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ENGINEERING PHYSICS II DIELECTRICS 4.5 VARIOUS POLARIZATION MECHANISMS IN DIELECTRICS Dielectric polarization is the displacement of charged particles under the action of the electric field. There are number of devices based on this concept. Those devices are rectifiers, resonators, amplifiers and transducers which converts electrical energy in to other forms of energy. In modern computers, memory devices are also based upon this concept. Several microscopic mechanisms are responsible for electric polarization. Specially, in the case of d.c. electric field, the macroscopic polarization vector P arises due to following four types of microscopic polarization mechanisms i. Electronic polarization ii. Ionic polarization iii. Orientation polarization iv. Space-charge polarization. 4.5.1 ELECTRONIC POLARIZATION Electronic polarization occurs due to the displacement of positively charged nucleus and negatively charged electrons in opposite directions, when an external electric field is applied and thereby creates a dipole moment in the dielectric. The induced dipole moment is given by µ=αe E Where, αe is the electronic polarizability. Monoatomic gases exhibit this kind of polarization. Electronic polarizability is proportional to the volume of the atoms and is independent of temperature. Calculation of electronic polarizability Without field Let us consider a classical model of an atom. Assume the charge of nucleus of that atom is Ze. The nucleus is surrounded by an electron cloud of charge –Ze, which is distributed in a sphere of radius R as shown in the fig The charge density of the charged sphere = −Ze 4πR3 3 75 ENGINEERING PHYSICS II (Or) DIELECTRICS charge density = −3Ze …(1) 4πR3 With field When the dielectric is placed in a d.c. electric field E, two phenomenon occur i. Lorentz force due to the electric field tends to separate the nucleus and the electron cloud from their equilibrium position. ii. After separation, an attractive coulomb force arises between the nucleus and the electron cloud which tries to maintain the original equilibrium position. Let x be the displacement made by the electron cloud from the positive core, as shown Since the core is heavy, it will not move when compared to the movement of the electron cloud. Here x <<R, where R is the radius of the atom. Since Lorentz and Coulomb forces are equal and opposite in nature, equilibrium is reached. At Equilibrium, Lorentz force = Coulomb force Lorentz force = Charge × Field … (2) = -ZeE The negative sign indicates the repulsive force. Coulomb force = Charge × Field = +Ze × 𝑄 4𝜋∈𝑜 𝑥 2 The positive sign indicates the attractive force. Coulomb force = Charge × Total negative charges (Q) enclosed in the sphere of radius x 4𝜋∈𝑜 𝑥 2 ….(3) The total number of negative charges (Q) enclosed in the sphere of radius ‘x’ = Charge density of electrons × volume of the sphere 76 ENGINEERING PHYSICS II DIELECTRICS Substitute the charge density from eqn (1), we get The total number of negative charges (Q) enclosed in the sphere of radius x −3Ze = 4 × π x3 4πR3 −𝑍𝑒 ∴Q= is 3 x3 ….. (4) R3 Substitute eqn (4) in (3) Coulomb force = Ze 4𝜋∈𝑜 𝑥 2 = . . −𝑍𝑒 x3 R3 −Z2 e2 x ..… (5) 4π∈o R3 At equilibrium position, eqn (2)= eqn (5) ∴ −ZeE = x = (Or) −Z 2 e2 x 4π ∈𝑜 R3 4π∈𝑜 R3 E .....(6) 𝑍𝑒 Therefore, the displacement of electron cloud (x) is proportional to the applied electric field E. Dipole moment Now the two electric charges +Ze and –Ze are displaced by a distance x under the influence of the field and form an induced dipole moment which is given by Induced dipole moment (µe) = Magnitude of charge × displacement = Zex Substituting the value of x from eqn (6), we get µe = Ze4π∈𝑜 R3 E 𝑍𝑒 µe = 4π∈𝑜 R3E µe α E µe = αe E Where, αe = 4π∈𝑜 R3 (Farad-m3) is called electronic polarization which is proportional to the volume of the atom. Relation between αe and Dielectric Constants We know, the induced electronic dipole moment is proportional to the applied field. This dipole moment per unit volume is called electronic polarization. This is independent of temperature. Electronic polarization Pe=N µe Where, N is the number of atoms 77 ENGINEERING PHYSICS II DIELECTRICS Pe=N αe E Since Pe E …….(8) = ∈𝑜 (∈𝑟 -1), we can write αe = ∈𝑜 (∈𝑟 −1) …….(9) 𝑁 4.5.2 IONIC POLARIZATION Ionic polarization arises due to the displacement of cations (+ve ions) and anions (- ve ions) from its original position, in opposite directions in the presence of electric field as shown in figure. Explanation Let us assume that there are one cation and one anion present in each unit cell of the ionic crystal NaCl. When the electric field is applied, let x1 and x2 be the distances to which positive and negative ions move from their equilibrium positions. The resultant dipole moment per unit cell, due to ionic displacement is given by Induced dipole moment = magnitude of charge × displacement ……… (1) µi = e(x1 + x2) Where, x1 is the shift of +ve ion and x2 is the shift of –ve ion, from their equilibrium positions. When the field is applied, the restoring force produced is proportional to the displacements. For +ve ion, Restoring force F α x1 or F = β1 x1 ……….(2) For –ve ion, Restoring force F α x2 or F = β2 x2 ……….(3) Here, β1 and β2 are restoring force constants, which depend on the masses of the ions and angular frequency of the molecule in which the ions are present. If ‘m’ is the mass of +ve ion and ‘M’ be the mass of the –ve ion and ωo is the angular frequency, then β1 =m ω2o ….….(4) β2 =M ω2o …….(5) Substituting for β1 in eqn (2), the restoring force for +ve ion can be written as F = mω2o x1 …….(6) 78 ENGINEERING PHYSICS II We know, DIELECTRICS …….(7) F = eE Equating eqn (6) and (7), we get eE = mω2o x1 x1 = Similarly for –ve ion we can write 𝑒𝐸 ……..(8) mω2o eE x2 = Mω2 ……..(9) o Adding eqn (8) and (9) we get x1 + x2 = 𝑒𝐸 ω2o 1 (m + 1 M ) ..…..(10) Substituting eqn (10) in (1), we get 𝑒 2𝐸 1 1 ( + ) 2 ωo m M µi = αi E µi = Or Where, αi is the ionic polarization given by αi = 𝑒2 ω2o 1 (m + 1 M ) so, the ionic polarizability αi is inversely proportional to the square of the natural frequency of the 1 ionic molecule and directly proportional to its reduced mass which is given by (m + 79 1 M )