IPR2023-00783
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`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
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`IPR2023-00783
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`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
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`IPR2023-00783
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`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
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`IPR2023-00783
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`Petitioner Intel Corp., Ex. 1038
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`Aeee
`IPR2023-00783
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`7
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`Petitioner Intel Corp., Ex. 1038
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`

`20.1
`
`Potential Difference and Electric Potential
`
`559
`
`
`
`Electric Potential and
`Capacitance
`
`the concept of electric potential is of great practical value. For example, the mea-
`sured voltage between any two points in anelectrical circuit is simply thedifference
`
`in electric potential between the points.
`
`This chapter is also concerned with the properties of capacitors, devices that
`store charge, Capacitors are commonly used in a variety of clectrical circuits. For
`instance, they are used to tune the frequencyof radio receivers, as filters
`in power
`
`supplies, to eliminate unwanted sparking in automobile ignition sys
`. and as
`energy-storing devices in electronic flash units.
`A capacitor basically consists of two conductors separated by an insulator. We
`
`shall see that the capacitanceof a given device depends on its geometry and on the
`material called a dielectric, separating the charged conductors.
`
`20.1 + POTENTIAL DIFFERENCE AND ELECTRIC POTENTIAL
`
`TessofpotentialenergywasintroducedinChapter7inconnec-
`
`ui
`ion with such conservative forces as the force ofgravity, the elastic force
`of a spring, and the electrostatic force. By using the principle ofcon-
`servation of energy, we were often able to avoid working directly with forces
`when solving mechanical problems. In this chapter we shall use the energy
`concept in our study of electricity. Because the electrostatic force (given by
`Coulomb's law) is conservative, electrostatic phenomena canconveniently be
`described in terms of an electric potential energy function. This concept ena-
`bles us to define a quantity called electric potential, which is a scalar function
`ofposition and thus leads toa simpler method of describing some electrostatic
`phenomena than the elec
`tric field method, Although
`the potential is clearly an
`easier path for many prob-
`lems, one often requires E.
`Tn some situations, a knowl
`edge of E provides the sim-
`plest path to calculation of
`the potential. As we shall
`see in subsequent chapters,
`
`andJim Lehman}
`
`Jennifer is holding on toa P
`charged sphere thet reaches.0
`potential of about 100 000
`volts. The device that generates
`this high potentiol is called
`Van de Graaff generator, Why
`do you supposeJennifer's hair
`stands on end ike the needles
`of a porcupine? Why is it im-
`portant that she stand ona
`pedestal insulated fram
`ground? [Courtesy ofManry Leap
`
`Whena test charge qis placed in an electrostatic field E, the electric force on the
`
`chargeis @E. This forceis the vector
`sum of the individual forces exerted on q@ by
`the various charges producing the field E.It follows that the force q,E is consery-
`ative because the individual forces governed by Coulomb's law are conservative.
`Whena charge is moved within an electric field at constant velocity, the work done
`on q) bythe electric field is equal to the negative of the work done by the external
`
`agent causing the displacement, For an infinitesimal displacement ds, the work
`donebythe electric field is F-ds = q,E- ds. This decreases the potential energyof
`the charge—field system by an amount dl! = — q,E- ds. For
`afinite displacement
`of the test charge between points A and 8, the change in potential energy is
`a
`AU=U,- =-0f, E-ds
`[20.1]
`The integral in Equation 20.1 is performed along the path by which g, moves
`from A to B and is called cither a path integral or a line integral. Because the
`force qE is conservative, this integral does not depend on the path taken be-
`tween A and B.
`The potential energy per unit charge, Ugo, is independentofthe value of gy
`
`and has a unique value at every point in anelectric field. The q
`tity U/ qo is called
`the electric potential (or simply the potential), Thus, the electric potential at
`any point in an electricfieldis
`
`[20.2]
`
`Vv
`V=—
`a
`Because potential energy isa scalar, electric potential is also a scalar quantity.
`The potential difference, AV = V,,—
`V,, between the points A and Bis de-
`fined as the changein potential energy divided by the test charge q:
`AU
`a
`[20.3]
`AV=—=— -f E-ds
`%
`A;
`Potential difference should not be confused with potential energy. The potcn
`ial difference is proportional to the potential cnergy difference, and we see from
`
`* Change in potential energy
`
`* Potential difference between
`twopoints
`
`CHAPTER OUTLINE
`Potential Difference and
`Electric Potential
`Potential Differences in a
`UniformElectric Field
`Electric Potential and
`Electric Potential Energy
`Due to Point Charges
`Obtaining E trom the
`Electric Potential
`Electric Potential Due to
`Continuous Charge
`Distributions
`Electric Potential of a
`Charged Conductor
`Capacitance
`Combinations of Capacitors
`Energy Storedin a Charged
`Capacitor
`20.10 Capacitors with Dielectrics
`
`
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`
`20.4
`
`Obtaming E from the Electric Potential
`
`(b)
`fa)
`Figure 20.8 Equipotential surfaces (dashed blue lines) and electric field lines (red lines) for
`(a) a uniform electric field produced by an infinite sheet of charge, (b) a point charge, and
`(c) an electric dipole. Inall cases, the equipotential surfaces are pependicularto the electric field
`lines at every point.
`
`567
`
`
`
`fe}
`
`for any displacement perpendicular to the electric field. This is consistent with the
`notion ofequipotential surfaces being perpendicular to the field, as in Figure 20.82.
`If the charge distribution has spherical symmetry, where the charge den-
`sity depends only on the radial distance, r, then the electric field is radial.
`In this case, E-ds = F, dr, and so we can express dV in the form dV = — £, dr
`Therefore,
`dV
`_=-——
`dr
`
`[20.16]
`
` eis liar result.
`
`Chapter 200
`
`Electric Potential and Capacitance
`
`Solution
`
`1o“* C}(—6.29 x 10° V)
`
`W= mVp= (3.00 *
`= -18.9% 10-5]
`The negative sign means that workis done by the field on the
`charge as it is displaced from infinity to P. Therefore, positive
`
`work would have to be done by an external agent to remove
`the charge from P back to infinity,
`EXERCISE 3 Find the total potential energy of the system
`of three charges in the configuration shown in Figure 20.7b.
`Answer —5.48 x 10°74
`
`
`
`th)
`fa)
`Figure 20.7
`(Exarmple 20.5) (a) The electric potential atthe point /* due wo the nwo point
`
`charges gq, and q t
`© algebraic sum of the potentials due to the individual charges. (b) What
`isthe potentia
`energy of the system of three changes?
`
`(a) Find the potential ata distance of 1.0 cm from a proton. (b) What is the
`EXERCISE 4
`potential difference between two points that are 1.0 em and 2.0 emfrom a proton? (Take
`V=Oatr==)
`Answer
`(a) 14X10°7V (b) -7.2% 108
`
`20.4 + OBTAINING E FROM THE ELECTRIC POTENTIAL
`
`
`For example, the potential of a point charge is V = h,q/r. Because Vis a function
`of ronly, the potential function has spherical symmetry. Applying Equation 20.16,
`wefind that the electric field due to the point cha
`kgf Pa Fan
`
`
`direc
`
`
`Note that the potential changes only in the
`radial direction, not
`
`Computer-generatedplot of the
`perpendicular to r. Thus, V
`(like £,)
`isa function only of r Again, this is consistent
`Theelectric field E and the electric potential Vare related by Equation 20,3, Both
`electric potential associated with
`with theidea that equipotential surfaces are perpendicular to field lines. In this
`quantities are determined by a specific charge distribution. We now show how to
`an eleciric dipole, The charges lie
`case the equipotential surfaces are a family of spheres concentric with the spheri-
`calculate the electric field if the electric potential is known im a certain region.
`in the horizontal plane, at the cen-
`
`cally symmetric charge distribution (Fig. 20.8b). The equipotential surfaces for the
`From Equation 20.3 we can express the potential difference dV between two
`ters of the potential spikes, The
`electric dipoleare sketchedin Figure 20.8c.
`points a distance ds apart as
`contour lines help visualize the size
`of the potential the values of which
`[20.14]
`dV= —E-ds
`(RiekRicher
`CONCEPTUAL PROBLEM 3ReCaeRe
`are plotted vertic
`
`nagreaphs,
`NYE
`Mose
`If the electric field has only ome component, F_, then E-ds = E, dx. Therefore,
`Can electric field lines ever cross? Why or why not? Can equipotential surfaces ever cross?
`Equation 20.14 becomes dV = —E, dx, o1
`Why or whynow
`
`dV
`[20.15]
`CONCEPTUAL PROBLEM 4
`ar
`ge
`reeen
`Su
`
`Thatis, the electric field is equal to the negative ofthe derivative of the electric
`ase you know the value of the electric potential
`at
`ome point. Can you find the electric
`field at that pomt fromthis information?
`potential with respect to some coordinate. Note that the potential change is zero
`
`
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`Petitioner Intel Corp., Ex. 1038
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`Petitioner Intel Corp., Ex. 1038
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`IPR2023-00783
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`Petitioner Intel Corp., Ex. 1038
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`IPR2023-00783
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`Petitioner Intel Corp., Ex. 1038
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`Petitioner Intel Corp., Ex. 1038
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`Chapter 200
`
`Electric Potential and Capacitance
`
`20.8
`
`Combinations of Capacitors
`
`579
`
`combination of ca-
`Thus, we see that the equivalent capacitance of a
`pacitors is larger than any ofthe individual capacitances and is the algebraic
`sum of the individual capacitances.
`EXERCISE 8 Two capacitors, C,; = 5.0 wF and C, = 12 uF, are connected in parallel, and
`and C,. In
`pacitors €,
`where AV, and AV, are the potential differences acro:
`the resulting combination is connected to a 9.0-V battery. (a) Whatis the value of the equi~
`
`general the potential difference across any number of
`icitors in series is equal
`
`alent capacitance of the combination? What is (b) the potential difference across each
`to the sum of the potential differences across the individual capacitors. Bec:
`
`capacitor and (c) the charge stored on each capacitor?=Answer (a)l7 uF (b) 9.0V
`(c) 45 wC and 108 aC
`Q = CAVcanbeapplied to each capacitor, the potential difference across each is
`
`where AVis the potential difference between the terminals of the battery and C,,,
`is the equivalent capacitance. From Figure 20.22a, we see that
`AV=AV,+ AY
`
`[20.26]
`
`Series Combination
`
`Now consider two capacitors connected in series, as illustrated in Figure 20.22a.
`For this series combination of capacitors, the magnitude of the charge must be
`the same on all the plates.
`To see why this must be true, let us consider the charge transfer process in
`some detail. We start with uncharged capacitors and follow what happensjust after
`a battery is connected to the circuit, When the connection is made,the right plate
`of C, and theleft plate of Cy form an isolated conductor. Thus, whatever negative
`charge enters one plate must be equal to the positive charge of the other plate, to
`maintain neutrality of the isolated conductor, As a result, both capacitors must have
`the same charge.
`Suppose an equivalent capacitor performs the same function as the series com-
`bination, After it is fully charged, the equivalent capacitor must end up with a
`charge of — Q on its right plate and + Q on its left plate. By applying the defi-
`nition ofcapacitance to the circuit shown in Figure 20.22b, we have
`ui
`AV= Cn
`
`aes
`
`2
`
`oo
`
`|
`
`+
`
`2-2
`
`
`OU w
`
`
`
`
`+
`
`Ls
`
`+
`
`=)
`
`a Bo
`
`
`Avy, = g iniG ‘
`
`
`Substituting these expressions into Equation 20.26, and noting that AV = Q/C,
`we have
`
`‘eq?
`
`22,2
`Coq
`G
`Cy
`
`Canceling Q, we arrive at the relationship
`1
`1
`1=—-+
`— 7 —
`Pec te
`If this analysis is applied to three or more capacitors connected in series, the equiv-
`alent capacitance is foundto be
`
`(series combination)
`
`[20.27]
`
`1
`1
`
`mor G + c eee
`
`(series combination)
`
`[20.28]
`
`This showsthat the equivalent capacitance ofa series combinationis always less
`than any individual capacitance in the combination and the inverse of the
`equivalent capacitance is the algebraic sum of the inverses of the individual
`capacitances.
`
`Example 20.9 Equivalent Capacitance
`Findtheequivalent capacitance between aand bior the com-
`
`bination of capacitors shown in Figure 20.2:
`| capaci-
`tances are in microfarads
`
`|
`1
`|
`€ te
`C..
`= 2.0 pF
`
`10 uF
`
`4 0 pe
`
`2.0 pF
`
`:
`reduce the
`Using Equations 20.25 and 20.28, we
`The
`combination step by step as indicated in the
`msists of Iwo
`Likewise, the lower branch in Figure 20.23b ¢
`ne ac
`
`1.0-F and 3.0-4F capacitors are in parallel and
`ib)
`(a)
`lent of
`
`B0-uF capacitors in series,
`nce is
`cording to C_.
`= G + C,. Their equivalent ¢
`F
`¢
`LciLOrs
`I
`c
`
`
`
`
`4.0 F. Likewise, alscoin=4.0 uF, Finally, the 2.0.8the 2.0-uF and 6.0-uF capacite
`Figure 20.22 A series combination of two capacitors. The charge on each capacitoris the
`
`1
`1
`I
`1
`nt apaciiance of
`
`
`parallel and have an equivalent capacitance «
`. The
`20.23c arein parallel and
`: of
`the circuit
`is
`
`same, and the equivalent capacitance can be calculated fromthe relationship —+—
`nq teats
`
`two 4,0yak
`upper branch in Figure 20.23b now con
`6.0 pF. Hence.
`the equiv
`Capacitors in series, which combine accord
`6.0 uF, as shown in Figure
`
`
`
`
`
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`585
`Electric Potential and Capacitance
`Chapter 20
`20.10©Capacitors with Dieleetrics
`
`The capacitance ofafilled «
`capacitor is greater than that
`ofan empty one by the
`factor «.
`
`[20.32]
`C= KC,
`where Cyis the capacitance in the absence ofthe dielectric. Thatis, the capacitance
`increases by the factor k when the dielectric completelyfills the region between the
`plates." For a parallel-plate capacitor, where Co = €)A/d, we can express the ca-
`pacitance whenthe capacitor is filled with a dielectric as
`eA
`d
`
`[20.33]
`
`c=«
`
`From this result, it would appear that the capacitance could be made very large
`by decreasing d, the distance between the plates. In practice, the lowest value of d
`is limited by the electrical discharge that could occur through the dielectric me
`dium separating the plates. For any given separation d, the maximum voltage that
`can be applied to a capacitor without causing a dischange depends on the dielectric
`strength (maximum electric field intensity) of the dielectric, which forair is equal
`to3 X 10° V/m. If the field strength in the medium exceeds the dielectric strength,
`the insulating properties break down and the medium begins to conduct. Most
`insulating materials have dielectric strengths and dielectric constants greater than
`those of air, as Table 20.1 indicates. Thus, we see that a dielectric provides the
`following advantages:
`*
`It increases the capacitance of a capacitor.
`*
`It increases the maximum operating voltage ofa capacitor.
`*
`It may provide mechanical support between the conducting plates.
`
`TABLE 20.1 Dielectric Constants and Dielectric Strengths
`of Various Materials at Room Temperature
`Dielectric
`Strength* (V/m)
`Dielectric Constant «
`Material
`—
`1.00000
`Vacuum
`3x 16
`1.00059
`Air (dry)
`24x 10°
`49
`Bakelite
`8x 106
`3.78
`Fused quartz
`14x 10°
`5.6
`Pyrex
`glass
`
`24% 10°
`256
`Polystyrene
`60 x 10%
`2.1
`Teflon
`12 x 10°
`6.7
`Neoprene rubber
`14 x 10°
`34
`Nvlon
`16 x 10°
`3.7
`Paper
`ax 10°
`233
`Strontiumtitanate
`_—
`a0
`oT
`Vv
`
`
`
`25Siliconeoil 15 x 10°
`
`field that can exist in a dielectric
`* The dielectric strength equals the maximumcle
`without electrical breakdown
`
`Ifanother experiment is performedin which the dielectric is introduced while the potential difference
`
`is held constant by means of a battery,
`t
`hargeincreases to the value Q = «Q,
`Theadditional charge
`is suppliedby the battery, and the capacitance still increases by thefactor «
`
`Types of Capacitors
`Commerical capacitors are often madeusing metal foil interlaced with a dielectric
`such as thin sheets of paraffin-impregnated paper. These alternate layers of metal
`
`foil and dielectric are then rolled into the shape of a cylinder to form a
`all
`package. High-voltage capacitors commonly consist of interwoven metal plates im-
`mersedin siliconeoil. Small capacitors are often constructed from ceramic mate-
`rials. Variable capacitors (typically 10 pF to 500 pF) usually consist of two interwoven
`sets of metal plates, one fixed and the other movable, with air as the dielectric.
`
`An electrolytic capacitor is often used to store large
`amounts of chargeat relatively
`
`lowvoltages. This device consists of a metal foil in contact with an electrolyte—a
`solution that conducts electricity by virtue of the motion of ions contained in the
`solution. When a voltageis applied between thefoil and the electrolyte, a thin layer
`of metal oxide (an insulator) is formed on thefoil, and this layer serves as the
`dielectric, Very large capacitance values can beattained because thedielectric layer
`is verythin.
`Whenelectrolytic capacitors are usedin circuits, the polarity must be installed
`properly.
`[f the polarity of the applied voltage is opposite whatis intended, the
`oxide layer will be removed and the capacitor will not be able to store charge.
`
`ThinkingPhysics 6
`
`Consider a parallel-plate capacitor with a dielectric material between theplates. Is the
`
`capacitance higher on a cold dayor a hot day
`fing [tis the polarization of the molecules in the dielectric that ine
`
`capacitance when the dielectric is added. As the temperatureincreases, there is more
`vibrati
`1 motion of the polarized molecules. This disturbs the orderly arrangement
`afthe polarized molecules, and thenet polarization decreases. Thus, the capacitance
`must decrease as the temperature increases.
`
`CONCEPTUAL PROBLEM 8
`
`It consists of a parallel-plate
`An electric stud finder will locate the studs in a wall in a home.
`capacitor, with the plates next to each other, as shown in Figure 20.26, How does this device
`detect the presence of a wooden stud in the wall?
`
`Figure 20.26
`stud
`
`ic stud finder
`(Conceptual Problem 8) An cle«
`over
`a wooden
`and (b)
`
`
`Sturt
`
` positioned (a) on a piece of wall |
`
`
`
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`

`587
`een
`and aslab of polystyrene (x = 2,56) is inserted between the
`EXERCISE 13 Supposethat the capacitancein the absence
`Example 20.11 A Paper-Filled Capacitor
`plates, calculate the energy difference U' — Lh.
`of a dielectric is 8.50 pF, and the capacitor is charged to a
`
`
`
`A parallel-plate capacitor has plates of dimensions 2.0cm * potentialdifference of12.0 V, If the batteryis disconnected==Answeris 1.0 mm,the maximum voltage that can be applied before 373 pl
`3.0 cm separated by a 1.0-mm thickness of paper. (a) Find—_breakdown is
`the capacitance of this device,
`<
`
`
`en Because x = 3.7 for paper (Table 20.1), we get (a) How much charge can beplaced onacapacitor with air between theAVrax = Emax = (1 = 1=)0a EXERCISE 14
`
`
`aA
`ct
`6.0 x 10-* m*®
`= 16x 10°V
`plates beforeit breaks down if the area of cach of the plates
`is 5.00 cm*? (b) Find the max-
`imumchargeif polystyrene isused betweenthe plates insteadofair.
`C=*«—= 3.7( B85 x 107! 3) are)
`Answer
`{a) 15.30C
`.
`naw a ee
`Hence, the maximum charge is
`(b) 272 nC
`= 20% 10-2 F = 90 pF
`Onan = CAVing, = (20 ¥ 107! FYE X 10° V)
`= 032 4c
`
`Chapter 20.
`
`Electric Potential and Capacitance
`
`(b) What is the maximum charge that can be placed on
`the capacitor?
`
`on From Table 20.1 we see that the dielectric strength
`of paper is 16
`10° Vm. Because the thickness of the paper
`
`EXERCISE 12. What is the maximum energy that can be
`stored in the capacitor?
`Answer
`2.5% 10734
`
`Example 20.12 Energy Stored Before and After Inserting a Dielectric
`sca
`:
`aeenaea
`‘siaks Gk wawinvial teat Ines a: Ginlecinic Goneikberied
`between the plates, as in Figure 20.27b. Find the energy
`stored in the capacitor before and after the dielectric is in-
`serted.
`
`Solution The energy stored in the capacitor in the absence
`of the dielectric is
`
`cea aeas
`Uy = GolAVo)
`Because AV, = Qy/C, this can be expressed as
`z
`u= Q*
`2Cy
`After the battery is removed and the dielectric is inserted
`betweenthe plates, the charge on the capacitor remains the same.
`Hence, the energy stored in the presence of the dielectric is
`
`
`
`Co Q
`
`F
`
`=
`| AV,
`(a)
`
`ee
`
`&
`*
`
`|
`
`=
`
`u
`
`:
`2c
`
`
`Figure 20.27
`
`(b)
`(Example 20.12)
`
`
`PROBLEM-SOLVING STRATEGY AND HINTS
`
`
`
`
`
`1. Be careful with your choice of units. To calculate capacitance in farads,
`
`make sure that distances are in meters and use
`SI value of €,. When
`checking consistency of units, remember that the units for electric fields
`can be either newtons per coulomb or volts per meter.
`2. Whentwo or more unequal capacitors are connectedin series, they carry
`the same charge, but their potential differences are not the same. The
`
`capacitances add as reciprocals, and the equivalent capacitance of the
`combinationis always less than the smallest individual capacitor.
`3. When two or more capacitors are connected in parallel, the potential
`differences across them are the same. The charge on each capacitoris
`proportionalto its capacitance; hence, the capacitances add directly to
`give the equivalent capacitance of the parallel combination.
`
`4. A dielectric increases capacitancebythefactor « (the dielectric constant)
`because induced surface charges on the dielectric reduce the electric field
`inside the material from Ey to Ey/«.
`This photographillustrates
`5. Be careful about problems in which you may be connecting or discon-
`dielectric breakdownin air. Sparks
`
`
`necting a battery to a capacitor. [1 is important to note whether modifi-
`are produced whenalarge alter-
`cationsto the capacitorare being madewhile thecapacitoris connected
`nating voltageis applied across the
`to the battery orafter it
`is disconnected,
`If the capacitor remains con- cee a high-voltage
`
`nectedto the battery, the voltage across the capacitor necessarily remains
`Cbeiden Fa Scie
`
`the same (equal to the battery voltage). and the charge is proportional
`to the capacitance, regardless of how it may be modified (say, by insertion of
`a dielectric), However, if you disconnect the capacitor from the battery
`
` before making any modifications to the capacitor, thenits charge remains
`the voltage across
`the same. In this case, as you vary the capacitance,
`the plates changes in inverse proportion to capacitance, according to
`AV= Q/¢
`
`But the capacitance in the presence of thedielectric is C =
`KC, and sobecomes
`
`Gu
`Qn?
`== =
`2x
`‘
`Because « > 1, we see that
`thefinal energy is less than the
`|
`,'x. This missing energy can be
`initial energy by the factor
`
`SUMMARY
`
`accounted for by noting that when the dielectricis inserted
`-
`5
`fe:
`i
`ba
`Breer
`Wasi’
`into the capacitor,it gets pulled into the device. An external
`agent must do negative work to keepthe slab from acceler- a ee ee eee poms 2 and Bin anelectrostatic ficld E,
`ating. This work is simply the difference — Uj.
`(Alterna
`the change in potential energyis
`tively, thepositive work done bythe system on the external
`agent is U, — C1)
`
`AL
`
`'s
`at
`
`E-
`
`ids
`
`(20.1)
`
`
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`Chapter 20
`
`Electric Potential and Capacitance
`
`Ir will also raise the potential of your bodyto 20-000 volts,
`because you are in contact with the car. This is not a prob-
`fem. If you step out of the car, your body at 20000 volts will
`make contact with the ground, which is at ground potential
`of zero volts, As a result, a current will flow through your
`body, and vouwill likely be injured. Thus, it is best to stay in
`the car until help arrives
`r often remains charged long after the voltage
`ronnected. This residual charge can be lethal,
`© can be safely handled after discharging the
`
`
`
`plates by short circuiting the device with a conductor, such
`as a screwdriver with an insulating handle.
`|. The capacitance of the device is measured by the electronics
`in the stud finder. When a (dielectric) wooden stud is
`broughtinto the electric field region, it has the same effect
`as introducing a dielectric between the plates of a parallel
`plate capacitor—the capacitance changes. The stud finder
`detects this change in capacitance and signals the presence
`of the stud,
`
`Zi
`
`Current and Direct
`
`Current Circuits
`
`[a far, our discussion ofelectrical phenomena has been confined to
`
`, for currents to exist outside a
`
`Photograph of o carbon fila-
`ment incandescent lamp. The
`resistance of such o lampis
`typically 10 ©, but its value
`changes with temperature.
`Mast modern lightbulbs use
`tungsten filaments, the resis-
`tance of which increases with
`increasing temperature.
`roxy of Contral Scientific Co.)
`
`
`
`charges at rest, or electrostatics. We shall nowconsider situations in-
`volving electric charges in motion. The term electric current, or simply
`current, is used to describe the rate offlow of charge through some region of
`space. Most practical applications of electricity involve electric currents. For
`example, the battery of a flashlight supplies current to the filament of the
`bulb when the switch is turned on. In these commonsituations, the flowof
`charge takes place in a conductor, such as a copper wire. It is also possible
`conductor, For
`instance,
`a
`beam of electrons in a televi-
`sion picture tube constitutes a
`current,
`In this chapter we shall
`first define current and cur-
`rent density. A microscopic de-
`scription of current will be
`given, and some of the factors
`that contribute to resistance to
`the flow of charge in conduc-
`tors will be discussed. Mecha-
`nisms responsible for the elec-
`tical
`resistances of various
`
`——.
`
`
`
`CH APTER OUTLINE
`Electric Current
`Resistance and Ohm's Law
`Superconductors
`A Model for Electrical
`Conduction
`5 Electrical Energy and Power
`§ Sources of emf
`Resistors in Series and in
`Parallel
`& Kirchholf’s Rules and Simple
`DC Circuits
`AC Circuits
`
`
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`
`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`
`IPR2023-00783
`
`
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`TABLE 21.2
`Critical Temperatures for
`
`Various Superconductors
`Material
`T,(K)
`YBa,Cu,0;
`92
`Bi-Sr-Ca-Cu-O
`105
`Ti-Ba-Ca-Cu-0
`125
`HgBa,CagGinyO,
`144
`Nb,Ge
`23.2
`Nb,Sn
`21.05
`Nb
`9.46
`7.18
`4.15
`3.72
`1.19
`0.88
`
`Hg
`
`
`
`Photographof a small permanent
`magnet levitated above a disk of
`the superconductor YBa,Ca.0-,
`which is at 77 K. This superconduc-
`tor has zero electric resistance at
`temperatures below 92 K and ex-
`pels any applied magnetic field.
`(Ctertesy of IBM Research Laboratory)
`
`Chapter 21
`
`Current and Direct Current Circuits
`
`One of the truly remarkable features of superconductors is the fact that once
`a currentis set up in them,it persists without any applied voltage (because R = 0). In
`fact, steady currents have been observed to persist in superconducting loops for
`several years with no apparent decay!
`An important recent development in physics that has created much excitement
`in the scientific community has been the discovery of high-temperature copper-
`oxide-based superconductors. The excitement began with a 1986 publication by
`Georg Bednorz and K, Alex Miiller, scientists at the IBM Zurich Research Labora-
`tory in Switzerland, in which they reported evidence for superconductivity at a
`temperature near 30 K in an oxide of barium, lanthanum, and copper. Bednorz
`and Miller were awardedthe Nobel Prize in 1987 for their remarkable discovery.
`Shortly thereafter, a mew family of compounds was openfor investigation, and re-
`search activity in the field ofsuperconductivity proceeded vigorously. In early 1987,
`groups at the University of Alabama at Huntsville and the University of Houston
`announced the discovery of superconductivity at about 92 K in an oxide ofyurium,
`barium, and copper (YBagCu,Q;). Late in 1987, teamsofscientists from Japan and
`the United States reported superconductivity at 105 K in an oxide of bismuth,stron-
`tium, calcium, and copper. More recently, scientists have reported superconductiv-
`ity at temperatures as high as 125 K in an oxide containing thallium. At this point
`one cannot rule out the possibility of room-temperature superconductivity, and the
`search for novel superconducting materials continues, It is an important search
`both for scientific reasons and because practical applications become more prob-
`able and widespread as the critical temperature is raised.
`An important and useful application is superconducting magnets in which the
`magnetic field strengths are about ten times greater than those of the best normal
`electromagnets. Such superconducting magnets are being considered as a means
`of storing energy. The idea of using superconducting power lines for transmitting
`power efficiently is also receiving some consideration, Modern superconducting
`electronic devices consisting of two thin-film superconductors separated by a thin
`insulator have been constructed. They include magnetometers (a magnetic-field
`measuring device) and various microwave devices.
`
`21.4 + A MODEL FOR ELECTRICAL CONDUCTION
`
`The classical model of electrical conduction in metals leads to Ohm's law and shows
`that resistivity can be related to the motion of electrons in metals.
`Consider a conductor as a regular array of atoms containing free electrons
`(sometimes called conduction electrons). Such electrons are free to move through
`the conductor (as we learned in our discussion of drift speed in Section 21.1) and
`are approximately equal in number to the atoms, In the absence ofan electric field,
`the free electrons move in random directions with average speeds on the order of
`10 m/s. Thesituation is similar to the motion of gas molecules confined in a vessel.
`In fact, some scientists refer to conduction electrons in a metal as an electron gas.
`The conduction electrons are nottotally free, because theyare confined to the
`interior of the conductor and undergo frequent collisions with the array of atoms.
`The collisions are the predominant mechanismfor the resistivity of a metal at nor-
`mal temperatures. Notethat there is no current through a conductorin the absence
`of an electric field, because the average velocity of the free electrons is zero. In
`
`214
`
`A Model for Electrical Conduction
`
`607
`
`&tb)P
`
`Figure 21.9 (a) A schematic diagram of the random motion of a charge carrier in a conductor
`in the absence ofan electric field. The diifi velocity is zero. (b} The motion of a charge carrier
`in a conductor in the presence of an electric field. Note that the random motion is modified by
`the field, and the charge carrier has a drift velocity.
`
`other words,just as many electrons move in one direction as in the opposite direc-
`tion, on the average, and so there is no net flow of charge.
`The situation is modified when an electric field is applied to the metal. In
`addition to random thermal motion, the free electrons drift slowly in a direction
`opposite that of the electric field, with an averagedrift speed of vy, which is much
`less (typically 10~* m/s; see Example 21.1) than the average speed betweencolli-
`sions (typically 10° m/s). Figure 21.9 provides a crude depiction of the motion of
`free electrons in a conductor. In the absence of an electric field, there is no net
`displacement after many collisions (Fig. 21.9). An electric field E modifies the
`random motion and causes the electrons to drift in a direction opposite that of E
`(Fig. 21.9b). The slight curvature in the paths in Figure 21.9b results from the
`acceleration of the electrons betweencollisions, caused by the applied field. One
`mechanical system somewhat analogousto this situation is a ball rol