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					                               Lecture No. #1
 DIELECTRICS
DIELECTRICS:-
   •   The dielectrics are materials, which contain no free electron, so that no current
       can flow through them. As a result the electric conductivity of a dielectric is poor
       and for an ideal dielectric, it is zero.


Energy band gap more than 3.0 eV is dielectric, according to the band theory
K (5 – 10) Glass, 3-6 (Plastic), 3 –6 (mica), 3-5 (rubber0, wood turpentine oil etc are
example of dielectric
                                          C                ε
                                     k=           or k =
                                          C0               ε0
The ratio of the capacitance of capacitor immersed in dielectric to that the same capacitor
in vacuum is called the dielectric constant,


NOTE:- C increases when dielectrics introduces,
It is independent of the size and shape of capacitor and is different for different materials.
Their value is 1 for vacuum and 1.0006 for air and nearly 6 for glass etc.



Dielectric on Atomic view:-
In any molecule there will be a distribution of protons and a distribution of electrons. The
protons can be considered to act as the equivalent positive electric charge at some
specific point at which negative charge is equivalent to the distributed electrons, the
center of gravity of electrons.

If two points coincide then the molecule is called non-polar

Example:

H2, O2 CO2, CH4, CCl4, C6H6, C6H12 etc.
It they are separated by a short distance then the molecule is called polar

Example: H2O, N2O, CHCl3, C6H5Cl………………. Etc

Note:- CO2 is non-polar molecule because center of gravity of positive and negative
charges coincide.

H2O is a polar molecule because in this molecule center of positive and negative charges
do not coincide.

THREE ELECTRIC VECTORS:

   (1) ELECTRIC FIELD STRENGTH (E):
         The electric field strength at any point in an electric field is defined as the force
   experience per unit positive charge (q0)

                                    F
                    E = Limit q=0
                                    q0

   The direction of E is along the direction of force

   (I) ELECTRIC POLARISATION:

   When a dielectric is placed in an external electric dipole moment and the dielectric is
   said to be polarized.

   The electric dipole moment induced per unit volume of the dielectric material is
   called the electric polarization of dielectric. It denoted by P.

    Let the surface charge density appearing at the end faces be + σ P & - σ P

   Therefore the charge induced on each face = + σ P A

   Total induced electric dipole moment = ql = + σ P Al

   Volume of block = Al
                                            l


                                                                      +σ P

                            A
                   −σ   P
                                              σ p . Al
Induced dipole moment per unit volume =                  =σp
                                                A.l

So by definition, this is polarization

                                    P =σ p


                                Lecture No.#2

(II) Electric Displacement Vector (D):-

Let σ Free be the surface density of free charges on capacitors plates and σ p that of

bound or polarization charges.

Therefore field strengths due to σ Free and σ p are given by

       σ free                   σp
E0 =            and      E' =      ………………………………………(2)
        ε0                      ε0

Let the net electric field within the dielectric is

                       σ free σ p
E = E0 − E ' ⇒               −
                        ε0     ε0

ε 0 E = σ free − σ P

σ free = ε 0 E + σ P ………………………………………………….(3)

But surface charge density is equal to polarization P = σ p
       σ free = ε 0 E + P …………………..(4)

      The quantity ( ε 0 E + P ) is the electric displacement vector…………………(5)

      Comparing (4) and (5), we note that magnitude of electric displacement vector
      surface density of free charges

                                        D = σ free


                                    D= ( ε 0 E + P )



        Equation (5) represents relation between three electric vectors D, E and P

.

POLARISATION:-
    When a polar or non-polar molecule is placed in an external electric field, the small
displacement of the orbital –electron will cause the distance between the centers of
gravity to alter. Thus non-polar molecule become induced dipoles whereas polar
molecules, which are already dipoles, will be oriented by the field and may have resultant
dipole moment modified since they possesses(1) permanent dipole moment (2) induced
dipole moment .

The orientation of the induced dipoles or of the permanent dipoles in an external electric
field will be such as to set r tend to set the axis of dipoles along the lines of force. This
phenomenon is called electric polarization.

Types of Polarization:

      (1) Electronic Polarization
      (2) Atomic or ionic polarization
      (3) Orientation or dipolar polarization
Molecualr Polarisability and Electric Susceptibility:
Polarization of Dielectric
If a material contains polar molecules, they will generally be in random orientations when
no electric field is applied. An applied electric field will polarize the material by orienting
the dipole moments of polar molecules.
This decreases the effective electric field between the plates and will increase the
capacitance of the parallel plate structure. The dielectric must be a good electric insulator
so as to minimize any DC leakage current through a capacitor.
                               Lecture No. #3
Parallel Plate with Dielectric
The capacitance of a set of charged parallel plates is increased by the insertion of a
dielectric material. The capacitance is inversely proportional to the electric field between
the plates, and the presence of the dielectric reduces the effective electric field. The
dielectric is characterized by a dielectric constant k, and the capacitance is multiplied by
that factor.




MOLECULAR POLARISABILITY:

When a non-polar molecule is placed in an electric field , the center of negative charge
is displaced relative to the center of positive charges. The molecule thus acquires induced
dipole moment for most molecules the induced dipole moment p is proportional to
applied electric field E i.e

                                          pα E

                                          p = αE

Where α is a constant and is called molecular polarizibility.

If there are n molecule per unit volume of dielectric , then total induced dipole moment
per unit volume = P = nαE

ELECTRIC SUSCEPTIBILITY:
               When a dielectric material is placed in an electric field, it becomes
electrically polarized. For most of materials the polarization is proportional to electric
field i.e.

                                                  pα E
                                                  P = ε0χeE

Where χ e is a constant, characteristics of materials, called electric susceptibility.

                                                           P
                                                   χe =
                                                          ε0E



That susceptibility may be defined as the ratio of polarization to permittivity of free space
time electric strength in the dielectric.

Note; the χ e Susceptibility of polar dielectrics depends on temperature while for non-

polar dielectric, it is independent of temperature.

Relation between Dielectric constant and Susceptibility:-

We have dielectric displacement vector

                                  D     =ε0E + P

                                            D = ε .E
                                            and P = ε 0 χ 0 E

                                        ε .E = ε 0 E + ε 0 χ 0 E



             or

                                  ε = ε0 + ε0χ0

                      ε
                         =1+ χ0
                      ε0
                                                           ε
                              k =1+ χ0         Where k =
                                                           ε0




                              Lecture No.#4


Gauss’s Law in Dielectrics:
The surface integral of displacement vector over a closed surface is equal to the net free
charge enclosed within the surface i.e.

                                 ∫ D.ds = q
                                  s



Proof. When a dielectric is placed in an electric field, there are two possibilities:

   (i)     The molecules of some dielectrics, like water, have permanent electric dipole
           moments. Such dielectrics are called polar dielectrics. In such dielectrics the
           electric dipole moments tend to align themselves along the external electric
           field. As the molecule are in constant thermal agitation, the degrees of
           alignment will not be complete but will increase as the applied electric field is
           increased or as the temperature is decreased.

   (ii)    The electric field tends to separate the negative and positive charges in the
           atoms or the molecules of the dielectric        whether      or not the atoms or
           molecules have permanent electric dipole moments. Thus an induced electric
           dipole moment is produced in the atoms or the molecules of the dielectric
           when placed in an electric field.

   Consider a parallel plate capacitor filled with a dielectric with a dielectric of
   permittivity ε carrying fixed charges +q and –q to provide a uniform external electric
   field. The function of the electric field is to separate the centre of the positive charge
   of the dielectric slightly from the centre of the negative charge.
Thus the applied electric field induces and negative charges in the dielectric, in such a
way that the electric field set up by them opposes external field. As the dielectric as a
whole remains electrically neutral, the positive induced surface charge must be equal
to the negative induced surface charge.

Let E0 and E be the electric field in the absence and the presence of the dielectic
between the plates respectively.

If no dielectric is present, Gauss’s law for a closed surface is given as

                                            q                  q
                                   ∫ E.ds = ε   0
                                                    orE0 A =
                                                               ε0

                           q
                   E0 =        …………………………………….(1)
                          ε0 A

If the dielectric is present , then net charge enclosed by Gaussian surface is q-q’ , q’
being induced charge due to polarization and so Gauss’s law gives
                   1
∫ E .dS       =
                  ε0
                        ( q − q ' )......... .......... .......... .......... .......... ( 2 )

          1
EA =              (q − q ' )
          ε0

          q             q'
E =               −        .......... .......... .......... .......... .......... ........( 3 )
       ε0A             ε0A
But we    have
 E   1
    = .......... .......... .......... .......... .......... .......... .......... ( 4 )
E0   k
    E0
E =
    K
      q
E =        .......... .......... .......... .......... .......... .......... .....( 5 )
    kε 0 A
Comparing    ( 3 ) and ( 5 ),
  q      q        q'
       =    −
kε 0 A   ε0    ε0A
                                                        1
or induced             ch arg e         q ' = q (1 −      )......... .......... .......... .......... .......... .......( 6 )
                                                        K




Obviously induced charge q’ is always less in magnitude than the free charge q and
equal to zero if no dielectric is present.

                                       q
From equation (6) q − q' =
                                       k

Substituting this in (2), the Gauss’s law in dielectric takes the form

                                                                  1      q
                                              ∫   E . dS =
                                                                ε    0   K
                                                                 q
                                              ∫   E . dS =
                                                                 ε
                                              ∫ ( ε E ) . dS     = q

                                              ∫ D . dS =         q
                               Lecture No.#5

   • Clausius Mossotti Relation:
Under the action of this field Em in the molecule displays an electric moment p in the
direction of the field which is dependent upon the field strength, i.e.,

p = αEm …………………………………………….(1)

Where α is the molecular polarizability. If there are n molecules per units volume of
the dielectric, then the polarization (electric moment/volume) is obtained as

                   P = np = nαEm
                                   P …………………………………..(2)
                   P = nα [ E +        ]
                                  3ε 0

        This equation may be rewritten in terms of the dielectric constant K, since

                                       P = ( K − 1)ε 0 E

                             In this way equation (2) becomes

                                                 ⎧     ( K − 1) E ⎫
                              ( K − 1)ε 0 E = nα ⎨ E +            ⎬
                                                 ⎩         3      ⎭
                              nα ( K − 1)
                                   =
                              3ε 0 ( K + 2)


                                                 3ε 0 ( K − 1)
                                     i.e., α =
                                                  n( K + 2)

     This equation is called Clausius Mossotti Relation and gives α , the molecular
                                        polarizability.
  • Problems
Practice


  1. Since capacitance is directly proportional to plate area, a lot of metal is needed to
     make a big capacitor. Since capacitance is also inversely proportional to plate
     separation, this metal foil should be separated by a very thin dielectric film. A
     thinner film means less metal foil is needed, but dielectric films can only be made
     so thin. Thus, big capacitances require big capacitors (capacitors with a large
     volume).

     That is, if one uses conventional materials and conventional designs. Enter the
     ultracapacitor (also known as the supercapacitor or electrochemical capacitor).
     Instead of two metal plates separated by a dielectric, an ultracapacitor uses a
     porous carbon electrode soaked in electrolytic paste. This effectively compresses
     a football field worth of surface area into a teaspoon of volume and shrinks plate
     separation down to the atomic scale.

     According to one manufacturer …

     An ultracapacitor gets its area from a porous carbon-based electrode material. The
     porous structure of this material allows its surface area to approach 2000 square
     meters per gram, much greater than can be accomplished using flat or textured
     films and plates. An ultracapacitor's charge separation distance is determined by
     the size of the ions in the electrolyte, which are attracted to the charged electrode.
     This charge separation (less than 10 angstroms) is much smaller than can be
     accomplished using conventional dielectric materials.
     Source: Maxwell Technologies.

     Determine the total …

         a. surface area,
         b. mass, and
         c. volume

     of the electrodes in a one farad ultracapacitor. (One angstrom is 10−10 m and the
     density of activated carbon is 0.50 g/cm3.)

     Solution …

         d. Answer it.
         e. Answer it.
         f. Answer it.

     Ultracapacitors can be used …
           o in place of rechargeable batteries for long periods in low current devices
             (like computer back up memory) and for shorter periods in high current
             devices (like power tools).
         o to provide a bridge current when power is switched from one source to
             another (when subway cars switch tracks, for example, or while waiting
             for a backup generator to come online during a blackout).
         o for load levelling high voltage, low energy devices connected to low
             voltage, high energy sources (remote radio transmitters connected to solar
             arrays, for example).
   2. Write something.
         o Answer it.
   3. Write something.
         o Answer it.
   4. Write something completely different.
         o Answer it.




Objective Questions:


Select the appropriate answer-


   1. Net outgoing electric flux through a closed surface enclosing charge qo is –
           (a) qo                                     (b)1/ qo
           (c) qo/ ε o                                (d) qo ε o


   2. Polarization of a dielectric represents-
           (a) Free surface charge density per unit electric field
           (b) Bound surface charge density per unit electric field
           (c) Free surface charge density
           (d) Bound surface charge density


   3. Which one is the wrong statement-
           (a) All dielectrics are insulators.
           (b) All insulators are dielectrics.
           (c) Symmetric molecules are non-polar.
       (d) Asymmetric molecules are polar.


4. Electric susceptibility of a dielectric is not depend upon-
       (a) Free surface charge density per unit electric field
       (b) Bound surface charge density per unit electric field
       (c) Magnetic moment per unit volume
       (d) Polarization


5. Which of the following polarization depend upon the temperature-
       (a) electronic polarization                 (b) ionic polarization
       (c) orientational polarization              (d) none


6. The orientational polarisability per molecule in a polyatomic gas is given by-
               μm                                         μm
                                                           2
       (a)                                         (b)
             3k B T 2                                    3k B T

               μm
                3
                                                           μm
       (c)                                         (d)
             3k B T                                      3k B T 2


7. In a dielectric, with increase in the external electric field, electronic polarisability
   increases and internal field-
       (a) increases                               (b) reduces
       (c) remains same                            (d) is always zero


8. For a dielectric with increase in temperature, the electronic polarisability will-
       (a) increase                                (b) decrease
       (c) remain same                             (d) increase/or decrease/or no change




9. The factor responsible for spontaneous polarization is
       (a) tightly bound electrons                 (b) permanent dipoles
          (c) free electrons                         (d) atoms


10. The losses in a dielectric subjected to a alternating electric field are determined
    by-
          (a) real part of the complex dielectric constant
          (b) imaginary part of the complex dielectric constant
          (c) both real and imaginary part of the complex dielectric constant
          (d) none of above


11. At frequencies around 5x1014 Hz, the ionic polarization becomes-
          (a) zero                                   (b) unity
          (c) infinite                               (d) negative


12. In a dielectric, the power loss is proportional to-
          (a) ω                                      (b) ω 2
                1                                          1
          (c)                                        (d)
                ω                                          ω2


13. In a dielectric, the polarization is -
          (a) linear function of applied electric field
          (b) square function of applied electric field
          (c) exponential function of applied electric field.
          (d) logarithmic function of applied electric field


14. The dipole moment per unit volume of a siolid is the sum of all the individual
    dipole moments and is called-
          (a) polarization of the solid
          (b) permittivity of the solid
          (c) electrstatic moment
          (d) none of these
   15. In a dielectric, the plarization P, applied electric field E and electric flux density D
         are related as-
             (a) E = ε o D + P                                   (b) D = E + ε o P

             (c) D = E ε o + P                                   (d) D = ε o (E + P)




Ans:         1. (c)              2. (d)          3. (b)          4. (a)         5. (b)
             6. (b)              7. (a)          8. (c)          9. (b)         10. (b)
             11. (a)             12. (a)         13. (a)         14 (a)         15. (c)




Short Answer Type Questions:
   1.        What are polar and non-polar molecules? Give some examples.
   2.        Does the dielectric constant for substance containing permanent dipoles vary
             with temperature?
   3.        Molecules consisting of dissimilar atoms are usually polar. Why?
   4.        Show that P = ε 0 ( ε r − 1) E , where P is polarization vector.
   5.        State and prove Gauss law in dielectrics.
   6.        What are dielectric losses and dielectric breakdown? Explain. What is the
             function of oil in transformers?
   7.        Write short note on dielectrics.
   8.        What do you mean by polarization of dielectric and dielectric susceptibility?
             Find the relationship between the two.
   9.        Write the Claussius Mossoti equation and explain the symbols.
   10.       How is dielectric loss utilized in cooking food in microwave oven? Explain.
Long Answer Type Questions:


 1.    Write the definition of three electric vectors E, D & P. Derive a relation
       between them.
 2.    Explain the terms - dielectric constant, permittivity, molecular polarisability and
       electric susceptibility. Derive a relation between dielectric constant and electric
       susceptibility.
 3.    Discuss the following phenomenon- (i) Electric Polarisation (ii) Ionic
       Polarisation (iii) Orientational Polarisation and (iv) Interfacial Polarisation.
 4.    A field of strength E is applied to a dielectric. Show that the energy stored per
       unit volume of the dielectric is (½) P.E, where P is the polarisation in the
       dielectric.
 5.    Prove that the induced charge varies with the dielectric constant K as
                   σp
        K = (1 −         ) −1 when a dielectric is placed in between the plates of a capacitor.
                     σ
       Also prove that the value of K tends to infinitive for a metal dielectric. Where
       σp and σ are the induced and free surface charge densities, respectively.
                                            1
 6.    Derive the relation, q ′ = q( 1 −      ) in the case of a capacitor with dielectric of
                                            K
       dielectric constant K in between its plate. Where, q ′ and q are the induced and
       free charges.
 7.    Write down the expression for total polarization P of a polyatomic gas. Explain
       how permanent dipole moment of molecules can be determined if P is known at
       different temperatures.
 8.    Derive the Claussius Mossoti relation expressing the dielectric constant of a
       solid.
 9.    Explain the frequency dependence of polarisation of a dielectric in an a.c.
       electric field.
 10.   What is meant by loss tangent and dipole relaxation of a dielectric? What is the
       importance of these terms for a dielectric material?
 11.     State and prove Gauss law in dielectrics. Deduce an expression for energy
         stored in dielectric in the electrostatic field.
 12.     What are the causes behind the dielectric losses occurring in RF, IR and visible
         region of the electromagnetic spectrum.
 13.     What are direct &inverse piezoelectric effects? Explain their importance and
         applications.




Numericals:


 14.     A parallel plate capacitor has a capacitance of 10μF. The dielectric has a
         permittivity εr=100 for an applied voltage of 1000 V, find the energy spent in
         polarising the dielectric.
(1 Joule, 0.99 Joule)


 15.     A parallel plate capacitor has a capacitance of 100 pF, a plate area of 100cm2
         and a mica dielectric (K=5.4). At 50V potential difference, calculate (a) E in
         mica (b) the magnitude of the free charge on the plates and (c) the magnitude of
         the induced surface charge on the mica.


 16.     Two parallel plates of area 100 cm2 are given charges of equal and opposite
         magnitude 8.9x 10-7 C. The electric field within the dielectric material filling the
         space between the plates is 1.4 x 106 V/m. Calculate- (a) the dielectric constant
         of the material (b) the magnitude of the charge induced on each dielectric
         surface.
(7.1, 7.6x10-6C)


 17.     A potential source of 100 Volts is connected across the plates (area=100cm2 ,
         separation= 1.0 cm) of a capacitor till the capacitor is fully charged and then
         disconnected. A dielectic slab of thickness 0.50 cm and dielectric constant K=
         7.0, is then inserted between the plates of the capacitor.          Calculate- (a)
           capacitance before the slab is inserted, (b) free charge on the plates, (c) electic
           field strength in the gap, (d) electric field strength in the dielectric, and (e)
           magnitude of vectors D and P in- (i) dielectric, (ii) air gap.
          ((a) 8.85x10-12 fd, (b) 8.85x10-10 fd, (c) 1.0x104Volt/m, (d) 0.14x104Volt/m, (e) (i)
          D= 8.85x10-8 C/m2, P=7.475x10-8 C/m2, (ii) D= 8.85x10-8 C/m2, P= 0 C/m2).


 18.       In previous question, if 100 Volt battery remains connected during the time the
           dielectric slab is being inserted, calculate- (a) the charge on the capacitor plates,
           (b) the electric field in the gap, (c) the electric field in the slab, (d) the
           capacitance.
          ((a) 61.95x10-10 C, (b) 7x104Volt/m, (c) 7x104Volt/m, (d)61.95x10-12 fd)


 19.       Calculate the capacity per cm of a capacitor formed from the two long co-axial
           cylinders of radii 6 cm & 8cm. The space between the two cylinders is filled
           with a material of dielectric constant K= 3.1.


 20.       Find the total polarisability of CO2 at 0°C and 1 atm pressure, if its
           susceptibility in this case is 9.85×10-4.
(3.2x10-40 Fd.m2)


 21.       The dielectric constant of a gas εr=1.006715 at 27°C and εr=1.005970 at 177°C.
           The molecular density is 2.5×1025 per m3, calculate the dipole moment of
           molecules and sum of electronic and ionic polarisabilities. (1 Debye unit =
           3.33x 10-30 Coul-mtr)
(0.95 Debye unit, 1.6x10-39 Fd.m2)


 22.       A solid contains 5×1028 identical atoms per m3, each with a polarisibility of
           3.6×10-40 Fd. m2 . Assuming that the internal field is given by Lorentz formula,
           calculate the ratio of the internal field to the applied field.
(3.105)
 23.       In a dielectric of relative permittivity εr is placed in a uniform external electric
           field. If the electric displacement vector is 25×10-9 C/m2. If the volume of the
           slab is 1.5 m3 and the magnitude of polarisation is 5×10-9 C/m2, find the value of
           εr and total dipole moment of the slab. (Given ε0=8.854×10-12 C2/N.m2).


 24.       If all the molecular dipoles in a 0.1 cm radius water drop are pointed in the same
           direction, calculate the intensity of polarization. Dipole moment of the water
           molecule is 6x10-30 C- m.
(8.4x10-7 C/m2)
 25.       Argon gas contains 2.7x1025atoms/m3 at 0 oC and at one atm. Pressure.
           Calculate the dielectric constant of argon gas at this temperature, if the diameter
           of argon atom is 0.384 nm.
(1.0024)

				
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