Capacitance and vkarpov/L05S.ch25.pdfآ  Capacitance and Dielectrics. Capacitance ... conductors, so:

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Text of Capacitance and vkarpov/L05S.ch25.pdfآ  Capacitance and Dielectrics. Capacitance ... conductors,...

  • Capacitance and Dielectrics

  • Capacitance

    General Definition:

    VqC /==== Special case for parallel

    plates:

    d

    A C 0

    εεεε ====

  • Potential Energy

    • I must do work to charge up a capacitor.

    • This energy is stored in the form of electric

    potential energy.

    • We showed that this is C

    Q U

    2

    2

    ====

    2

    02

    1 Eu εεεε====

    • Then we saw that this energy is stored in the

    electric field, with a volume energy density

  • Potential difference and

    Electric field Since potential difference is work per unit charge,

    ∫∫∫∫====∆∆∆∆ b

    a EdxV

    For the parallel-plate capacitor E is uniform, so

    EdV ====

    Also for parallel-plate case Gauss’s Law gives

    d

    A

    V

    Q CsoVd

    A

    Q E 0

    0

    0/ εεεε

    εεεε εεεεσσσσ ====================

  • Spherical example A spherical capacitor has inner radius a = 3mm,

    outer radius b = 6mm. The charge on the inner

    sphere is q = 2 C. What is the potential difference?

    From Gauss’s Law or the Shell

    Theorem, the field inside is 2r

    kq E ====

    From definition of

    potential difference

    V 33

    33

    99 103

    6

    1

    3

    1 1018

    106

    1

    103

    1 102109 ××××====

    

      

     −−−−××××====

     

     ××××

    −−−− ××××

    ××××××××××××==== −−−−−−−− −−−−

    ∫∫∫∫  

    

     −−−−======== b

    a ba kqdr

    r

    kq V

    11 2

    What is the capacitance?

    FVCVQC 4

    107.6)3000/()2(/ −−−−××××============

  • Q.25-1

    (1) 3 µF

    (2) 6 µF

    (3) 12 µF

    (4) 24 µF

    A capacitor has capacitance C = 6 µF

    and charge Q = 2 nC. If the charge is

    increased to 4 nC what will be the

    new capacitance?

  • Q. 25 - 1

    A capacitor has capacitance C = 6 µF and a

    charge Q = 2 nC. If the charge is increased to 4

    nC what will be the new capacitance?

    (1) 3 µF (2) 6 µF (3) 12 µF (4) 24 µF

    Solution:

    Capacitance depends on the structure of

    the capacitor, not on its charge.

  • Q.25-2

    1) 10,000 V/m

    2) 900 V/m

    3) 90 nV/m

    A parallel-plate capacitor has plates of area

    0.1 m2 and carries a charge of 9 nC.

    What is the electric field between the plates?

  • Q. 25 - 2

    Solution: Field at surface

    of conductor:

    mVmVmV /109)3(/900)2(/101)1( 84 −−−−××××××××

    ?

    109

    1.0

    9

    2

    ====

    ××××====

    ==== −−−−

    E

    CQ

    mA

    mVE

    AQ

    E

    /101 109

    109

    109 101

    109 /

    /

    4

    12

    8

    8

    1

    9

    0

    ××××==== ×××× ××××

    ====

    ××××==== ×××× ××××

    ========

    ====

    −−−−

    −−−−

    −−−− −−−−

    −−−−

    σσσσ

    εεεεσσσσ

  • Capacitors in Parallel

    If several capacitors are

    connected in parallel as

    shown, we can consider

    the arrangement as a

    single capacitor.

    L++= 21 qqqtot By charge conservation:

  • L============ 21 VVVtot

    The wires are good

    conductors, so:

    tottottot

    tot

    CVq V

    qqq

    V

    q

    V

    q

    V

    q CC

    ======== ++++++++++++

    ====

    ++++++++++++====++++++++

    /321

    3

    3

    2

    2

    1

    1 21

    L

    LL

    Therefore:

  • Capacitors in Series

    What if several capacitors are

    connected in series as shown?

    Now charge conservation

    gives:

    L============ 21 qqqtot

    And now it is the voltages that add:

    L++++++++==== 21 VVVtot

  • So now the voltage sum,

    when written in terms of the

    capacitances, gives the result

    for the series case:

    L

    L

    L

    ++++++++====

    ++++++++====

    ++++++++====

    21

    2211

    21

    /1/1/1

    ///

    CCC

    CqCqCq

    VVV

    tot

    tottot

    tot

  • Results for Capacitor Combinations

    • Capacitors in parallel:

    L++++++++==== 21 CCCtot

    • Capacitors in series:

    L++++++++==== 21 /1/1/1 CCCtot

  • Example: Problem 25-8

    What is effective capacitance of the device?

  • C12 FC

    FC

    FC

    µµµµ

    µµµµ

    µµµµ

    0.4

    0.5

    10

    3

    2

    1

    ====

    ====

    ====

    FC

    CCC

    µµµµ3.33/10

    10/35/110/1/1/1/1

    12

    2112

    ========

    ====++++====++++====

    FCCCtot µµµµ3.743.3312 ====++++====++++====

  • Dielectric Materials

    • Applied electric field polarizes material

    • This produces an induced surface charge

    • This reduces field within material

    • This reduces potential difference

    • This increases capacitance

    Effect of placing a dielectric between the

    plates of a charged capacitor:

  • E=0

    electronnucleus

    Polarization of atoms and molecules

    Non-polar molecules:

    induced dipoles

    Polar molecules:

    preferential orientation

  • Dielectric materials

    Dielectric materials polarize under external electric field E0

    Polarization creates the opposing electric field: –E’

    Total field E=E0-E ’=E0/κ < E0

    κ is called dielectric constant

    It shows how the field in a dielectric is weaker than in vacuum

  • Dielectric Constant

    • So the effect of a dielectric is to

    decrease the field and increase the

    capacitance.

    • The factor by which E is decreased

    and C is increased is defined to be the

    dielectric constant κ.

    0CC κκκκ==== κκκκ/0EE ====

  • Example: Problem 25-48

    Given a parallel-plate capacitor, dielectric-

    filled, with area A = 100 cm2, charge Q = 890

    nC, and electric field E = 1.4 kV/mm.

    (a) Find dielectric constant of the material.

    (b) Find the induced charge.

  • (a)

    (((( )))) 00 // εεεεAqE ====

    If there were no dielectric the field

    between the plates would be given

    by Gauss’s Law as

    mVE /1001.1 1085.8

    10/109.8 7 12

    27

    0 ××××====×××× ××××

    ==== −−−− −−−−−−−−

    2.7104.1/1001.1/ 67

    0 ====××××××××======== EESoκκκκ

    κ/0EE =But by definition:

  • 0

    /

    εεεε Aq

    E

    Gauss

    tot====

    ⇒⇒⇒⇒

    nCnCnCqqqso

    qqqBut

    tot

    tot

    766124890'

    '

    ====−−−−====−−−−====

    −−−−====

    nCC

    EAqSo tot

    124104.1210104.11085.8 82612

    0

    ====××××====××××××××××××××××====

    ==== −−−−−−−−−−−−

    εεεε

    (b)