Capacitance and Dielectrics - lwillia2/old42web/  · Chapter 26 HW: 7th/6th 4/6, ... Capacitance

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Transcript of Capacitance and Dielectrics - lwillia2/old42web/  · Chapter 26 HW: 7th/6th 4/6, ... Capacitance

  • Chapter 26 HW: 7th/6th4/6, 9/11, 11/13 (on 9 and 11 start from Gausss Law!), 17/21, 19/23, 25/29, 28/33, 32/40,39/45,41/49,42/50, 44/53,45/73,49/58, 58/68, 59/67

    Capacitance

    and

    Dielectrics

  • Recall: The Infinite Charged Plane

    02E

    =

  • Potential Difference in a Uniform Field

    Electric field lines always point in the direction of decreasing electric potentialWhen the electric field is directed downward, point B is at a lower potential than point AWhen a positive test charge moves from A to B, the charge-field system loses potential energy

    B B

    B A A AV V V d E d Ed = = = = E s s

    r r

  • Parallel Plates: In the center we assume the electric field is constant and uniform and that the potential difference between any two points depends on the distance d between those points!

    0

    E

    =

    V Ed =

  • Parallel Plate Assumptions

    The assumption that the electric field is uniform is valid in the central region, but not at the ends of the platesIf the separation between the plates is small compared with the length of the plates, the effect of the non-uniform field can be ignoredThe charge density on the plates is = Q/A

    A is the area of each plate, which are equalQ is the charge on each plate, equal with opposite signs

  • A capacitor consists of two conductorsThese conductors are called platesWhen the conductor is charged, the plates carry charges of equal magnitude and opposite directions

    A potential difference exists between the plates due to the chargeCapacitance will always be a positive quantityThe capacitance of a given capacitor is constantThe farad is a large unit, typically you will see microfarads (F) and picofarads (pF)A capacitor stores electrical energy

    Makeup of a Capacitor

  • Capacitors

    Capacitors are devices that store electric charge and energyExamples of where capacitors are used include:

    radio receiversfilters in power suppliesenergy-storing devices in electronic flashes

  • Some Uses of Capacitors

    DefibrillatorsWhen fibrillation occurs, the heart produces a rapid, irregular pattern of beatsA fast discharge of electrical energy through the heart can return the organ to its normal beat pattern

    In general, capacitors act as energy reservoirs that can be slowly charged and then discharged quickly to provide large amounts of energy in a short pulse

  • Charging a Parallel Plate Capacitor

    Each plate is connected to a terminal of the batteryIf the capacitor is initially uncharged, the battery establishes an electric field in the connecting wires

  • Active Figure 26.10

    AF_2604.html

  • CapacitanceThe capacitance, C, of a capacitor is defined as the ratio of the magnitude of the charge on either conductor to the potential difference between the conductors

    The capacitance is a measure of the capacitors ability to store chargeThe SI unit of capacitance is the farad (F)

    QCV

    =

  • CapacitanceThe capacitance is proportional to the area of its plates and inversely proportional to the distance between the plates

    QCV

    =

    ( )= = = = =

    /o

    o

    o

    AQ Q Q QC V Ed Q A d dd

    = o ACd

  • Circular Plates

    (plate radius = 10 cm) 0= 8.85x10 12 Farad/m

    = o ACd

    QCV

    =

  • Capacitors in SeriesWhen a battery is connected to the circuit, electrons are transferred from the left plate of C1 to the right plate of C2through the batteryAs this negative charge accumulates on the right plate of C2, an equivalent amount of negative charge is removed from the left plate of C2, leaving it with an excess positive chargeAll of the right plates gain charges of Q and all the left plates have charges of +Q thus:

    QCV

    =

    = =1 2Series: Q Q Q

  • Capacitors in SeriesThe potential differences add up to the battery voltageV = V1 + V2 +

    The equivalent capacitance of a series combination is always less than any individual capacitor in the combination

    1 2

    1 1 1

    eqC C C= + +K

    =eq

    QVC

  • Active Figure 26.08Capacitors in Series

  • Capacitors in ParallelThe total charge is equal to the sum of the charges on the capacitors

    Qtotal = Q1 + Q2The potential difference across the capacitors is the same

    And each is equal to the voltage of the battery:

    V = V1 = V2

  • The capacitors can be replaced with one capacitor with a capacitance of Ceq

    Qtotal = Q1 + Q2

    Capacitors in Parallel

    Essentially, the areas are combined

    = total eqQ C V

    Ceq = C1 + C2 +

    = + 1 2eqC V C V C V

  • Active Figure 26.07Capacitors in Parallel

  • Capacitor Summary

    = + + = =

    = + +K

    1 2

    1 2

    1 2

    Capacitors in Series:

    Q1 1 1eq

    V V VQ Q

    C C C

    = =

    = +

    = + +K

    1 2

    1 2

    1 2

    Capacitors in Parallel:

    toteq

    V V VQ Q QC C C

  • Equivalent CapacitanceDraw the reduced circuit it in stages!

  • Fig P26-22, p.824

    Threecapacitorsareconnectedtoabatteryasshown.TheircapacitancesareC1=3C,C2=C,andC3=5C.(a)Whatistheequivalentcapacitanceofthissetofcapacitors?(b)Statetherankingofthecapacitorsaccordingtothechargetheystore,fromlargesttosmallest.(c)Rankthecapacitorsaccordingtothepotentialdifferencesacrossthem,fromlargesttosmallest.

    = = = +

    = + +K

    1 2

    1 2

    1 2

    Capacitors in Parallel:

    toteq

    V V VQ Q QC C C

    = + + = =

    = + +K

    1 2

    1 2

    1 2

    Capacitors in Series:

    Q1 1 1

    eq

    V V VQ Q

    C C C

  • Fig P26-27

    ( )( )

    1

    1

    2

    1

    1 1 3.33 F5.00 10.02 3.33 2.00 8.66 F

    2 10.0 20.0 F

    1 1 6.04 F8.66 20.0

    s

    p

    p

    eq

    C

    C

    C

    C

    = + =

    = + =

    = =

    = + =

    FindtheequivalentcapacitancebetweenpointsaandbforthegroupofcapacitorsconnectedasshowninFigureP26.27.TakeC1=5.00F,C2=10.0F,andC3=2.00F.

    You Try:

  • P26.72 Assume a potential difference across a and b, and notice that the potential differenceacross the 8.00 F capacitor must be zero by symmetry. Then the equivalent capacitance can bedetermined from the following circuit:

    FIG. P26.72

    Series? Parallel? Neither Both?

  • Find the Equivalent Capacitance

    P26.75 By symmetry, the potential difference across 3C is zero, so the circuit reduces to

    11 1 8 42 4 6 3eq

    C C CC C

    = + = =

    .

    You Try:

  • Energy in a CapacitorConsider the circuit to be a systemBefore the switch is closed, the energy is stored as chemical energy in the batteryWhen the switch is closed, the energy is transformed from chemical to electric potential energy

  • Energy Stored in a CapacitorAssume the capacitor is being charged and, at some point, has a charge q on itThe work needed to transfer a charge from one plate to the other is

    The total work required is

    The work done in charging the capacitor appears as electric potential energy U:

    qdW Vdq dqC

    = =

    2

    0 2Q q QW dq

    C C= =

    221 1 ( )

    2 2 2QU Q V C VC

    = = =

  • This applies to a capacitor of any geometryThe energy stored increases as the charge increases and as the potential difference increasesThe energy can be considered to be stored in the electric field The total work done by a battery charging a capacitor is QV. Half the energy is either dissipated as heat or radiated as electromagnetic waves. For a parallel-plate capacitor, the energy can be expressed in terms of the field as U = (oAd)E2It can also be expressed in terms of the energy density (energy per unit volume) uE = oE2

    221 1 ( )

    2 2 2QU Q V C VC

    = = =

    Energy Stored in a Capacitor

  • Energy ProblemDetermine the energy stored in C2 when C1 = 15 F, C2 = 10 F, C3 = 20 F, and V0 = 18 V.

    a. 0.72 mJb. 0.32 mJc. 0.50 mJd. 0.18 mJe. 1.60 mJ

    221 1 ( )

    2 2 2QU Q V C VC

    = = =

  • If VA VB = 50 V, how much energy is stored in the 36-F capacitor?

    a. 50 mJb. 28 mJc. 13 mJd. 8.9 mJe. 17 mJ

    You Try:2

    21 1 ( )2 2 2QU Q V C VC

    = = =

  • Example 26.4: Rewiring Two Charged Capacitors

    Fig. 26-12, p. 733

  • Energy Problem

    A 3.0-F capacitor charged to 40 V and a 5.0-F capacitor charged to 18 V are connected to each other, with the positive plate of each connected to the negative plate of the other. What is the final charge on the 3.0-F capacitor?a. 11 Cb. 15 Cc. 19 Cd. 26 Ce. 79 C

    221 1 ( )

    2 2 2QU Q V C VC

    = = =

  • ProblemWhat is the maximum voltage that can be sustainedbetween 2 parallel plates separated by 2.5 cm of dry air?Dry air supports max field strength of 3x 106 V/m .

    V Ed=6(3 10 / )(.025 )x V m m=

    47.5 10x V=

    75kV=

    More than this and the air breaks down and becomes a conductor. LIGHTENING!

  • The dielectric breakdown strength of dry air, at Standard Temperature and Pressure (STP), between spherical electrodes is approximately 33 kV/cm.

    http://upload.wikimedia.org/wikipedia/commons/0/03/Plasma-filaments.jpghttp://en.wikipedia.org/wiki/Image:Sparkplug.jpg

  • Dielectric Breakdown of Air

  • DielectricsA dieletric is an insulator that increases the capacitance.

    Reduced E field prevents breakdown & discharge between plates.

    0 ACd

    =

  • The dieletric constant is the ratio of the net electric field magnitude without the dielectric,E0, and with the dielectric, E.