Capacitance

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Definition of Capacitance
Calculating Capacitance
Combinations of Capacitors
Energy Stored in a Charged Capacitor
Capacitors with Dielectrics
An Atomic Description of a Dielectric
Capacitance and Dielectric


Capacitors: Device that store electric charge
A capacitor consists of two conductors
separated by an insulator.
Capacitance: Depends on its geometry and on
the material, called a dielectric, that separates
the conductors.
Definition of Capacitance


A capacitor consists of two
conductors (known as plates)
carrying charges of equal
magnitude but opposite sign.
A potential difference DV exists
between the conductors due to the
presence of the charges.
What is the capacity of the device
for storing charge at particular
value of DV?



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Definition of Capacitance
Experiments show the quantity of electric charge
Q on a capacitor is linearly proportional to the
potential difference between the conductors, that
is Q ~ DV. Or we write Q = C DV

The capacitance C of a capacitor is the ratio of the
magnitude of the charge on either conductor to
the magnitude of the potential difference between
them:

                       Q
            C=
                      DV
SI Unit: farad (F), 1F = 1 C/V
Typical device have capacitances ranging from microfarad to picofarad.


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Parallel - Plate Capacitors
                  A parallel-plate capacitor consists
                  of two parallel conducting plates,
                  each of area A, separated by a
                  distance d. When the capacitor is
                  charged, the plates carry equal
                  amounts of charge. One plate
                  carries positive charge, and the
                  other carries negative charge.


 The plates are charged by connection to a battery.
 Describe the process by which the plates get
 charged up.


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Parallel-Plate Capacitors                                           d

Two parallel metallic plates of equal
area A separated by a distance d as
shown.
One plate carries a charge Q and
the other carries a charge –Q. And                              A
surface charge density of each plate
is s = Q/A.
 Variation with A
If plates are large, then charges can distribute
themselves over a substantial area, and the amount of
charge that can be stored on a plate for a given
potential diff increases as A is increased.
Thus we expect C to be proportional to A
C~A
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Parallel-Plate Capacitors                                    d


Variation with d

Potential difference DV constant
across, E field increases as d
decreases.                                               A

Imagine d decreases and consider
situation before any charges have
had a chance to move in response
to this change.
Because no charge move  E the
same but over a shorter distance.


DV = Ed means that DV decreases.
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Parallel-Plate Capacitors                                                   d
Variation with d (cont’d)

 The difference between this new capacitor
 voltage and the terminal voltage of the
 battery now exists as a potential difference
 across the wires connecting the battery to the
 capacitor.                                                             A
 A E field result in the wires that drives more
 charge onto the plates, increasing the
 potential diff. DV until it matches that of the
 battery.  potential diff. Across wire = 0 
 flow of charges stop.
 More charges has accumulated at the
 capacitor as a result.
 We have d decrease, Q increases. Similarly d
 increases  Q decreases.
 Capacitance inversely proportional to d.
  C ~ 1/d

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Parallel-Plate Capacitors




(a) The electric field between the plates of a parallel-plate
    capacitor is uniform near the center but nonuniform near
    the edges.
(b) Electric field pattern of two oppositely charged conducting
    parallel plates.



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Parallel-Plate Capacitors
Assume electric field uniform between the plates, we have
(see lecture on Gauss’s Law)

           s     Q
   E=         =
           eo   eoA


                   Qd
   DV = E d =
                   eoA

           Q              Q
   C=            =
           DV           Qd /eoA

      eoA
   C=
       d
          (As we have argued before)
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Example
 SB 26 Q8 A 1-megabit computer memory chip
 contains many 60.0-fF capacitors. Each
 capacitor has a plate area of 21.0 x 10-12 m2.
 Determine the plate separation of such a
 capacitor (assume a parallel-plate
 configuration). The characteristic atomic
 diameter is 10-10 m =0.100nm. Express the
 plate separation in nanometers.
     e 0 A
 C         60.0  1015 F
      d
     e 0 A 18.85  1012 21.0  1012 
d            
        C                     60.0  1015
                    9
d  3.10  10            m     3.10 nm


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What is a lightning discharge ?
                       Friction forces between the air
                       molecules within a cloud result in
                       positively charged molecules
                       moving to the lower surface, and
                       the negative charges moving to
                       the upper surface .
                       The lower surface induces a high
                       concentration of negative charges
                       in the earth beneath.
                       The resultant electric field is very
                       strong, containing large amounts
                       of charge and energy.
                       If the electric field is greater than
                       than breakdown strength of air, a
                       lightning discharge occurs, in
                       which air molecules are ripped
                       apart, forming a conducting path
                       between the cloud and the earth.
Example: Lightning
SB 26 Q6 Regarding the Earth and a
cloud layer 800 m above the Earth as the
“plates” of a capacitor, calculate the
capacitance if the cloud layer has an area
of 1.00 x 1.00 km2. Assume that the air
between the cloud and the ground is
pure and dry.
Assume that charge builds up on the
cloud and on the ground until a uniform
electric field of 3.00 x 106 N/C
throughout the space between them
makes the air break down and conduct
electricity as a lightning bolt. What is the
maximum charge the cloud can hold?

C= eoA/d = 8.85x10-12 x (1000)2/800 = 11.1 nF
Potential between ground and cloud is
DV = Ed = 3.0 x106 x 800 = 2.4 x 109 V
Q = C(DV) = 26.6 C
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Cylindrical Capacitors



             L




A solid cylindrical conductor of radius a and charge Q is
coaxial with a cylindrical shell of negligible thickness,
radius b > a, and charge –Q. Find the capacitance of this
cylindrical capacitor if its length is L.




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Cylindrical Capacitors
Assume that L is >> a and b,
neglect the end effects.
E is perpendicular to the long axis                       L
of the cylinders and is confined to
the region between them.
Potential difference between the
two cylinders is given by

                b

Vb-Va = -           E . ds   Where E is the E field in the region a < r < b.

            a


  Our discussion on Gauss’s Law  Er = 2kl/r where l is the
  linear charge density of the cylinder.
  Note that the charge on outer cylinders does not contribute to
  E field inside it.
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Cylindrical Capacitors
                     b                         b
                                                   dr                    b
  Vb-Va = -              Er dr = - 2kl              r
                                                        = - 2kl ln(
                                                                         a
                                                                           )

                 a                         a


  Using l = Q/L , we have


        Q                      Q                             L
  C=         =
        DV           2kQ             b
                               ln(     )
                      L              a


                 L
   C=
                         b
            2k ln(         )
                         a


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Cylindrical Capacitors

                L
    C=                                              L
                  b
           2k ln(   )
                  a


What is the capacitance per unit length ?


Example
Co-axial Cable. Read the cable, typically 50 pF/m. Is this sensible ?
Typically a  0.5 mm, b  1.5 mm

                1
C/L                         50pF/m
      2  8.99  10  ln(3)
                   9




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Cylindrical Capacitors vs Parallel Plate Capacitors




                L




What are the advantages of cylindrical capacitor
over that of a parallel plate capacitor?




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Capacitance of an isolated sphere
Calculate the capacitance of an isolated
spherical conductor of radius R and
charge Q by assuming that the second
conductor making up the capacitor is a
concentric hollow sphere of infinite radius.

Electric potential of the sphere of radius R is
                                                                Q
kQ/R and V= 0 at infinity, we have

       Q          Q         R
 C=         =          =        = 4peo R
       DV       kQ/R        k


 C is proportional to its radius and independent of
 both the charge on the sphere and the potential
 difference.

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Example
SB 26 Q4
(a) If a drop of liquid has capacitance 1.00 pF, what is its radius ?
(b) If another drop has radius 2.00 mm, what is its capacitance ?
(c) What is the charge on the smaller drop if its potential is 100V ?


  C = 4peo R
  R = (8.99 x 109 N · m2/C2)(1.00 x 10–12 F) = 8.99 mm
  C = 4p (8.85 x 10-12) x 2.0x10-3 = 0.222 pF
  Q =CV = 0.222 pF x 100 V = 2.22 x 10-11 C




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The Spherical Capacitors




A spherical capacitor consists of a spherical conducting
shell of radius b and charge –Q concentric with a
smaller conducting sphere of radius a and charge Q.
Find the capacitance of this device.




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The Spherical Capacitors
Gauss’s law  E field outside a
spherically symmetric charge distribution
is radial and given by Er = kQ/r2.
The potential difference between the
spheres is
                   b                      b
                                              dr            1          1
  Vb-Va = -            Er dr = - kQ            r2
                                                    = kQ
                                                            b          a
               a                      a


           Q                 ab
    C=                 =
         Vb-Va             k (b-a)


 What happens to the capacitance of this system when the radius of
 the outer sphere approaches infinity?
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 Example
SB 26 Q12 A 20.0 mF spherical capacitor is composed
of two metallic spheres, one having a radius twice as
large as the other. If the region between the spheres is
a vacuum, determine the volume of this region.


Let the radii be b and a with b = 2a. Put charge Q on the inner
conductor and –Q on the outer. Electric field exists only in the
volume between them. The potential of the inner sphere is
Va=kQ/a; that of the outer is Vb=kQ/b. Then

         kQ kQ     Q  ba                          Q      4pe 0 ab
Va Vb =       =                             C=        =
          a   b   4p e 0  ab                     Va Vb    ba
    4pe 0 2a 2                                 C
 C=            =8p e 0 a              a=
       a                                     8pe 0
                                                                               C3
                           4 p b 3  4 p a 3  74 pa 3   74 p  3 3 3
The intervening volume is 3
                                                                         8 p e0
                                      3           3                3

            7C 3
    V               = 2.13 x 1016 m3
         384p 2e 0
                   3
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Example

 SB 26 Q16 What is the capacitance of the Earth ?
 Think of Earth spherical conductor and the outer
 conductor of the “spherical capacitor” may be
 considered as a conducting sphere at infinity
 where V approaches zero.
 C  4p e 0 R
                             
    4p 8.85  10 12 C N  m 2 6.37  106 m    
  C= 7.08 x 10-4 F

  A large capacitor !




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Combinations of Capacitors
Parallel Combination




The individual potential differences across capacitors connected
in parallel are all the same and are equal to the potential
difference applied across the combination.
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Combinations of Capacitors
Parallel Combination




  When the capacitors are first connected, electrons transfer
  between wires and plates. Leave left plates positively charged
  and right plates negatively charged.
  Energy source for this charge transfer is internal chemical
  energy stored in the battery.
  Flow of charges ceases when the voltage across the capacitors
  is equal to that across the battery terminals.
  Capacitors reach their maximum charge when the flow of
  charges ceases.
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Combinations of Capacitors
Parallel Combination




  Let the maximum charges on the two capacitors Q1 and Q2.
  Total charge Q stored by two capacitors is Q = Q1+Q2.
  Voltage across are the same  Q1=C1DV, Q2=C2DV
  Define an equivalent capacitor having Ceq s.t. Q = CeqDV
  We have CeqDV = C1DV + C2DV
  And hence Ceq = C1 + C2     (for parallel combination)
  In general Ceq = C1 + C2+ C3+ ………….. (for parallel combination)
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Combinations of Capacitors
Series Combination




 Start with uncharged situation and follow what happen just after a
 battery is connected to the circuit.
 When a battery is connected, electrons transferred out of the left
 plate of C1 and into the right plate of C2.
 As this charge accumulates on the right plate of C2, an equivalent
 amount of negative charge is forced off the left plate of C2 and
 this left plate therefore has an excess positive charge. (cont’d)
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Combinations of Capacitors
Series Combination




 The negative charge leaving the left plate of C2 travels through
 the connecting wire and accumulates on the right plate of C1.
 As a result, all right plates end up with a charge –Q and all the left
 plates end up with a charge +Q.
 Thus the charges on capacitors connected in series are the same.


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Combinations of Capacitors
Series Combination




 Voltage DV across battery terminals is split between two capacitors.
                 DV = DV1 + DV2
 Where DV1 and DV2 are potential diff across capacitors C1 and C2.
 Suppose we have equivalent capacitor Ceq = Q/DV
 For each capacitor, we have DV1 = Q/C1 and DV2=Q/C2 
                 Q/Ceq = Q/C1 + Q/C2 
                 1/Ceq = 1/C1 + 1/C2    (series combination)
 In general 1/Ceq = 1/C1 + 1/C2 + 1/C3 + ….. (series combination)

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Example: Equivalent Capacitance
           In series use 1/C=1/C1+1/C2

                                          2.50 mF


                                                         20.00 mF




                                         6.00 mF
In series use 1/C=1/C1+1/C2
      8.50 mF                            In parallel use C=C1+C2



                20.00 mF                      5.965 mF

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Example: Equivalent Capacitance

                       In parallel use C=C1+C2




 In parallel use C=C1+C2

                  In series use 1/C=1/C1+1/C2




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Example: Equivalent Capacitance 26.22




                               In parallel use Ceq=C+C/2+C/3

  In series use 1/CA=1/C+1/C                   C


                                                C/2




                                            C/3
   In series use 1/CB=1/C+1/C+1/C

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Example

26.10 A variable air capacitor used in tuning
circuits is made of N semicircular plates each of
radius R and positioned a distance d from each
other.
As shown, a second identical set of plates is
enmeshed with its plates halfway between
those of the first set. The second set can rotate
as a unit. Determine the capacitance as a
function of the angle of rotation q, where q=0
corresponds to the maximum capacitance.




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Example

S&B 26.10

With qp , the plates are out of mesh and the
overlap area is zero. With q0, the overlap
area is that of a semi-circle, pR2/2. By
proportion, the effective area of a single sheet
of charge is (pqR2/2.

When there are two plates in each comb, the number of
adjoining sheets of positive and negative charge is 3, as
shown in the sketch. When there are N plates on each
comb, the number of parallel capacitors is 2N-1 and the total
capacitance is
            e 0 Aeffective       2 N  1e 0 p  q R 2 2
C  2 N  1   distance
                             
                                           d 2


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Energy Stored in a Charged Capacitor
Suppose q is the charge on the capacitor at some instant during
the charging process.
At the same instance, the potential difference across the capacitor
is DV=q/C.
The work necessary to transfer an increment of charge dq from the
plate carrying charge –q to the plate carrying charge q (which is at
the higher electric potential) is
                         q
        dW = DV dq =        dq
                         C

The total work required to charge the capacitor from q = 0 to
some final charge q = Q is

               Q                  Q
                   q          1                Q2
        W=             dq =           q dq =
                   C          C                2C
              0                   0
Energy Stored in a Charged Capacitor
Work done in charging the capacitor = electric potential energy
U stored in the capacitor.

                Q2         1              1
         U=           =        QDV =          C (DV)2
                 2C        2              2

This result applies to any capacitors, regardless of its geometry.




  Is there a limit to the maximum energy (or charge)
  that can be stored in a capacitor? If so, what is the
  limiting factor?
Energy Stored in a Charged Capacitor

                  A plot of potential difference versus
                  charge for a capacitor is a straight line
                  having a slope 1/C. The work required
                  to move charge dq through the
                  potential difference DV across the
                  capacitor plates is given by the area of
                  the shaded rectangle. The total work
                  required to charge the capacitor to a
                  final charge Q is the triangular area
                  under the straight line, W = QDV/2.
                  1V = 1 J/C hence the unit for the area
                  is joule J.




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Energy Density Stored in a Charged Capacitor
For parallel plate capacitor, DV = E d, C = eoA/d we have

                 1    eoA                     1
          U=                 (E d)2 =             (eoA d) E   2
                 2      d                     2

  Ad = Volume occupied by the E field. This lead to a new
  quantity known as Energy Density
  u = U/Volume = U/Ad
                               1
                        u=         eo E   2
                               2
  Although above equation was derived for parallel-plate capacitor, the
  expression is generally valid.

 Energy Density in any electric field is proportional to the
 square of the magnitude of the electric field at a given point.
Example
26.33: A parallel-plate capacitor is charged and then
disconnected from a battery. By what fraction does the
stored energy change (increase or decrease) when the plate
separation is doubled ?



 U = Q2/2C and C = eoA/d and d2 = 2 d1 then
 C2= C1/2 and the energy stored doubles.




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Example
26.35: A parallel-plate capacitor has a charge Q and plates
of area A. Show that the force exerted on each plate by the
other is F = Q2/2eoA. Is this force attractive or repulsive.
Does this equation follow from common sense?


             W  U   F dx
         dU d  Q  d  Q x      2                        2

  F=                 2e A 
                                
         dx dx  2C  dx  0 

           F = Q2/2eoA
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Rewiring Two Charged Capacitors




Two capacitors C1 and C2 (where C1 > C2) are charged to the
same initial potential difference DVi, but with opposite polarity.
The charged capacitors are removed from the battery, and
their plates are connected as shown. (a) Find the final
potential difference DVf between a and b after the switches
are closed. (b) Find the total energy stored in the capacitors
before and after the switches are closed and the ratio of the
final energy to the initial energy.



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Rewiring Two Charged Capacitors

Before switches are closed:
Q1i = C1 DVi and
Q2i = - C2 DVi (negative sign for plate 2)
Total Q = Q1i + Q2i = (C1-C2) DVi


After switches are closed:
Total charge in system remain the same
Total Q = Q1f + Q2f
Charges redistribute until the entire
system is at the same potential DVf. And
this potential is the same across both the
capacitors.
Q1f = C1 DVf and Q2f = C2 DVf

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Rewiring Two Charged Capacitors
After switches are closed (cont’d):
Q = Q1f + Q2f
Q1f / Q2f = C1/C2  Q1f = [ C1/C2 ] Q2f
Hence Q = Q1f + Q2f =[ 1+ C1/C2 ] Q2f

                         C2                                   C1
We have Q2f = Q                         and       Q1f = Q
                      C2 + C1                               C2 + C1

                               C1
                     Q
           Q1f               C2 + C1                 Q
  DV1f =         =                            =             = DV2f = DVf
           C1                  C1                 C2 + C1

And
                         Q              (C1-C2) DVi
         DVf =                      =
                     C2 + C1              C2 + C1
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Rewiring Two Charged Capacitors
Energy
Before switches are closed:
Ui = C1 (DVi)2/2 + C2 (DVi)2/2 = ( C1 + C2 ) (DVi)2/2
After switches are closed:
Uf = C1 (DVf)2/2 + C2 (DVf)2/2 = ( C1 + C2 ) (DVf)2/2
                                         2                 Q2
        1                      Q                    1
Uf =        ( C1 + C2 )                       =
        2                  C2 + C1                  2   C2 + C1



            1   (C1-C2)2 (DVi )2                    1
 Uf =                                        Ui =       ( C1 + C2 ) (DVi)2
            2      C2 + C1                          2

                                             2
 We have             Uf        C1 - C2
                           =
                      Ui       C1+ C2
Example
S&B Chapter 26 Q 23
Consider the circuit as shown, where C1 = 6.00mF and
C2= 3.00 mF and DV =20.0V. Capacitor C1 is first charged
by closing of switch S1. Switch S1 is then opened and the
charged capacitor is connected to the uncharged capacitor
by the closing of S2. Calculate the initial charge acquired
by C1 and the final charge on each.




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Example
S&B Chapter 26 Q 23


S1 close, S2 open 

C=Q/V  Q=120 mC

After S1 open, S2 close 

Q1+Q2=120 mC

Same potential  Q1/C1=Q2/C2  (120-Q2)/C1= Q2/C2
(120-Q2)/6 = Q2/ 3  Q2 = 40 mC
 Q1 = 80 mC




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Capacitors with Dielectrics
A dielectric is a nonconducting material, such as rubber,
glass, or waxed paper.
When a dielectric is inserted between the plates of a
capacitor, the capacitance increases.
If the dielectric completely fills the space between the
plates, the capacitance increases by a dimensionless factor
 , which is called the dielectric constant.
Dielectric constant is a property of a material and varies
from one material to another.




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Capacitors with Dielectrics




A charged capacitor (a) before and (b) after insertion of a
dielectric between the plates. The charge on the plates remains
unchanged, but the potential difference decreases from DVo to
DV = DVo/. Thus the capacitance increases from Co to  Co.
Note no battery is involved in this example.
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Capacitors with Dielectrics

 Capacitance increases by the factor  when dielectric
 completely fills the region between the plates.


 If the dielectric is introduced while the potential difference is
 being maintained constant by a battery, the charge increases
 to a value Q =  Qo . The additional charge is supplied by the
 battery and the capacitance again increases by the factor .


 For parallel plate capacitor:

                      eoA
                  C=
                       d


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Dielectric Strength

 For any given separation d, the maximum voltage that can be
 applied to a capacitor without causing a discharge depends on
 the dielectric strength (maximum electric field) of the
 dielectric.
 If magnitude of the electric field in the dielectric exceeds the
 dielectric strength, then the insulating properties break down
 and the dielectric begins to conduct.




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Dielectric Constant and Dielectric Strength of Various
Materials at Room Temperature
 Material             Dielectric Constant    Dielectric Strength (V/m)
 Air (dry)                  1.00059                    3 x 106
 Bakelite                      4.9                    24 x 106
 Fused quartz                 3.78                     8 x 106
 Neoprene rubber               6.7                    12 x 106
 Nylon                         3.4                    14 x 106
 Paper                         3.7                    16 x 106
 Polystyrene                  2.56                    24 x 106
 Polyvinyl Chloride            3.4                    40 x 106
 Porcelain                     6                      12 x 106
 Pyrex Glass                   5.6                    14 x 106
 Silicone Oil                  2.5                    15 x 106
 Strontium Titanate           233                      8 x 106
 Teflon                        2.1                    60 x 106
 Vacuum                     1.00000                       -
 Water                         80                         -



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Capacitors with Dielectric Material

What are the advantages of dielectric material in a capacitor?


 • Increase the capacitance
 • Increase the maximum operating voltage
 • Possible mechanical support between the plates, which
 allows the plates to be close together without touching,
 thereby decreasing d and increasing C.




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Types of Capacitors




 (a) A tubular capacitor, whose plates are separated by paper and
 then rolled into a cylinder. (b) A high-voltage capacitor
 consisting of many parallel plates separated by insulating oil. (c)
 An electrolytic capacitor.




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Energy Stored Before and After




 A parallel-plate capacitor is charged with a battery to a charge
 Qo as shown (a). The battery is then removed and a slab of
 material that has a dielectric constant  is inserted between the
 plates as shown (b). Find the energy stored in the capacitor
 before and after the dielectric is inserted.


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Energy Stored Before and After

 Before
                Qo2
       Uo =
                2 Co

 Charge on the capacitor the same
  before and after. (Why?)
                                                  (a)
          Qo2          Qo2          Uo
  U=              =             =
          2C           2  Co       

 Energy reduced, where does the
  “missing” energy go to?
Dielectric, when inserted, gets pulled into
the device. External agent do negative
work to keep dielectric from accelerating.
Work = U-Uo
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Capacitors with Dielectric




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




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The H2O Molecule (Example of a Polar Molecule)
                              Molecules are said to be polarized when a
                              separation exists between the average
                              position of the negative charges and the
                              average position of the positive charges.


                 Center of negative charges (why not at the center of the oxygen atom?)
                 (Hint think about the effect of the Hydrogen atoms)

Center of positive charges




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How does a microwave work?

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




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Induced polarization
                   This molecule is non-polar since the
                   center of negative charges coincide
                   with the center of positive charges.




                   The presence of an external E field
                   causes the charges to change their
                   positions and such that the center of
                   negative charges does not coincide with
                   the center of positive charges. We say
                   that a dipole moment has been induced
                   due to the presence of the E field.
                   This is called induced polarization.




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An Atomic Description of Dielectrics
Potential difference DVo between the plates of a capacitor is
reduced to DVo/ when a dielectric is introduced.
How is this possible?


Think about the E field, if Eo is the E field without the dielectric,
then the field in the presence of E field is E = Eo/.
I.e. The field is reduced.
How is this possible?




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An Atomic Description of Dielectrics




(a) Polar molecules are randomly oriented in the absence of an
    external electric field.
(b) When an external field is applied (to the right as shown),
    the molecules partially align with the field. (dielectric is
    polarized!)
What is the effective E field inside the dielectric?



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An Atomic Description of Dielectrics




(a) When a dielectric is polarized, the dipole moments of the
molecules in the dielectric are partially aligned with the external
field Eo. (b) This polarization causes an induced charge on the
opposite side. This separation of charge results in a reduction in
the net electric field within the dielectric.
The net effect on the dielectric is the formation of an induced
positive surface charge density sind on the right face and an
equal negative surface charge density –sind on the left face.

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An Atomic Description of Dielectrics




The induced surface charge give rise to an induced electric field
Eind in the direction opposite the external field Eo.
Therefore, the net electric field E in the dielectric has a
magnitude
                            E = Eo-Eind




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What is the magnitude of the induced charge
density?
                        For parallel plate capacitor,

                        External field Eo =   s/ eo
                        Induced Field Eind =   sind/ eo
                        And E = Eo/ =     s/  eo
                        Substitute into E = Eo-Eind gives
                             s     s          sind
                                 =
                             eo   eo          eo
                                        1
                     And    sind =                   s
                                         
   Induced charge on a dielectric placed between the plates of a
   charged capacitor. Note that the induced charge density on the
   dielectric is less than the charge density on the plates.
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Effect of a Metallic Slab between the plates




A parallel-plate capacitor has a plate separation d and plate area
A. An uncharged metallic slab of thickness a is inserted midway
between the plates. (a) Find the capacitance of the device. (b)
Show that the capacitance is unaffected if the metallic slab is
infinitesimally thin.
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Effect of a Metallic Slab between the plates
(a)




Charge on one plate must induce a charge of equal magnitude but
opposite sign on the near side of the metallic slab.
Net charge on the slab is zero, electric field inside the slab is zero.
Capacitor = two capacitor in series, each having a plate separation
of (d-a)/2 as shown.

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 Effect of a Metallic Slab between the plates
(a)
                  1/C = 1/C1+ 1/C2

                      1             1                  1
                          =                   +
                      C             eoA            eoA
                                (d-a)/2           (d-a)/2


                                        eoA
                              C =
                                     (d-a)


  What happens when a  d? Why?

                                eoA
(b) As a  0, d-a  d and C =             The original capacitance
                                d



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Effect of a Metallic Slab between the plates




What are the differences in the result as compared to the
previous example if we insert the metallic slab as shown?


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A Partially Filled Capacitor




A parallel-plate capacitor with a plate separation d has a
capacitance Co in the absence of a dielectric. What is the
capacitance when a slab of dielectric material of dielectric
constant  and thickness d/3 is inserted between the plates as
shown.


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A Partially Filled Capacitor
                   Two capacitors in series
                   1/C = 1/C1 + 1/C2 where

                           eoA                        eoA
                   C1 =           and       C2 =
                           d/3                         2d/3

                    1       d/3           2d/3
                      =           +
                    C      eoA           eoA
                            d         1
                       =                    +2
                           3eoA       

                           3     eoA
                  C=
                       2+ 1       d


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