . ... The capacitance C of a capacitor is deÞned as the ratio

individual capacitors. That is, the equivalent capacitor must store Q units of chargewhen connected to the battery. We can see from Figure 26.9c that the voltage acrossthe equivalent capacitor also is !V because the equivalent capacitor is connecteddirectly across the battery terminals. Thus, for the equivalent capacitor,

Substituting these three relationships for charge into Equation 26.7, we have

If we extend this treatment to three or more capacitors connected in parallel, we findthe equivalent capacitance to be

(26.8)

Thus, the equivalent capacitance of a parallel combination of capacitors is thealgebraic sum of the individual capacitances and is greater than any of the indi-vidual capacitances. This makes sense because we are essentially combining the areasof all the capacitor plates when we connect them with conducting wire, and capaci-tance of parallel plates is proportional to area (Eq. 26.3).

Series Combination

Two capacitors connected as shown in Figure 26.10a and the equivalent circuitdiagram in Figure 26.10b are known as a series combination of capacitors. The left plateof capacitor 1 and the right plate of capacitor 2 are connected to the terminals of abattery. 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 tohave zero net charge. To analyze this combination, let us begin by considering theuncharged capacitors and follow what happens just after a battery is connected to thecircuit. When the battery is connected, electrons are transferred out of the left plate ofC1 and into the right plate of C 2. As this negative charge accumulates on the rightplate of C 2, an equivalent amount of negative charge is forced off the left plate of C2,and this left plate therefore has an excess positive charge. The negative charge leaving

Ceq # C 1 & C 2 & C 3 & + + + (parallel combination)

Ceq # C1 & C 2 (parallel combination)

C eq ∆V # C 1 ∆V &C 2 !V

Q # C eq ∆V

804 C H A P T E R 2 6 • Capacitance and Dielectrics

(b)∆V

Q1 = Q2 = QC1 C2

∆V1 ∆V2

–+ + –

(c)∆V

Ceq C1 C2

1 1 1= +

(a)

+ –

C2

∆V

C1∆V1 ∆V2

+Q – Q +Q – Q

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

1Ceq

#1

C 1&

1C 2

.

At the Active Figures linkat http://www.pse6.com, youcan adjust the battery voltageand the individual capacitancesto see the resulting chargesand voltages on the capacitors.You can combine up to fourcapacitors in series.

Capacitors in parallel

Page 11: Chapter 26 Capacitance and Dielectrics - … · Capacitance and Dielectrics ... whic h store electric charge and energy . ... The capacitance C of a capacitor is deÞned as the ratio

the left plate of C 2 causes negative charges to accumulate on the right plate of C1. As aresult, all the right plates end up with a charge $ Q , and all the left plates end up witha 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 splitbetween the two capacitors:

(26.9)

where !V1 and !V2 are the potential differences across capacitors C1 and C2, respectively.In general, the total potential difference across any number of capacitors connectedin 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 effecton the circuit as the series combination when it is connected to the battery. After it isfully charged, the equivalent capacitor must have a charge of $ Q on its right plate anda charge of & Q on its left plate. Applying the definition of capacitance to the circuit inFigure 26.10c, we have

Because we can apply the expression Q # C !V to each capacitor shown in Figure26.10b, the potential differences across them are

Substituting these expressions into Equation 26.9, we have

Canceling Q , we arrive at the relationship

When this analysis is applied to three or more capacitors connected in series, therelationship for the equivalent capacitance is

(26.10)

This shows that the inverse of the equivalent capacitance is the algebraic sum of theinverses of the individual capacitances and the equivalent capacitance of a seriescombination is always less than any individual capacitance in the combination.

1C eq

#1

C 1&

1C 2

&1

C 3& + + + (series combination)

1C eq

#1

C 1&

1C 2

(series combination)

QCeq

#QC1

&QC 2

∆V1 #QC1

∆V2 #QC 2

∆V #Q

Ceq

∆V # ∆V1 & ∆V2

S ECT I O N 26 . 3 • Combinations of Capacitors 805

Quick Quiz 26.3 Two capacitors are identical. They can be connected inseries or in parallel. If you want the smallest equivalent capacitance for the combina-tion, do you connect them in (a) series, in (b) parallel, or (c) do the combinationshave the same capacitance?

Quick Quiz 26.4 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 poten-tial difference across the combination, do you connect them in (a) series, in (b) paral-lel, or (c) do the combinations have the same potential difference?

Capacitors in series

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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 potentialdifference across each is the same. The charge on each capacitor isproportional to its capacitance; hence, the capacitances can be addeddirectly to give the equivalent capacitance of the parallel combination. The equivalent capacitance is always larger than the individualcapacitances.

• When two or more capacitors are connected in series, they carry the same charge, and the sum of the potential differences equals the totalpotential difference applied to the combination. The sum of thereciprocals of the capacitances equals the reciprocal of the equivalentcapacitance, which is always less than the capacitance of the smallestindividual capacitor.

Example 26.4 Equivalent Capacitance

Find the equivalent capacitance between a and b for thecombination of capacitors shown in Figure 26.11a. Allcapacitances are in microfarads.

Solution Using Equations 26.8 and 26.10, we reduce thecombination step by step as indicated in the figure. The 1.0-%Fand 3.0-%F capacitors are in parallel and combine accordingto the expression Ceq # C1 & C2 # 4.0 %F. The 2.0-%F and6.0-%F capacitors also are in parallel and have an equivalentcapacitance of 8.0 %F. Thus, the upper branch in Figure26.11b consists of two 4.0-%F capacitors in series, whichcombine as follows:

The lower branch in Figure 26.11b consists of two 8.0-%Fcapacitors in series, which combine to yield an equiva-lent capacitance of 4.0 %F. Finally, the 2.0-%F and 4.0-%Fcapacitors in Figure 26.11c are in parallel and thus have an

equivalent capacitance of 6.0 %F.

C eq # 2.0 %F

1

C eq#

1C 1

&1

C 2#

14.0 %F

&1

4.0 %F#

12.0 %F

Figure 26.11 (Example 26.4) To find the equivalent capacitance of the capacitors inpart (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.

4.04.0

8.08.0

ba

(b)

4.0

ba

(c)

2.0

6.0 ba

(d)

4.0

8.0

ba

(a)

2.0

6.0

3.0

1.0

Practice reducing a combination of capacitors to a single equivalent capacitance at the Interactive Worked Example link athttp://www.pse6.com.

Interactive

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S ECT I O N 26 . 4 • Energy Stored in a Charged Capacitor 807

3 We shall use lowercase q for the time-varying charge on the capacitor while it is charging, to distin-guish it from uppercase Q , which is the total charge on the capacitor after it is completely charged.

26.4 Energy Stored in a Charged Capacitor

Almost everyone who works with electronic equipment has at some time verified that acapacitor can store energy. If the plates of a charged capacitor are connected by a con-ductor, such as a wire, charge moves between each plate and its connecting wire untilthe capacitor is uncharged. The discharge can often be observed as a visible spark. Ifyou should accidentally touch the opposite plates of a charged capacitor, your fingersact as a pathway for discharge, and the result is an electric shock. The degree of shockyou 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 ofa television set. Because the charges can be stored in a capacitor even when the set isturned off, unplugging the television does not make it safe to open the case and touchthe components inside.

To calculate the energy stored in the capacitor, we shall assume a charging processthat is different from the actual process described in Section 26.1 but which gives thesame final result. We can make this assumption because the energy in the final configu-ration does not depend on the actual charge-transfer process. We imagine that thecharge is transferred mechanically through the space between the plates. We reachin and grab a small amount of positive charge on the plate connected to the negativeterminal and apply a force that causes this positive charge to move over to the plateconnected to the positive terminal. Thus, we do work on the charge as we transfer itfrom one plate to the other. At first, no work is required to transfer a small amount ofcharge dq from one plate to the other.3 However, once this charge has been trans-ferred, a small potential difference exists between the plates. Therefore, work must bedone to move additional charge through this potential difference. As more and morecharge is transferred from one plate to the other, the potential difference increases inproportion, and more work is required.

Suppose that q is the charge on the capacitor at some instant during the chargingprocess. 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 chargedq from the plate carrying charge $ q to the plate carrying charge q (which is at thehigher electric potential) is

This is illustrated in Figure 26.12. The total work required to charge the capacitorfrom q # 0 to some final charge q # Q is

The work done in charging the capacitor appears as electric potential energy U storedin the capacitor. Using Equation 26.1, we can express the potential energy stored in acharged capacitor in the following forms:

(26.11)

This result applies to any capacitor, regardless of its geometry. We see that for a givencapacitance, the stored energy increases as the charge increases and as the potentialdifference 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 ultimatelyoccurs between the plates. For this reason, capacitors are usually labeled with amaximum operating voltage.

U #Q

2

2C# 1

2Q ∆V # 12C (∆V )2

W # "Q

0

qC

dq #1C

"Q

0 q dq #

Q 2

2C

dW # ∆V dq #qC

dq

V

dq

q

Q

Figure 26.12 A plot of potentialdifference versus charge for acapacitor is a straight line having aslope 1/C. The work required tomove charge dq through thepotential difference !V existingat the time across the capacitorplates is given approximately bythe area of the shaded rectangle.The total work required to chargethe capacitor to a final charge Qis the triangular area under thestraight line, . (Don’tforget that 1 V # J/C; hence, theunit for the triangular area is thejoule.)

W # 12Q !V

Energy stored in a chargedcapacitor

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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 electricfield created between the plates as the capacitor is charged. This description is reason-able because the electric field is proportional to the charge on the capacitor. For aparallel-plate capacitor, the potential difference is related to the electric field throughthe relationship !V # Ed. Furthermore, its capacitance is C # )0A/d (Eq. 26.3). Substi-tuting these expressions into Equation 26.11, we obtain

(26.12)

Because the volume occupied by the electric field is Ad, the energy per unit volumeuE # U/Ad, known as the energy density, is

(26.13)

Although Equation 26.13 was derived for a parallel-plate capacitor, the expressionis generally valid, regardless of the source of the electric field. That is, the energydensity in any electric field is proportional to the square of the magnitude ofthe electric field at a given point.

uE # 12

) 0 E 2

U # 12

) 0Ad

(E2d2) # 12()0Ad )E2

Quick Quiz 26.5 You have three capacitors and a battery. In which of thefollowing combinations of the three capacitors will the maximum possible energy bestored when the combination is attached to the battery? (a) series (b) parallel (c) Bothcombinations will store the same amount of energy.

Quick Quiz 26.6 You charge a parallel-plate capacitor, remove it from thebattery, and prevent the wires connected to the plates from touching each other. Whenyou 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) energystored in the capacitor.

Quick Quiz 26.7 Repeat Quick Quiz 26.6, but this time answer the ques-tions for the situation in which the battery remains connected to the capacitor whileyou pull the plates apart.

Example 26.5 Rewiring Two Charged Capacitors

Two capacitors C1 and C 2 (where C1 , C 2) are chargedto the same initial potential difference !V i . The chargedcapacitors are removed from the battery, and their platesare connected with opposite polarity as in Figure 26.13a.The switches S1 and S2 are then closed, as in Figure26.13b.

(A) Find the final potential difference !Vf between a and bafter the switches are closed.

Solution Figure 26.13 helps us conceptualize the initialand final configurations of the system. In Figure 26.13b, itmight appear as if the capacitors are connected in parallel,but there is no battery in this circuit that is applying a volt-age across the combination. Thus, we cannot categorize thisas a problem in which capacitors are connected in parallel.We can categorize this as a problem involving an isolatedsystem for electric charge—the left-hand plates of the capac-

Figure 26.13 (Example 26.5) (a) Two capacitors are charged tothe same initial potential difference and connected together withplates of opposite sign to be in contact when the switches areclosed. (b) When the switches are closed, the charges redistribute.

+ –

Q1i+

ba

(a)

–C1

Q 2i– +

C2

S1 S2

+

ba

(b)

S1 S2

Q1fC1

Q 2f C2

Interactive

! PITFALL PREVENTION 26.4 Not a New Kind of

EnergyThe energy given by Equation26.13 is not a new kind of energy.It is familiar electric potentialenergy associated with a system ofseparated source charges. Equa-tion 26.13 provides a new interpre-tation, or a new way of modelingthe energy, as energy associatedwith the electric field, regardlessof the source of the field.

itors form an isolated system because they are not con-nected to the right-hand plates by conductors. To analyze

Energy density in an electricfield

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S ECT I O N 26 . 4 • Energy Stored in a Charged Capacitor 809

the problem, note that the charges on the left-hand platesbefore the switches are closed are

The negative sign for Q 2i is necessary because the charge onthe left plate of capacitor C2 is negative. The total charge Qin the system is

After the switches are closed, the total charge Q in thesystem remains the same but the charges on the individualcapacitors change to new values Q 1f and Q 2f . Because thesystem is isolated,

The charges redistribute until the potential difference is thesame across both capacitors, !Vf . To satisfy this requirement,the charges on the capacitors after the switches are closed are

Dividing the first equation by the second, we have

Combining Equations (2) and (3), we obtain

Using Equations (3) and (4) to find Q1f in terms of Q , we have

Finally, using Equation 26.1 to find the voltage across eachcapacitor, we find that

As noted earlier, !V1f # !V 2f # !Vf .To express !Vf in terms of the given quantities C1, C2,

and !Vi , we substitute the value of Q from Equation (1) intoeither Equation (6) or (7) to obtain

$ C1 $ C 2

C1 & C 2% ∆Vi∆Vf #

(7) ∆V2f #Q 2f

C 2#

Q [C 2/(C1 & C 2)]C 2

#Q

C1 & C 2

(6) ∆V1f #Q 1f

C1#

Q [C1/(C1 & C 2)]C1

#Q

C1 & C 2

# Q $ C 1

C1 & C 2%

(5) Q 1f #C1

C 2 Q 2f #

C1

C 2 Q $ C 2

C1 & C 2%

(4) Q 2f # Q $ C 2

C1 & C 2%

Q # Q 1f & Q 2f #C 1

C 2 Q 2f & Q 2f # Q 2f $1 &

C 1

C 2%

(3) Q 1f #C1

C 2 Q 2f

Q 1f # C1 ∆Vf and Q 2f # C 2 ∆Vf

(2) Q # Q 1f & Q 2f

(1) Q # Q 1i & Q 2i # (C1 $ C 2)∆Vi

Q 1i # C1 ∆Vi and Q 2i # $C 2 ∆Vi

(B) Find the total energy stored in the capacitors before andafter the switches are closed and the ratio of the final energyto the initial energy.

Solution Before the switches are closed, the total energystored in the capacitors is

After the switches are closed, the total energy stored in thecapacitors is

Using the results of part (A), we can express this as

Therefore, the ratio of the final energy stored to the initialenergy stored is

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 conservationhas been violated, but this is not the case. The “missing’’ en-ergy is transferred out of the system of the capacitors by themechanism of electromagnetic waves, as we shall see inChapter 34.

What If? What if the two capacitors have the same capaci-tance? What would we expect to happen when the switchesare closed?

Answer The equal-magnitude charges on the two capaci-tors should simply cancel each other and the capacitors willbe uncharged afterward.

Let us test our results to see if this is the case mathemati-cally. In Equation (1), because the charges are of equal mag-nitude and opposite sign, we see that Q # 0. Thus,Equations (4) and (5) show us that Q1f # Q 2f # 0, consis-tent with our prediction. Furthermore, Equations (6) and(7) show us that !V1f # !V2f # 0, which is consistentwith uncharged capacitors. Finally, if C1 # C2, Equation(8) shows us that Uf # 0, which is also consistent withuncharged capacitors.

$ C 1 $ C 2

C 1 & C 2%2

#

(8) Uf

Ui#

12 (C 1 $ C 2)2(∆Vi)2/(C1 & C 2)

12(C1 & C2)(∆Vi)2

12 (C 1 $ C 2)2(∆Vi)2

(C 1 & C 2)Uf #

Uf # 12C 1(∆Vf )2 & 1

2C 2(∆Vf )2 # 12(C 1 & C 2)(!Vf )2

12(C1 & C 2)(∆Vi)2Ui # 12C 1(∆Vi)2 & 12C 2(∆Vi)2 #

At the Interactive Worked Example link at http://www.pse6.com, explore this situation for various initial values of the volt-age and the capacitances.

Page 16: Chapter 26 Capacitance and Dielectrics - … · Capacitance and Dielectrics ... whic h store electric charge and energy . ... The capacitance C of a capacitor is deÞned as the ratio

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 isfully charged. The defibrillator can deliver all this energy to a patient in about 2 ms. (Thisis roughly equivalent to 3 000 times the power delivered to a 60-W lightbulb!) Under theproper conditions, the defibrillator can be used to stop cardiac fibrillation (random con-tractions) 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 theorgan to its normal beat pattern. Emergency medical teams use portable defibrillatorsthat contain batteries capable of charging a capacitor to a high voltage. (The circuitryactually permits the capacitor to be charged to a much higher voltage than that of thebattery.) The stored energy is released through the heart by conducting electrodes, calledpaddles, that are placed on both sides of the victim’s chest. The paramedics must wait be-tween applications of the energy due to the time necessary for the capacitors to becomefully charged. In this case and others (e.g., camera flash units and lasers used for fusionexperiments), capacitors serve as energy reservoirs which can be slowly charged and thendischarged 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 energystored is much less than that stored in a defibrillator. After the flash unit’s capacitor ischarged, tripping the camera’s shutter causes the stored energy to be sent through aspecial lightbulb that briefly illuminates the subject being photographed.

26.5 Capacitors with Dielectrics

A dielectric is a nonconducting material, such as rubber, glass, or waxed paper. Whena dielectric is inserted between the plates of a capacitor, the capacitance increases. Ifthe dielectric completely fills the space between the plates, the capacitance increasesby a dimensionless factor 1, which is called the dielectric constant of the material.The dielectric constant varies from one material to another. In this section, we analyzethis 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 microscopicorigin of these changes.

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 beingrevived 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 torestore the heart’s normal rhythm pattern.

Adam

Har

t-Dav

is/SP

L/Cu

stom

Med

ical S

tock

Page 17: Chapter 26 Capacitance and Dielectrics - … · Capacitance and Dielectrics ... whic h store electric charge and energy . ... The capacitance C of a capacitor is deÞned as the ratio

S ECT I O N 26 . 5 • Capacitors with Dielectrics 811

4 If the dielectric is introduced while the potential difference is held constant by a battery, thecharge increases to a value Q # 1Q 0. The additional charge comes from the wires attached to thecapacitor, and the capacitance again increases by the factor 1.

We can perform the following experiment to illustrate the effect of a dielectric ina capacitor. Consider a parallel-plate capacitor that without a dielectric has a chargeQ 0 and a capacitance C0. The potential difference across the capacitor is !V0 #Q 0/C 0. Figure 26.15a illustrates this situation. The potential difference is measured bya voltmeter, which we shall study in greater detail in Chapter 28. Note that no battery isshown in the figure; also, we must assume that no charge can flow through an idealvoltmeter. Hence, there is no path by which charge can flow and alter the charge onthe 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 1 as follows:

Because !V / !V0, we see that 1 , 1.Because the charge Q 0 on the capacitor does not change, we conclude that the

capacitance must change to the value

(26.14)

That is, the capacitance increases by the factor 1 when the dielectric completely fills theregion between the plates.4 For a parallel-plate capacitor, where C0 # )0 A/d (Eq. 26.3),we can express the capacitance when the capacitor is filled with a dielectric as

(26.15)

From Equations 26.3 and 26.15, it would appear that we could make the capaci-tance very large by decreasing d, the distance between the plates. In practice, thelowest value of d is limited by the electric discharge that could occur through thedielectric medium separating the plates. For any given separation d, the maximum volt-age that can be applied to a capacitor without causing a discharge depends on the

C # 1 )0Ad

C # 1C 0

C #Q 0

∆V #

Q 0

∆V0/1# 1

Q 0

∆V0

∆V #∆V0

1

C0 Q 0

+–

C Q 0

Dielectric

∆V∆V0

+–

(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 potentialdifference decreases from !V0 to !V # !V0/1. Thus, the capacitance increases fromC0 to 1C0.

Capacitance of a capacitor filledwith a material of dielectricconstant #

! PITFALL PREVENTION 26.5 Is the Capacitor

Connected to aBattery?

In problems in which you aremodifying a capacitor (by inser-tion of a dielectric, for example),you must note whether modi-fications to the capacitor arebeing made while the capacitor isconnected to a battery or after it isdisconnected. If the capacitorremains connected to the battery,the voltage across the capacitornecessarily remains the same. Ifyou disconnect the capacitor fromthe battery before making anymodifications to the capacitor, thecapacitor is an isolated system andits charge remains the same.

Page 18: Chapter 26 Capacitance and Dielectrics - … · Capacitance and Dielectrics ... whic h store electric charge and energy . ... The capacitance C of a capacitor is deÞned as the ratio

dielectric strength (maximum electric field) of the dielectric. If the magnitude of theelectric field in the dielectric exceeds the dielectric strength, then the insulating prop-erties break down and the dielectric begins to conduct. Figure 26.16 shows the effect ofexceeding the dielectric strength of air. Sparks appear between the two wires, due toionization of atoms and recombination with electrons in the air, similar to the processthat produced corona discharge in Section 25.6.

Physical capacitors have a specification called by a variety of names, including work-ing voltage, breakdown voltage, and rated voltage. This parameter represents the largestvoltage that can be applied to the capacitor without exceeding the dielectric strengthof the dielectric material in the capacitor. Consequently, when selecting a capacitorfor a given application, you must consider the capacitance of the device along with theexpected voltage across the capacitor in the circuit, making sure that the expectedvoltage will be smaller than the rated voltage of the capacitor. You can see the ratedvoltage on several of the capacitors in the opening photograph for this chapter.

Insulating materials have values of 1 greater than unity and dielectric strengthsgreater than that of air, as Table 26.1 indicates. Thus, we see that a dielectric providesthe following advantages:

• Increase in capacitance

• Increase in maximum operating voltage

• Possible mechanical support between the plates, which allows the plates to be closetogether without touching, thereby decreasing d and increasing C.

Types of Capacitors

Commercial capacitors are often made from metallic foil interlaced with thin sheets ofeither paraffin-impregnated paper or Mylar as the dielectric material. These alternatelayers 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

812 C H A P T E R 2 6 • Capacitance and Dielectrics

Figure 26.16 Dielectric breakdownin air. Sparks are produced whenthe high voltage between the wirescauses the electric field to exceedthe dielectric strength of air.

©Lo

ren

Win

ters

/Visu

als U

nlim

ited

a The dielectric strength equals the maximum electric field that can exist in a dielectric withoutelectrical breakdown. Note that these values depend strongly on the presence of impurities andflaws in the materials.

Dielectric Strengtha

Material Dielectric Constant # (106 V/m)

Air (dry) 1.000 59 3Bakelite 4.9 24Fused quartz 3.78 8Mylar 3.2 7Neoprene rubber 6.7 12Nylon 3.4 14Paper 3.7 16Paraffin-impregnated 3.5 11

paperPolystyrene 2.56 24Polyvinyl chloride 3.4 40Porcelain 6 12Pyrex glass 5.6 14Silicone oil 2.5 15Strontium titanate 233 8Teflon 2.1 60Vacuum 1.000 00 —Water 80 —

Approximate Dielectric Constants and Dielectric Strengths of Various Materials at Room Temperature

Table 26.1

Page 19: Chapter 26 Capacitance and Dielectrics - … · Capacitance and Dielectrics ... whic h store electric charge and energy . ... The capacitance C of a capacitor is deÞned as the ratio

metallic plates immersed in silicone oil (Fig. 26.17b). Small capacitors are often con-structed from ceramic materials.

Often, an electrolytic capacitor is used to store large amounts of charge at relatively lowvoltages. This device, shown in Figure 26.17c, consists of a metallic foil in contact withan electrolyte—a solution that conducts electricity by virtue of the motion of ions con-tained in the solution. When a voltage is applied between the foil and the electrolyte, athin layer of metal oxide (an insulator) is formed on the foil, and this layer serves as thedielectric. Very large values of capacitance can be obtained in an electrolytic capacitorbecause 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 apolarity, which is indicated by positive and negative signs marked on the device. Whenelectrolytic capacitors are used in circuits, the polarity must be aligned properly. If thepolarity of the applied voltage is opposite that which is intended, the oxide layer isremoved and the capacitor conducts electricity instead of storing charge.

Variable capacitors (typically 10 to 500 pF) usually consist of two interwoven sets ofmetallic 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.

S ECT I O N 26 . 5 • Capacitors with Dielectrics 813

Metal foil

Paper

Plates

Oil

Electrolyte

Case

Metallic foil + oxide layer

Contacts

(a) (b) (c)

Figure 26.17 Three commercial capacitor designs. (a) A tubular capacitor, whose platesare separated by paper and then rolled into a cylinder. (b) A high-voltage capacitorconsisting of many parallel plates separated by insulating oil. (c) An electrolytic capacitor.

Figure 26.18 A variable capacitor.When one set of metal plates isrotated so as to lie between a fixedset of plates, the capacitance of thedevice changes.

Geor

ge S

empl

e

Quick Quiz 26.8 If you have ever tried to hang a picture or a mirror, youknow 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 sideinstead of facing one another, as shown in Figure 26.19. When the device is moved overa stud, does the capacitance increase or decrease?

Capacitorplates

Stud-finder

Wall board

Stud

(b)(a)

Figure 26.19 (Quick Quiz 26.8) A stud-finder. (a) The materials between the plates ofthe capacitor are the wallboard and air. (b) When the capacitor moves across a stud inthe wall, the materials between the plates are the wallboard and the wood. The changein the dielectric constant causes a signal light to illuminate.

Page 20: Chapter 26 Capacitance and Dielectrics - … · Capacitance and Dielectrics ... whic h store electric charge and energy . ... The capacitance C of a capacitor is deÞned as the ratio

814 C H A P T E R 2 6 • Capacitance and Dielectrics

Quick Quiz 26.9 A fully charged parallel-plate capacitor remains con-nected to a battery while you slide a dielectric between the plates. Do the followingquantities increase, decrease, or stay the same? (a) C ; (b) Q ; (c) E between the plates;(d) !V.

Example 26.6 A Paper-Filled Capacitor

A parallel-plate capacitor has plates of dimensions 2.0 cm by3.0 cm separated by a 1.0-mm thickness of paper.

(A) Find its capacitance.

Solution Because 1 # 3.7 for paper (see Table 26.1), we have

(B) What is the maximum charge that can be placed on thecapacitor?

20 pF# 20 * 10$12 F #

# 3.7 $ (8.85 * 10$12 C2/N+m2)(6.0 * 10$4 m2)1.0 * 10$3 m %

C # 1 )0Ad

Solution From Table 26.1 we see that the dielectricstrength of paper is 16 * 106 V/m. Because the thickness ofthe paper is 1.0 mm, the maximum voltage that can beapplied before breakdown is

Hence, the maximum charge is

0.32 %C#

Q max # C !Vmax # (20 * 10$12 F )(16 * 103 V )

# 16 * 103 V∆Vmax # Emax d # (16 * 106 V/m)(1.0 * 10$3 m)

Example 26.7 Energy Stored Before and After

A parallel-plate capacitor is charged with a battery to acharge Q 0, as shown in Figure 26.20a. The battery is thenremoved, and a slab of material that has a dielectricconstant 1 is inserted between the plates, as shown in Figure26.20b. Find the energy stored in the capacitor before andafter the dielectric is inserted.

Solution From Equation 26.11, we see that the energy storedin the absence of the dielectric is

After the battery is removed and the dielectric inserted, thecharge on the capacitor remains the same. Hence, the energystored in the presence of the dielectric is

But the capacitance in the presence of the dielectric isC # 1C0, so U becomes

Because 1 , 1, the final energy is less than the initialenergy. We can account for the “missing’’ energy by notingthat the dielectric, when inserted, is pulled into thedevice (see Section 26.7). An external agent must do neg-ative work to keep the dielectric from accelerating. Thiswork is simply the difference U $ U 0. (Alternatively, thepositive work done by the system on the external agent isU 0 $ U.)

U0

1U #

Q 02

21C 0#

U #Q 0

2

2C

Q 02

2C 0U0 #

–+

Q 0C 0

∆V 0

(a)

Dielectric

–+Q 0

(b)

Figure 26.20 (Example 26.7) (a) A battery charges up a paral-lel-plate capacitor. (b) The battery is removed and a slab of di-electric material is inserted between the plates.

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26.6 Electric Dipole in an Electric Field

We have discussed the effect on the capacitance of placing a dielectric between the platesof 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 electricdipole that we began in Section 23.4 (see Example 23.6). The electric dipole consists oftwo charges of equal magnitude and opposite sign separated by a distance 2a, as shown inFigure 26.21. The electric dipole moment of this configuration is defined as the vector pdirected from $q toward &q along the line joining the charges and having magnitude2aq :

(26.16)

Now suppose that an electric dipole is placed in a uniform electric field E, asshown in Figure 26.22. We identify E as the field external to the dipole, distinguishing itfrom the field due to the dipole, which we discussed in Section 23.4. The field E isestablished by some other charge distribution, and we place the dipole into this field.Let us imagine that the dipole moment makes an angle 2 with the field.

The electric forces acting on the two charges are equal in magnitude (F # qE ) andopposite in direction as shown in Figure 26.22. Thus, the net force on the dipole iszero. However, the two forces produce a net torque on the dipole; as a result, thedipole rotates in the direction that brings the dipole moment vector into greateralignment with the field. The torque due to the force on the positive charge about anaxis through O in Figure 26.22 has magnitude Fa sin 2, where a sin 2 is the momentarm of F about O. This force tends to produce a clockwise rotation. The torque aboutO on the negative charge is also of magnitude Fa sin 2; here again, the force tends toproduce a clockwise rotation. Thus, the magnitude of the net torque about O is

Because F # qE and p # 2aq, we can express 3 as

(26.17)

It is convenient to express the torque in vector form as the cross product of the vectorsp and E:

(26.18)

We can determine the potential energy of the system—an electric dipole in an exter-nal 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 dipolethrough an angle so as to cause the dipole moment vector to become less aligned with thefield. The work done is then stored as potential energy in the system. The work dWrequired to rotate the dipole through an angle d2 is dW # 3 d2 (Eq. 10.22). Because3 # pE sin 2 and because the work results in an increase in the potential energy U, we findthat for a rotation from 2i to 2f the change in potential energy of the system is

The term that contains cos 2i is a constant that depends on the initial orientation ofthe dipole. It is convenient for us to choose a reference angle of 2i # 90°, so thatcos 2i # cos 90° # 0. Furthermore, let us choose Ui # 0 at 2i # 90° as our referenceof potential energy. Hence, we can express a general value of U # Uf as

(26.19)

We can write this expression for the potential energy of a dipole in an electric field asthe dot product of the vectors p and E:

U # $pE cos 2

# pE [$cos 2]2i

2f # pE(cos 2i $ cos 2f )

Uf $ Ui # "2f

2i 3 d2 # "2f

2i

pE sin 2 d2 # pE"2f

2i

sin 2 d2

$ " p % E

3 # 2aqE sin 2 # p4 sin 2

3 # 2Fa sin 2

p ! 2aq

S ECT I O N 26 . 6 • Electric Dipole in an Electric Field 815

+ q

– q

2a

p–

+

Figure 26.21 An electric dipoleconsists of two charges of equalmagnitude and opposite signseparated by a distance of 2a. Theelectric dipole moment p isdirected from $ q toward & q.

+ q

θ

– q

F

E– F

O

+

Figure 26.22 An electric dipole ina uniform external electric field.The dipole moment p is at anangle 2 to the field, causing thedipole to experience a torque.

Torque on an electric dipole inan external electric field

Page 22: Chapter 26 Capacitance and Dielectrics - … · Capacitance and Dielectrics ... whic h store electric charge and energy . ... The capacitance C of a capacitor is deÞned as the ratio

(26.20)

To develop a conceptual understanding of Equation 26.19, compare this expres-sion with the expression for the potential energy of the system of an object in the gravi-tational field of the Earth, U # mgh (see Chapter 8). The gravitational expressionincludes 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—itsdipole moment p. The gravitational expression includes the magnitude of the gravita-tional 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 gravita-tional expression, the potential energy depends on how high we lift the object, mea-sured by h. In Equation 26.19, the potential energy depends on the angle 2 throughwhich we rotate the dipole. In both cases, we are making a change in the configurationof the system. In the gravitational case, the change involves moving an object in a trans-lational sense, whereas in the electrical case, the change involves moving an object in arotational sense. In both cases, however, once the change is made, the system tends toreturn to the original configuration when the object is released: the object of mass mfalls back to the ground, and the dipole begins to rotate back toward the configurationin which it is aligned with the field. Thus, apart from the type of motion, the expres-sions for potential energy in these two cases are similar.

Molecules are said to be polarized when a separation exists between the averageposition of the negative charges and the average position of the positive charges in themolecule. In some molecules, such as water, this condition is always present—suchmolecules are called polar molecules. Molecules that do not possess a permanentpolarization are called nonpolar molecules.

We can understand the permanent polarization of water by inspecting the geome-try of the water molecule. In the water molecule, the oxygen atom is bonded to thehydrogen 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, andthe center of the positive charge distribution lies at a point midway along the linejoining the hydrogen atoms (the point labeled * in Fig. 26.23). We can model thewater molecule and other polar molecules as dipoles because the average positions ofthe positive and negative charges act as point charges. As a result, we can apply ourdiscussion of dipoles to the behavior of polar molecules.

Microwave ovens take advantage of the polar nature of the water molecule. Whenin operation, microwave ovens generate a rapidly changing electric field that causesthe polar molecules to swing back and forth, absorbing energy from the field in theprocess. Because the jostling molecules collide with each other, the energy they absorbfrom the field is converted to internal energy, which corresponds to an increase intemperature of the food.

Another household scenario in which the dipole structure of water is exploited iswashing with soap and water. Grease and oil are made up of nonpolar molecules, whichare generally not attracted to water. Plain water is not very useful for removing this type ofgrime. Soap contains long molecules called surfactants. In a long molecule, the polaritycharacteristics 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 apolar molecule. The nonpolar end can attach to a grease or oil molecule, and the polarend can attach to a water molecule. Thus, the soap serves as a chain, linking the dirt andwater 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 polariza-tion can be induced by placing the molecule in an electric field. A field directed to theleft, as shown in Figure 26.24b, would cause the center of the positive charge distribu-tion to shift to the left from its initial position and the center of the negative chargedistribution to shift to the right. This induced polarization is the effect that predominatesin most materials used as dielectrics in capacitors.

U # $p+E

816 C H A P T E R 2 6 • Capacitance and Dielectrics

O

HH 105°

−−

+ +*

Figure 26.23 The water molecule,H2O, has a permanent polarizationresulting from its nonlineargeometry. The center of thepositive charge distribution is atthe point *.

E

(a)

(b)

++ −

−+ − +

Figure 26.24 (a) A linear symmet-ric molecule has no permanentpolarization. (b) An externalelectric field induces a polarizationin the molecule.

Potential energy of the systemof an electric dipole in anexternal electric field

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S ECT I O N 26 . 7 • An Atomic Description of Dielectrics 817

Example 26.8 The H2O Molecule

The water (H2O) molecule has an electric dipole momentof 6.3 * 10$30 C + m. A sample contains 1021 water mole-cules, with the dipole moments all oriented in the directionof an electric field of magnitude 2.5 * 105 N/C. How muchwork is required to rotate the dipoles from this orientation(2 # 0°) to one in which all the moments are perpendicularto the field (2 # 90°)?

Solution The work required to rotate one molecule 90° isequal to the difference in potential energy between the 90°orientation and the 0° orientation. Using Equation 26.19,

we obtain

Because there are 1021 molecules in the sample, the totalwork required is

1.6 * 10$3 JWtotal # (1021)(1.6 * 10$24 J) #

# 1.6 * 10$24 J

# pE # (6.3 * 10$30 C+m)(2.5 * 105 N/C)

W # U905 $ U05 # ($pE cos 905) $ ($pE cos 05)

26.7 An Atomic Description of Dielectrics

In Section 26.5 we found that the potential difference !V0 between the plates of acapacitor is reduced to !V0/1 when a dielectric is introduced. The potential differenceis reduced because the magnitude of the electric field decreases between the plates. Inparticular, if E0 is the electric field without the dielectric, the field in the presence of adielectric is

(26.21)

Let us first consider a dielectric made up of polar molecules placed in the electricfield between the plates of a capacitor. The dipoles (that is, the polar molecules mak-ing up the dielectric) are randomly oriented in the absence of an electric field, asshown in Figure 26.25a. When an external field E0 due to charges on the capacitorplates is applied, a torque is exerted on the dipoles, causing them to partially align withthe field, as shown in Figure 26.25b. We can now describe the dielectric as being polar-ized. The degree of alignment of the molecules with the electric field depends ontemperature and on the magnitude of the field. In general, the alignment increaseswith decreasing temperature and with increasing electric field.

If the molecules of the dielectric are nonpolar, then the electric field due to theplates produces some charge separation and an induced dipole moment. These induceddipole 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 themolecules are polar or nonpolar.

E #E0

1

E0(b) (c)

– + – +–

+ –+

–+–+–+

– +

–+ –+

– + –+

– + – + – +

– +– + – +

– + – + – +– + – + – +

– +

(a)

E0

Eind

– indσ indσ

+

+

+

+

+

+

+

+

+

+

+

+

Figure 26.25 (a) Polar molecules are randomly oriented in the absence of an externalelectric field. (b) When an external electric field is applied, the molecules partiallyalign with the field. (c) The charged edges of the dielectric can be modeled as anadditional pair of parallel plates establishing an electric field E ind in the directionopposite to that of E0.

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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 theplates of a capacitor so that it is in a uniform electric field E0, as shown in Figure26.25b. The electric field due to the plates is directed to the right and polarizes thedielectric. The net effect on the dielectric is the formation of an induced positivesurface charge density ' ind on the right face and an equal-magnitude negative surfacecharge density $ ' ind on the left face, as shown in Figure 26.25c. Because we canmodel these surface charge distributions as being due to parallel plates, the inducedsurface charges on the dielectric give rise to an induced electric field Eind in thedirection opposite the external field E0. Therefore, the net electric field E in thedielectric has a magnitude

(26.22)

In the parallel-plate capacitor shown in Figure 26.26, the external field E0 is relatedto the charge density ' on the plates through the relationship E0 # '/)0. The inducedelectric field in the dielectric is related to the induced charge density 'ind through therelationship E ind # ' ind/)0. Because E # E0/1 # '/1)0, substitution into Equation26.22 gives

(26.23)

Because 1 , 1, this expression shows that the charge density ' ind induced on thedielectric is less than the charge density ' on the plates. For instance, if 1 # 3 we seethat the induced charge density is two-thirds the charge density on the plates. If nodielectric is present, then 1 # 1 and ' ind # 0 as expected. However, if the dielectric isreplaced by an electrical conductor, for which E # 0, then Equation 26.22 indicatesthat E0 # E ind; this corresponds to ' ind # '. That is, the surface charge induced onthe conductor is equal in magnitude but opposite in sign to that on the plates, result-ing in a net electric field of zero in the conductor (see Fig. 24.16).

We can use the existence of the induced surface charge distributions on the dielec-tric to explain the result of Example 26.7. As we saw there, the energy of a capacitor notconnected to a battery is lowered when a dielectric is inserted between the plates; thismeans that negative work is done on the dielectric by the external agent inserting thedielectric into the capacitor. This, in turn, implies that a force must be acting on thedielectric that draws it into the capacitor. This force originates from the nonuniformnature of the electric field of the capacitor near its edges, as indicated in Figure 26.27.

'ind # $ 1 $ 11

%'

'

1)0#

'

)0$

' ind

)0

E # E0 $ E ind

+

+

+

+

+

+

++

+

+

+

+

+

+

+

+

+

+

+

+

σ – σ–

––

–σ indσ σindσ σ

Figure 26.26 Induced charge on adielectric placed between theplates of a charged capacitor. Notethat the induced charge density onthe dielectric is less than the chargedensity on the plates.

+Q

–Q

+

+

+

+

+

+

+

Figure 26.27 The nonuniform electric field near the edges of a parallel-platecapacitor causes a dielectric to be pulled into the capacitor. Note that the field acts onthe induced surface charges on the dielectric, which are nonuniformly distributed.

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S ECT I O N 26 . 7 • An Atomic Description of Dielectrics 819

(b)

(d – a)/2

(d – a)/2

(a)

d a

(d – a)/2

(d – a)/2σ–

σ

–– – – – –

– – – – –

+ + + + +

+ + + + +

σ

σ Figure 26.28 (Example 26.9) (a) A parallel-plate capacitor ofplate separation d partially filled with a metallic slab of thick-ness a. (b) The equivalent circuit of the device in part (a) con-sists of two capacitors in series, each having a plate separation(d $ a)/2.

Example 26.9 Effect of a Metallic Slab

A parallel-plate capacitor has a plate separation d and platearea A. An uncharged metallic slab of thickness a is insertedmidway between the plates.

(A) Find the capacitance of the device.

Solution We can solve this problem by noting that anycharge that appears on one plate of the capacitor mustinduce a charge of equal magnitude and opposite signon 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, thecapacitor is equivalent to two capacitors in series, eachhaving a plate separation (d $ a)/2, as shown in Figure26.28b.

Using Eq. 26.3 and the rule for adding two capacitors inseries (Eq. 26.10), we obtain

Note that C approaches infinity as a approaches d. Why?

)0Ad $ a

C #

1C

#1C1

&1C2

#1

' )0A(d $ a)/2 (

&1

' )0A(d $ a)/2 (

(B) Show that the capacitance of the original capacitor isunaffected by the insertion of the metallic slab if the slab isinfinitesimally thin.

Solution In the result for part (A), we let a : 0:

which is the original capacitance.

What If? What if the metallic slab in part (A) is notmidway between the plates? How does this affect thecapacitance?

Answer Let us imagine that the slab in Figure 26.27a ismoved upward so that the distance between the upper edgeof the slab and the upper plate is b. Then, the distancebetween the lower edge of the slab and the lower plate isd $ b $ a. As in part (A), we find the total capacitance ofthe series combination:

This is the same result as in part (A). It is independent ofthe value of b, so it does not matter where the slab is located.In Figure 26.28b, when the central structure is moved up ordown, the decrease in plate separation of one capacitor iscompensated by the increase in plate separation for theother.

C #) 0 A

d $ a

# b

) 0 A&

d $ b $ a) 0 A

#d $ a) 0 A

1C

#1

C 1&

1C 2

#1

() 0 A/b)&

1) 0 A/(d $ b $ a)

C # lima : 0

) 0 A

d $ a#

) 0 Ad

Example 26.10 A Partially Filled Capacitor

A parallel-plate capacitor with a plate separation d has acapacitance C0 in the absence of a dielectric. What is thecapacitance when a slab of dielectric material of dielectricconstant 1 and thickness is inserted between the plates(Fig. 26.29a)?

13 d

Solution In Example 26.9, we found that we could insert ametallic slab between the plates of a capacitor and considerthe combination as two capacitors in series. The resultingcapacitance was independent of the location of the slab.Furthermore, if the thickness of the slab approaches zero,

The horizontal component of this fringe field acts on the induced charges on the surfaceof the dielectric, producing a net horizontal force directed into the space between thecapacitor plates.

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820 C H A P T E R 2 6 • Capacitance and Dielectrics

13– d

23– d d

(a)

κ

(b)

C 1

C 2

13– d

23– d

κ

Figure 26.29 (Example 26.10) (a) A parallel-plate capacitor ofplate separation d partially filled with a dielectric of thicknessd/3. (b) The equivalent circuit of the capacitor consists of twocapacitors connected in series.

then the capacitance of the system approaches the capaci-tance when the slab is absent. From this, we conclude thatwe can insert an infinitesimally thin metallic slab anywherebetween the plates of a capacitor without affecting thecapacitance. Thus, let us imagine sliding an infinitesimallythin metallic slab along the bottom face of the dielectricshown in Figure 26.29a. We can then consider this system tobe the series combination of the two capacitors shown inFigure 26.29b: one having a plate separation d/3 and filledwith a dielectric, and the other having a plate separation2d/3 and air between its plates.

From Equations 26.15 and 26.3, the two capacitancesare

Using Equation 26.10 for two capacitors combined in series,we have

Because the capacitance without the dielectric is C0 # )0 A/d,we see that

$ 31

21 & 1 % C 0C #

C # $ 31

21 & 1 % ) 0A

d

1C

#d

3 ) 0 A $ 1

1& 2% #

d3 ) 0 A

$ 1 & 21

1 %

1C

#1

C 1&

1C 2

#d/3

1) 0 A&

2d/3) 0 A

C 1 #1) 0 Ad/3

and C 2 #) 0 A2d/3

A capacitor consists of two conductors carrying charges of equal magnitude andopposite sign. The capacitance C of any capacitor is the ratio of the charge Q oneither conductor to the potential difference !V between them:

(26.1)

The capacitance depends only on the geometry of the conductors and not on an exter-nal 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 ofcapacitors is

(26.8)

If two or more capacitors are connected in series, the charge is the same on all ofthem, and the equivalent capacitance of the series combination is given by

C eq # C 1 & C 2 & C 3 & + + +

C ! Q

∆V

S U M M A RY

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Page 27: Chapter 26 Capacitance and Dielectrics - … · Capacitance and Dielectrics ... whic h store electric charge and energy . ... The capacitance C of a capacitor is deÞned as the ratio

Questions 821

(26.10)

These two equations enable you to simplify many electric circuits by replacing multiplecapacitors with a single equivalent capacitance.

Energy is stored in a capacitor because the charging process is equivalent to thetransfer of charges from one conductor at a lower electric potential to another conduc-tor at a higher potential. The energy stored in a capacitor with charge Q is

(26.11)

When a dielectric material is inserted between the plates of a capacitor, the capaci-tance increases by a dimensionless factor 1, called the dielectric constant:

(26.14)

where C0 is the capacitance in the absence of the dielectric. The increase in capaci-tance is due to a decrease in the magnitude of the electric field in the presence of thedielectric. The decrease in the magnitude of E arises from an internal electric fieldproduced by aligned dipoles in the dielectric.

The electric dipole moment p of an electric dipole has a magnitude

(26.16)

The direction of the electric dipole moment vector is from the negative charge towardthe positive charge.

The torque acting on an electric dipole in a uniform electric field E is

(26.18)

The potential energy of the system of an electric dipole in a uniform external electric field E is

(26.20)U # $p & E

$ " p % E

p ! 2aq

C # 1C 0

U #Q2

2C# 1

2Q ∆V # 12C(∆V )2

1C eq

#1

C1&

1C2

&1

C 3& + + +

Geometry Capacitance Equation

Isolated sphere of radius R 26.2(second spherical conductor assumed to have infinite radius)

Parallel-plate capacitor of plate 26.3area A and plate separation d

Cylindrical capacitor of length 26.4! and inner and outer radii aand b, respectively

Spherical capacitor with inner 26.6and outer radii a and b,respectively

C # 4()0R

C #ab

k e (b $ a)

C #!

2k e ln(b/a)

C # ) 0 Ad

Capacitance and Geometry

Table 26.2

1. The plates of a capacitor are connected to a battery. Whathappens to the charge on the plates if the connectingwires are removed from the battery? What happens to thecharge if the wires are removed from the battery and con-nected to each other?

2. A farad is a very large unit of capacitance. Calculate thelength of one side of a square, air-filled capacitor that hasa capacitance of 1 F and a plate separation of 1 m.

3. A pair of capacitors are connected in parallel while anidentical pair are connected in series. Which pair would be

Q U E S T I O N S

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822 C H A P T E R 2 6 • Capacitance and Dielectrics

more dangerous to handle after being connected to thesame battery? Explain.

4. If you are given three different capacitors C 1, C 2, C 3,how many different combinations of capacitance can youproduce?

5. What advantage might there be in using two identicalcapacitors in parallel connected in series with anotheridentical parallel pair, rather than using a single capacitor?

6. Is it always possible to reduce a combination of capacitorsto one equivalent capacitor with the rules we have devel-oped? Explain.

7. The sum of the charges on both plates of a capacitor iszero. What does a capacitor store?

8. Because the charges on the plates of a parallel-platecapacitor are opposite in sign, they attract each other.Hence, it would take positive work to increase the plateseparation. What type of energy in the system changesdue to the external work done in this process?

9. Why is it dangerous to touch the terminals of a high-voltage capacitor even after the applied potential differ-ence has been turned off? What can be done to make thecapacitor safe to handle after the voltage source has beenremoved?

10. Explain why the work needed to move a charge Qthrough a potential difference !V is W # Q !V whereasthe energy stored in a charged capacitor is Where does the factor come from?

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 firstcharging a group of capacitors connected in parallel andthen activating a switch arrangement that in effect dis-

11.

12

U # 12 Q !V.

connects the capacitors from the charging source andfrom each other and reconnects them in a series arrange-ment. The group of charged capacitors is thendischarged in series. What is the maximum potentialdifference that can be obtained in this manner by usingten capacitors each of 500 %F and a charging source of800 V?

13. Assume you want to increase the maximum operatingvoltage of a parallel-plate capacitor. Describe how youcan do this for a fixed plate separation.

14. An air-filled capacitor is charged, then disconnectedfrom the power supply, and finally connected to avoltmeter. Explain how and why the potential differencechanges when a dielectric is inserted between the platesof the capacitor.

15. Using the polar molecule description of a dielectric,explain how a dielectric affects the electric field inside acapacitor.

16. Explain why a dielectric increases the maximum operat-ing voltage of a capacitor although the physical size ofthe capacitor does not change.

17. What is the difference between dielectric strength andthe 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 itscapacitance change? (Ignore thermal expansion andassume that the dipole orientations are temperature-dependent.)

If you were asked to design a capacitor where small sizeand large capacitance were required, what factors wouldbe important in your design?

20.

Section 26.1 Definition of Capacitance

1. (a) How much charge is on each plate of a 4.00-%F capaci-tor when it is connected to a 12.0-V battery? (b) If thissame capacitor is connected to a 1.50-V battery, whatcharge is stored?

2. Two conductors having net charges of & 10.0 %C and$ 10.0 %C have a potential difference of 10.0 V betweenthem. (a) Determine the capacitance of the system.(b) What is the potential difference between the two con-ductors if the charges on each are increased to & 100 %Cand $ 100 %C?

Section 26.2 Calculating Capacitance

An isolated charged conducting sphere of radius 12.0 cmcreates an electric field of 4.90 * 104 N/C at a distance

3.

21.0 cm from its center. (a) What is its surface charge den-sity? (b) What is its capacitance?

4. (a) If a drop of liquid has capacitance 1.00 pF, what is itsradius? (b) If another drop has radius 2.00 mm, what is itscapacitance? (c) What is the charge on the smaller drop ifits potential is 100 V?

5. Two conducting spheres with diameters of 0.400 m and1.00 m are separated by a distance that is large comparedwith the diameters. The spheres are connected by a thinwire and are charged to 7.00 %C. (a) How is this totalcharge shared between the spheres? (Ignore any chargeon the wire.) (b) What is the potential of the system ofspheres when the reference potential is taken to be V # 0at r # 0?

6. Regarding the Earth and a cloud layer 800 m abovethe Earth as the “plates’’ of a capacitor, calculate the

1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide

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= paired numerical and symbolic problems

P R O B L E M S

Page 29: Chapter 26 Capacitance and Dielectrics - … · Capacitance and Dielectrics ... whic h store electric charge and energy . ... The capacitance C of a capacitor is deÞned as the ratio

Problems 823

d

θ

R

Figure P26.10

CC

C

C CC

Figure P26.18

capacitance. Assume the cloud layer has an area of1.00 km2 and that the air between the cloud and theground is pure and dry. Assume charge builds up on thecloud and on the ground until a uniform electric field of3.00 * 106 N/C throughout the space between themmakes the air break down and conduct electricity as alightning bolt. What is the maximum charge the cloudcan hold?

An air-filled capacitor consists of two parallel plates,each with an area of 7.60 cm2, separated by a distance of1.80 mm. A 20.0-V potential difference is applied to theseplates. Calculate (a) the electric field between the plates,(b) the surface charge density, (c) the capacitance, and(d) the charge on each plate.

8. A 1-megabit computer memory chip contains many 60.0-f Fcapacitors. Each capacitor has a plate area of21.0 * 10$12 m2. Determine the plate separation of such acapacitor (assume a parallel-plate configuration). Theorder of magnitude of the diameter of an atom is10$10 m # 0.1 nm. Express the plate separation innanometers.

When a potential difference of 150 V is applied to theplates of a parallel-plate capacitor, the plates carry a surfacecharge density of 30.0 nC/cm2. What is the spacingbetween the plates?

10. A variable air capacitor used in a radio tuning circuit ismade of N semicircular plates each of radius R and posi-tioned a distance d from its neighbors, to which it is elec-trically connected. As shown in Figure P26.10, a secondidentical set of plates is enmeshed with its plates halfwaybetween those of the first set. The second set can rotateas a unit. Determine the capacitance as a function of theangle of rotation 2, where 2 # 0 corresponds to the maxi-mum capacitance.

9.

7.

other. The region between the spheres is a vacuum. Deter-mine the volume of this region.

An air-filled spherical capacitor is constructed with innerand outer shell radii of 7.00 and 14.0 cm, respectively.(a) Calculate the capacitance of the device. (b) Whatpotential difference between the spheres results in acharge of 4.00 %C on the capacitor?

14. A small object of mass m carries a charge q and is sus-pended by a thread between the vertical plates of aparallel-plate capacitor. The plate separation is d. If thethread makes an angle 2 with the vertical, what is thepotential difference between the plates?

15. Find the capacitance of the Earth. (Suggestion: The outerconductor of the “spherical capacitor’’ may be consideredas a conducting sphere at infinity where V approacheszero.)

Section 26.3 Combinations of Capacitors

16. Two capacitors, C1 # 5.00 %F and C 2 # 12.0 %F, areconnected in parallel, and the resulting combination isconnected to a 9.00-V battery. (a) What is the equivalentcapacitance of the combination? What are (b) the poten-tial difference across each capacitor and (c) the chargestored on each capacitor?

17. What If? The two capacitors of Problem 16 are nowconnected in series and to a 9.00-V battery. Find (a) theequivalent capacitance of the combination, (b) the poten-tial difference across each capacitor, and (c) the charge oneach capacitor.

18. Evaluate the equivalent capacitance of the configurationshown in Figure P26.18. All the capacitors are identical,and each has capacitance C.

13.

A 50.0-m length of coaxial cable has an inner conduc-tor that has a diameter of 2.58 mm and carries a charge of8.10 %C. The surrounding conductor has an inner diame-ter of 7.27 mm and a charge of $ 8.10 %C. (a) What is thecapacitance of this cable? (b) What is the potential differ-ence between the two conductors? Assume the regionbetween the conductors is air.

12. A 20.0-%F spherical capacitor is composed of two concen-tric metal spheres, one having a radius twice as large as the

11.

19. Two capacitors when connected in parallel give an equiva-lent capacitance of 9.00 pF and give an equivalent capaci-tance of 2.00 pF when connected in series. What is thecapacitance of each capacitor?

20. Two capacitors when connected in parallel give an equiva-lent capacitance of C p and an equivalent capacitance of Cswhen connected in series. What is the capacitance of eachcapacitor?

Four capacitors are connected as shown in FigureP26.21. (a) Find the equivalent capacitance betweenpoints a and b. (b) Calculate the charge on each capacitorif !Vab # 15.0 V.

21.

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824 C H A P T E R 2 6 • Capacitance and Dielectrics

22. Three capacitors are connected to a battery as shown inFigure P26.22. Their capacitances are C 1 # 3C, C 2 # C,and C 3 # 5C. (a) What is the equivalent capacitance ofthis set of capacitors? (b) State the ranking of the capaci-tors according to the charge they store, from largest tosmallest. (c) Rank the capacitors according to the poten-tial differences across them, from largest to smallest.(d) What If? If C 3 is increased, what happens to thecharge stored by each of the capacitors?

line with capacitance 29.8 %F between A and B. What addi-tional capacitor should be installed in series or in parallelin that circuit, to meet the specification?

A group of identical capacitors is connected first in seriesand then in parallel. The combined capacitance in parallelis 100 times larger than for the series connection. Howmany capacitors are in the group?

26. Consider three capacitors C1, C 2, C 3, and a battery. If C1 isconnected to the battery, the charge on C1 is 30.8 %C. NowC1 is disconnected, discharged, and connected in serieswith C 2. When the series combination of C 2 and C1 is con-nected across the battery, the charge on C 1 is 23.1 %C. Thecircuit is disconnected and the capacitors discharged.Capacitor C 3, capacitor C1, and the battery are connectedin series, resulting in a charge on C1 of 25.2 %C. If, afterbeing disconnected and discharged, C 1, C 2, and C3 areconnected 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 forthe group of capacitors connected as shown in FigureP26.27. Take C1 # 5.00 %F, C2 # 10.0 %F, and C 3 # 2.00 %F.

25.

6.00 µF

20.0 µF

3.00 µF15.0 µF

a b

µ µ

µ

µFigure P26.21

C2 C3

C1

Figure P26.22

C1 C2

S2S1

∆V

Figure P26.23

C2 C2

C1 C1

C2 C2

C3

b

a

Figure P26.27 Problems 27 and 28.

ba

6.0 µF

5.0 µF

7.0 µF

4.0 µFµ

µ

µ

µ

Figure P26.29

Consider the circuit shown in Figure P26.23, whereC 1 # 6.00 %F, C 2 # 3.00 %F, and !V # 20.0 V. CapacitorC 1 is first charged by the closing of switch S1. Switch S1 isthen opened, and the charged capacitor is connected tothe uncharged capacitor by the closing of S2. Calculatethe initial charge acquired by C 1 and the final charge oneach capacitor.

23.

24. According to its design specification, the timer circuitdelaying the closing of an elevator door is to have a capaci-tance of 32.0 %F between two points A and B. (a) Whenone circuit is being constructed, the inexpensive butdurable capacitor installed between these two points isfound to have capacitance 34.8 %F. To meet the specifica-tion, one additional capacitor can be placed between thetwo points. Should it be in series or in parallel with the34.8-%F capacitor? What should be its capacitance?(b) What If? The next circuit comes down the assembly

28. For the network described in the previous problem, if thepotential difference between points a and b is 60.0 V, whatcharge is stored on C3?

29. Find the equivalent capacitance between points a and b inthe combination of capacitors shown in Figure P26.29.

30. Some physical systems possessing capacitance continuouslydistributed over space can be modeled as an infinite arrayof discrete circuit elements. Examples are a microwavewaveguide and the axon of a nerve cell. To practice analy-

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Problems 825

sis of an infinite array, determine the equivalent capaci-tance C between terminals X and Y of the infinite set ofcapacitors represented in Figure P26.30. Each capacitorhas capacitance C 0. Suggestion: Imagine that the ladder iscut at the line AB, and note that the equivalent capaci-tance of the infinite section to the right of AB is also C.

person. A particular device can be destroyed by adischarge releasing an energy of 250 %J. To what voltageon the body does this correspond?

36. A uniform electric field E # 3 000 V/m exists within acertain region. What volume of space contains an energyequal to 1.00 * 10$7 J? Express your answer in cubicmeters and in liters.

A parallel-plate capacitor has a charge Q and plates ofarea A. What force acts on one plate to attract it towardthe other plate? Because the electric field between theplates is E # Q /A)0, you might think that the force isF # QE # Q 2/A)0. This is wrong, because the field E in-cludes contributions from both plates, and the field cre-ated by the positive plate cannot exert any force on thepositive plate. Show that the force exerted on each plate isactually F # Q 2/2)0A. (Suggestion: Let C # )0A/x for anarbitrary plate separation x ; then require that the workdone in separating the two charged plates be W # )F dx.)The force exerted by one charged plate on another issometimes used in a machine shop to hold a workpiecestationary.

38. The circuit in Figure P26.38 consists of two identical paral-lel metal plates connected by identical metal springs to a100-V battery. With the switch open, the plates areuncharged, are separated by a distance d # 8.00 mm, andhave a capacitance C # 2.00 %F. When the switch is closed,the distance between the plates decreases by a factor of0.500. (a) How much charge collects on each plate and(b) what is the spring constant for each spring? (Sugges-tion: Use the result of Problem 37.)

37.

C0

C0

C0

X

Y

A

B

C0

Figure P26.30

+ –

kk

d

∆V

S

Figure P26.38

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 thecapacitor had been connected to a 6.00-V battery, howmuch energy would have been stored?

32. The immediate cause of many deaths is ventricular fibrilla-tion, uncoordinated quivering of the heart as opposed toproper beating. An electric shock to the chest can causemomentary paralysis of the heart muscle, after which theheart will sometimes start organized beating again. A defib-rillator (Fig. 26.14) is a device that applies a strong electricshock 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 pad-dles, about 8 cm across and coated with conducting paste,are held against the chest on both sides of the heart. Theirhandles are insulated to prevent injury to the operator, whocalls, “Clear!’’ and pushes a button on one paddle to dis-charge the capacitor through the patient’s chest. Assumethat an energy of 300 J is to be delivered from a 30.0-%F ca-pacitor. To what potential difference must it be charged?

33. Two capacitors, C1 # 25.0 %F and C2 # 5.00 %F, are con-nected in parallel and charged with a 100-V power supply.(a) Draw a circuit diagram and calculate the total energystored in the two capacitors. (b) What If? What potentialdifference would be required across the same two capaci-tors connected in series in order that the combinationstores the same amount of energy as in (a)? Draw a circuitdiagram of this circuit.

34. A parallel-plate capacitor is charged and then discon-nected from a battery. By what fraction does the storedenergy change (increase or decrease) when the plate sepa-ration is doubled?

35. As a person moves about in a dry environment, electriccharge accumulates on his body. Once it is at high voltage,either positive or negative, the body can discharge viasometimes noticeable sparks and shocks. Consider ahuman body well separated from ground, with the typicalcapacitance 150 pF. (a) What charge on the body willproduce a potential of 10.0 kV? (b) Sensitive electronicdevices can be destroyed by electrostatic discharge from a

39. Review problem. A certain storm cloud has a potential of1.00 * 108 V relative to a tree. If, during a lightning storm,50.0 C of charge is transferred through this potential dif-ference and 1.00% of the energy is absorbed by the tree,how much sap in the tree can be boiled away? Model thesap as water initially at 30.0°C. Water has a specific heat of4 186 J/kg°C, a boiling point of 100°C, and a latent heat ofvaporization of 2.26 * 106 J/kg.

40. Two identical parallel-plate capacitors, each with capaci-tance C , are charged to potential difference !V and con-nected in parallel. Then the plate separation in one of thecapacitors is doubled. (a) Find the total energy of thesystem of two capacitors before the plate separation isdoubled. (b) Find the potential difference across eachcapacitor after the plate separation is doubled. (c) Find the

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826 C H A P T E R 2 6 • Capacitance and Dielectrics

total energy of the system after the plate separation is dou-bled. (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 sphereof radius R and charge Q surrounded by a vacuum isU # ke Q 2/2R .

42. Consider two conducting spheres with radii R1 and R2.They are separated by a distance much greater than eitherradius. A total charge Q is shared between the spheres,subject to the condition that the electric potential energyof the system has the smallest possible value. The totalcharge Q is equal to q1 & q2, where q1 represents thecharge on the first sphere and q 2 the charge on the sec-ond. Because the spheres are very far apart, you canassume that the charge of each is uniformly distributedover its surface. You may use the result of Problem 41.(a) Determine the values of q1 and q2 in terms of Q , R1,and R2. (b) Show that the potential difference betweenthe spheres is zero. (We saw in Chapter 25 that two con-ductors joined by a conducting wire will be at the samepotential in a static situation. This problem illustrates thegeneral principle that static charge on a conductor willdistribute itself so that the electric potential energy of thesystem is a minimum.)

Section 26.5 Capacitors with Dielectrics

43. Determine (a) the capacitance and (b) the maximumpotential difference that can be applied to a Teflon-filledparallel-plate capacitor having a plate area of 1.75 cm2 andplate separation of 0.040 0 mm.

44. (a) How much charge can be placed on a capacitor withair between the plates before it breaks down, if the area ofeach of the plates is 5.00 cm2? (b) What If? Find the maxi-mum charge if polystyrene is used between the platesinstead of air.

A commercial capacitor is to be constructed as shown inFigure 26.17a. This particular capacitor is made from twostrips of aluminum separated by a strip of paraffin-coatedpaper. Each strip of foil and paper is 7.00 cm wide. Thefoil is 0.004 00 mm thick, and the paper is 0.025 0 mmthick and has a dielectric constant of 3.70. What lengthshould the strips have, if a capacitance of 9.50 * 10$8 F isdesired before the capacitor is rolled up? (Adding asecond strip of paper and rolling the capacitor effectivelydoubles its capacitance, by allowing charge storage onboth sides of each strip of foil.)

46. The supermarket sells rolls of aluminum foil, of plasticwrap, and of waxed paper. Describe a capacitor made fromsupermarket materials. Compute order-of-magnitude esti-mates for its capacitance and its breakdown voltage.

47. A parallel-plate capacitor in air has a plate separation of1.50 cm and a plate area of 25.0 cm2. The platesare charged to a potential difference of 250 V anddisconnected from the source. The capacitor is thenimmersed in distilled water. Determine (a) the chargeon the plates before and after immersion, (b) the capaci-tance and potential difference after immersion, and(c) the change in energy of the capacitor. Assume theliquid is an insulator.

45.

48. A wafer of titanium dioxide (1 # 173) of area 1.00 cm2 hasa thickness of 0.100 mm. Aluminum is evaporated on theparallel faces to form a parallel-plate capacitor. (a) Calcu-late the capacitance. (b) When the capacitor is chargedwith a 12.0-V battery, what is the magnitude of charge de-livered to each plate? (c) For the situation in part (b),what are the free and induced surface charge densities?(d) What is the magnitude of the electric field?

49. Each capacitor in the combination shown in Figure P26.49has a breakdown voltage of 15.0 V. What is the breakdownvoltage of the combination?

20.0 µF

10.0 µF

20.0 µF

20.0 µF

20.0 µF

µ

µ

µ

µ

µ

Figure P26.49

Section 26.6 Electric Dipole in an Electric Field

50. A small rigid object carries positive and negative 3.50-nCcharges. It is oriented so that the positive charge has coor-dinates ($ 1.20 mm, 1.10 mm) and the negative charge isat the point (1.40 mm, $ 1.30 mm). (a) Find the electricdipole moment of the object. The object is placed in anelectric field E # (7 800 i $ 4 900 j) N/C. (b) Find thetorque acting on the object. (c) Find the potential energyof the object–field system when the object is in this orien-tation. (d) If the orientation of the object can change, findthe difference between the maximum and minimumpotential energies of the system.

51. A small object with electric dipole moment p is placed in anonuniform electric field E # E(x) i. That is, the field is inthe x direction and its magnitude depends on the coordi-nate x. Let 2 represent the angle between the dipolemoment and the x direction. (a) Prove that the dipolefeels a net force

in the direction toward which the field increases. (b) Con-sider a spherical balloon centered at the origin, withradius 15.0 cm and carrying charge 2.00 %C. EvaluatedE/dx at the point (16 cm, 0, 0). Assume a water droplet atthis point has an induced dipole moment of 6.30 i nC + m.Find the force on it.

Section 26.7 An Atomic Description of Dielectrics

52. A detector of radiation called a Geiger tube consists of aclosed, hollow, conducting cylinder with a fine wire along itsaxis. Suppose that the internal diameter of the cylinder is2.50 cm and that the wire along the axis has a diameter of0.200 mm. The dielectric strength of the gas between thecentral wire and the cylinder is 1.20 * 106 V/m. Calculatethe maximum potential difference that can be appliedbetween the wire and the cylinder before breakdown occursin the gas.

F # p $ dEdx % cos 2

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Problems 827

53. The general form of Gauss’s law describes how a chargecreates an electric field in a material, as well as in vacuum.It is

where ) # 1)0 is the permittivity of the material. (a) Asheet with charge Q uniformly distributed over its area A issurrounded by a dielectric. Show that the sheet creates auniform electric field at nearby points, with magnitudeE # Q /2A). (b) Two large sheets of area A, carrying oppo-site charges of equal magnitude Q , are a small distanced apart. Show that they create uniform electric field inthe 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 isA)/d # 1A)0/d.

Additional Problems

54. For the system of capacitors shown in Figure P26.54, find(a) the equivalent capacitance of the system, (b) thepotential across each capacitor, (c) the charge on eachcapacitor, and (d) the total energy stored by the group.

* E+d A #q)

capacitance of the three-plate system P1P2P3? (b) What isthe charge on P2? (c) If P4 is now connected to the positiveterminal of the battery, what is the capacitance of the four-plate system P1P2P3P4? (d) What is the charge on P4?

56. One conductor of an overhead electric transmission line isa long aluminum wire 2.40 cm in radius. Suppose that at aparticular moment it carries charge per length 1.40 %C/mand is at potential 345 kV. Find the potential 12.0 m belowthe wire. Ignore the other conductors of the transmissionline and assume the electric field is everywhere purelyradial.

57. Two large parallel metal plates are oriented horizontallyand separated by a distance 3d. A grounded conductingwire joins them, and initially each plate carries no charge.Now a third identical plate carrying charge Q is insertedbetween the two plates, parallel to them and located a dis-tance d from the upper plate, as in Figure P26.57. (a) What induced charge appears on each of the two origi-nal plates? (b) What potential difference appears betweenthe middle plate and each of the other plates? Each platehas area A.

4.00 µF2.00 µF

6.00 µF3.00 µF

90.0 V

µ µ

µ µ

Figure P26.54

12.0 V

P2 P3 P4P1

d d d

Figure P26.55

2d

d

Figure P26.57

55. Four parallel metal plates P1, P2, P3, and P4, each of area7.50 cm2, are separated successively by a distance d #1.19 mm, as shown in Figure P26.55. P1 is connected to thenegative terminal of a battery, and P2 to the positive termi-nal. The battery maintains a potential difference of 12.0 V.(a) If P3 is connected to the negative terminal, what is the

58. A 2.00-nF parallel-plate capacitor is charged to an initialpotential difference !Vi # 100 V and then isolated. Thedielectric material between the plates is mica, with adielectric constant of 5.00. (a) How much work is requiredto withdraw the mica sheet? (b) What is the potentialdifference of the capacitor after the mica is withdrawn?

A parallel-plate capacitor is constructed using adielectric material whose dielectric constant is 3.00and whose dielectric strength is 2.00 * 108 V/m. Thedesired capacitance is 0.250 %F, and the capacitor mustwithstand a maximum potential difference of 4 000 V. Findthe minimum area of the capacitor plates.

60. A 10.0-%F capacitor has plates with vacuum between them.Each plate carries a charge of magnitude 1 000 %C. Aparticle with charge $ 3.00 %C and mass 2.00 * 10$16 kgis fired from the positive plate toward the negative platewith an initial speed of 2.00 * 106 m/s. Does it reach thenegative plate? If so, find its impact speed. If not, whatfraction of the way across the capacitor does it travel?

61. A parallel-plate capacitor is constructed by fillingthe space between two square plates with blocks of threedielectric materials, as in Figure P26.61. You may assumethat ! ,, d. (a) Find an expression for the capacitance ofthe device in terms of the plate area A and d, 11, 12, and13. (b) Calculate the capacitance using the valuesA # 1.00 cm2, d # 2.00 mm, 11 # 4.90, 12 # 5.60, and13 # 2.10.

59.

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828 C H A P T E R 2 6 • Capacitance and Dielectrics

62. A 10.0-%F capacitor is charged to 15.0 V. It is nextconnected in series with an uncharged 5.00-%F capacitor.The series combination is finally connected across a 50.0-Vbattery, as diagrammed in Figure P26.62. Find the newpotential differences across the 5-%F and 10-%F capacitors.

sent the potential difference. (c) Find the direction andmagnitude of the force exerted on the dielectric, assuminga 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 dielec-tric is glass (1 # 4.50). (Suggestion: The system can be con-sidered 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 mayassume 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 distancex into the capacitor. The charges on the plates are notdisturbed as the block slides in. In a static situation, ametal prevents an electric field from penetrating inside it.The metal can be thought of as a perfect dielectric, with1 : 0. (a) Calculate the stored energy as a function of x.(b) Find the direction and magnitude of the force thatacts on the metallic block. (c) The area of the advancingfront face of the block is essentially equal to !d. Consider-ing the force on the block as acting on this face, find thestress (force per area) on it. (d) For comparison, expressthe energy density in the electric field between the capaci-tor 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 impor-tant parameter. Using the following data, compare theenergy per unit mass ( J/kg) for gasoline, lead–acid batter-ies, and capacitors. (The ampere A will be introduced inthe 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 battery: 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.

An isolated capacitor of unknown capacitance has beencharged to a potential difference of 100 V. When thecharged capacitor is then connected in parallel to anuncharged 10.0-%F capacitor, the potential differenceacross the combination is 30.0 V. Calculate the unknowncapacitance.

68. To repair a power supply for a stereo amplifier, an elec-tronics technician needs a 100-%F capacitor capable ofwithstanding a potential difference of 90 V between theplates. The only available supply is a box of five 100-%Fcapacitors, each having a maximum voltage capability of50 V. Can the technician substitute a combination of thesecapacitors that has the proper electrical characteristics? Ifso, what will be the maximum voltage across any of thecapacitors used? (Suggestion: The technician may not haveto use all the capacitors in the box.)

A parallel-plate capacitor of plate separation d is chargedto a potential difference !V0. A dielectric slab of thicknessd and dielectric constant 1 is introduced between theplates while the battery remains connected to the plates.(a) Show that the ratio of energy stored after the dielectricis introduced to the energy stored in the empty capacitoris U/U0 # 1. Give a physical explanation for this increasein stored energy. (b) What happens to the charge on thecapacitor? (Note that this situation is not the same as in

69.

67.

d/2

!/2

d

!

1κ2κ

Figure P26.61

5.00 Fµ

50.0 V

∆Vi = 15.0 V

–+10.0 Fµ

Figure P26.62

xd

!

κ

Figure P26.64 Problems 64 and 65.

63. (a) Two spheres have radii a and b and their centers are adistance d apart. Show that the capacitance of this system is

provided that d is large compared with a and b. (Suggestion:Because the spheres are far apart, assume that the poten-tial of each equals the sum of the potentials due to eachsphere, and when calculating those potentials assume thatV # keQ /r applies.) (b) Show that as d approaches infinitythe above result reduces to that of two spherical capacitorsin series.

64. A capacitor is constructed from two square plates of sides !and separation d. A material of dielectric constant 1is inserted a distance x into the capacitor, as shown inFigure P26.64. Assume that d is much smaller than x.(a) Find the equivalent capacitance of the device. (b) Cal-culate the energy stored in the capacitor, letting !V repre-

C #4()0

1a

&1b

$2d

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Answers to Quick Quizzes 829

Example 26.7, in which the battery was removed from thecircuit before the dielectric was introduced.)

70. A vertical parallel-plate capacitor is half filled with a dielec-tric for which the dielectric constant is 2.00 (Fig. P26.70a).When this capacitor is positioned horizontally, whatfraction of it should be filled with the same dielectric (Fig.P26.70b) in order for the two capacitors to have equalcapacitance?

75. Determine the equivalent capacitance of the combinationshown in Figure P26.75. (Suggestion: Consider the symme-try involved.)

(b)(a)

Figure P26.70

a

b2.00 µF

4.00 µF

2.00 µF 4.00 µF8.00 µF

µ

µ µ

µ

µ

Figure P26.72

C

C

3C

2C

2CFigure P26.75

71. Capacitors C1 # 6.00 %F and C2 # 2.00 %F are charged asa parallel combination across a 250-V battery. The capaci-tors are disconnected from the battery and from eachother. They are then connected positive plate to negativeplate and negative plate to positive plate. Calculate theresulting charge on each capacitor.

72. Calculate the equivalent capacitance between the points aand b in Figure P26.72. Note that this is not a simple seriesor parallel combination. (Suggestion: Assume a potentialdifference !V between points a and b. Write expressionsfor !V ab in terms of the charges and capacitances for thevarious possible pathways from a to b, and require conser-vation of charge for those capacitor plates that are con-nected to each other.)

The inner conductor of a coaxial cable has a radius of0.800 mm, and the outer conductor’s inside radius is3.00 mm. The space between the conductors is filled withpolyethylene, which has a dielectric constant of 2.30 and adielectric strength of 18.0 * 106 V/m. What is the maxi-mum potential difference that this cable can withstand?

74. You are optimizing coaxial cable design for a major manu-facturer. Show that for a given outer conductor radius b,maximum potential difference capability is attained whenthe radius of the inner conductor is a # b/e where e is thebase of natural logarithms.

73.

76. Consider two long, parallel, and oppositely charged wiresof radius d with their centers separated by a distance D.Assuming the charge is distributed uniformly on thesurface of each wire, show that the capacitance per unitlength of this pair of wires is

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 moreeffective in terms of increasing the capacitance than increas-ing 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 systemand does not vary with applied voltage. According toEquation 26.1, if the voltage is doubled, the charge isdoubled.

26.2 (a). When the key is pressed, the plate separation isdecreased and the capacitance increases. Capacitancedepends only on how a capacitor is constructed and noton the external circuit.

26.3 (a). When connecting capacitors in series, the inversesof the capacitances add, resulting in a smaller overallequivalent capacitance.

26.4 (a). When capacitors are connected in series, the volt-ages add, for a total of 20 V in this case. If they are com-bined in parallel, the voltage across the combination isstill 10 V.

26.5 (b). For a given voltage, the energy stored in a capacitoris proportional to C : U # C(!V )2/2. Thus, you want tomaximize the equivalent capacitance. You do this byconnecting the three capacitors in parallel, so that thecapacitances add.

26.6 (a) C decreases (Eq. 26.3). (b) Q stays the same becausethere is no place for the charge to flow. (c) E remainsconstant (see Eq. 24.8 and the paragraph following it).(d) !V increases because !V # Q /C, Q is constant (partb), and C decreases (part a). (e) The energy stored inthe capacitor is proportional to both Q and !V (Eq.

C!

#()0

ln[(D $ d)/d]

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830 C H A P T E R 2 6 • Capacitance and Dielectrics

26.11) and thus increases. The additional energy comesfrom the work you do in pulling the two plates apart.

26.7 (a) C decreases (Eq. 26.3). (b) Q decreases. The batterysupplies a constant potential difference !V ; thus, chargemust flow out of the capacitor if C # Q /!V is todecrease. (c) E decreases because the charge density onthe plates decreases. (d) !V remains constant because ofthe presence of the battery. (e) The energy stored in thecapacitor decreases (Eq. 26.11).

26.8 Increase. The dielectric constant of wood (and of allother insulating materials, for that matter) is greater than1; 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 thebattery maintains a constant !V, Q must increase if Cincreases. (c) E between the plates remains constantbecause !V # Ed and neither !V nor d changes. Theelectric field due to the charges on the plates increasesbecause more charge has flowed onto the plates. Theinduced surface charges on the dielectric create a fieldthat opposes the increase in the field caused by thegreater number of charges on the plates (see Section26.7). (d) The battery maintains a constant !V.