Dielectrics 9.1 INTRODUCTION 9.2 DIELECTRIC CONSTANT

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1 9 Dielectrics 9.1 INTRODUCTION A dielectric is an insulating material in which all the electrons are tightly bound to the nuclei of the atoms and there are no free electrons available for the conduction of current. Therefore, the electrical conductivity of a dielectric is very low. The conductivity of an ideal dielectric is zero. On the basis of band theory, the forbidden gap ( g ) is very large in dielectrics. Materials such as glass, polymers, mica, oil and paper are examples of dielectrics. They prevent flow of current through them. Therefore, they can be used for insulating purposes. 9. DILCTRIC CONSTANT It is found experimentally that the capacitance of a capacitor is increased if the space between its plates is filled with a dielectric material. To understand this fact, Faraday took two identical capacitors, one was evacuated and the other was filled with dielectric material, as shown in Fig Then these two capacitors were charged with a battery of same potential difference. He found that the charge on the capacitor filled with dielectric is larger than that of the other filled with air. If C be the capacitance in vacuum and C the capacitance when the space is filled with a dielectric material, then the dielectric constant of the material C K = C I Fig. 9.1 Thus, the dielectric constant of a material is the ratio of the capacitance of a given capacitor completely filled with that material to the capacitance of the same capacitor in vacuum. In other words, the ratio of permittivity of medium to that of the vacuum is also known as dielectric constant,

2 9. ngineering Physics i.e., ε K = = ε ε r This is also known as relative permittivity (e r ). It is found to be independent of the shape and dimension of the capacitor. 9.3 TYPS OF DILCTRICS A molecule is a neutral system in which the algebraic sum of all the charges is zero. Based on the dipole moment, the molecules of dielectrics are termed as non-polar and polar molecules. Accordingly these dielectrics are referred to as non-polar and polar dielectrics Non-polar Dielectrics A non-polar molecule is the one in which the centre of gravity of the positive (protons) and negative charges (electrons) coincide. So such molecule does not have any permanent dipole moment, as shown in Fig. 9.a. Few common examples of non-polar molecules are oxygen (O ), nitrogen (N ) and hydrogen (H ). As mentioned earlier, the dielectrics having non-polar molecules are known as non-polar dielectrics Polar Dielectrics A polar molecule is the one in which the centre of gravity of the positive charges is separated by finite distance from that of the negative charges. Unbalanced electric charges, usually valency electrons, of such molecules result in a dipole moment and orientation. Therefore, these molecules possess permanent electric (a) (b) dipole (Fig. 9.b). Few examples of polar molecules are N O, H O and HCl. The dielectrics having polar molecules are known as polar dielectrics. Fig POLARISATION OF DILCTRICS When an electric field is applied to a dielectric material; it exerts a force on each charged particle and pushes the positive charge in its own direction while the negative charge is displaced in opposite direction, as shown in Fig Consequently, the centres of positive and negative charges of each atom are displaced from their equilibrium positions. Such a molecule (or atom) is then called as induced electric dipole and this process is known as dielectric polarisation. We consider a parallel plate capacitor which has vacuum initially between its plates. When it is charged with a battery, the electric field of strength is set up between the plates of the capacitor (Fig. 9.4a). If s and s are the surface charge densities of the two plates of the capacitor, then the electric field developed between the plates is given by Fig. 9.3 = σ (i) ε

3 Dielectrics 9.3 If now a slab of dielectric material is placed between the two plates of the capacitor (Fig. 9.4b), then it becomes electrically polarised. Hence, its molecules become electric dipole oriented in the direction of the field. Because of this the centre of positive and negative charges gets displaced from each other. Therefore, in the interior of the dielectric as marked by dotted lines these charges cancel. However, the polarisation charges on the opposite faces of the dielectric slab are not cancelled. These charges produces their own electric field p, which opposes the external applied field. Under this situation, the net electric filed in the dielectric is given by Fig = p (ii) Polarisation Density The induced dipole moment developed per unit volume in a dielectric slab on placing it inside an electric field is known as polarisation density. It is denoted by a symbol P. If p is induced dipole moment of individual atom and N is the number of atoms in a unit volume, then polarisation density is P = Np (iii) The induced dipole moment of an individual atom is found to be proportional to the applied electric field and is given by P = αε (iv) where a is the proportionality constant and is also known as atomic polarisability. From qs. (iii) and (iv), we get P = Nαε (v) Suppose S is the area of each plate of the capacitor and d is the separation between them (Fig. 9.5). Then the volume of the dielectric slab is Sd. Since q i and +q i are the induced charges developed on the two faces of the dielectric slab, the total dipole moment of the slab will be equal to q i d. From the definition of the polarisation density P = Total dipole moment Volume of slab qid qi = = = Sd S σ p \ P = s p (vi)

4 9.4 ngineering Physics On placing the dielectric material between the two plates of the capacitor, the reduced value of the electric field may be evaluated as follows σ p p = σ σ σ = ε ε ε (vii) or P = ε (viii) Fig. 9.5 From q. (vii) ε = σ σ σ P [ p = P = σp ] or s = e + P (ix) The quantity (e + P) is of special significance and is known as the electric displacement vector D given by D = ε + P (x) 9.4. Relation between Dielectric Constant and lectric Susceptibility The polarisation density of a dielectric is proportional to the effective value of electric field and is given by P = χε (xi) where χ is constant of proportionality and is known as susceptibility of dielectric material. By using q. (viii), we get χε = = χ ε or = (1 + χ) or / = 1 + χ or K = 1 + χ K = 9.5 TYPS OF POLARISATION The important types of polarisation are categorised as under. [ ] lectronic Polarisation Under the action of an external field, the electron clouds of atoms are displaced with respect to heavy fixed nuclei to a distance less than the dimensions of the atom (Fig. 9.6). This is called electronic polarisation, which does not depend on temperature. The electronic polarisation is represented as below P = Nα (i) e e

5 Dielectrics Ionic Polarisation This type of polarisation occurs in ionic crystals, for example in sodium chloride crystal. In the presence of an external electric field, the positive and negative ions are displaced in opposite directions until ionic bonding forces stop the process (Fig. 9.7). This way, the dipoles get induced. The ionic polarisation does not depends upon temperature. Fig Orientation Polarisation This types of polarisation is applicable in polar dielectrics. In the absence of an external electric field, the permanent dipoles are oriented randomly such that they cancel the effects of each other (Fig. 9.8a). When the electric field is applied, these dipoles tend to rotate and align in the direction of the applied filed (Fig. 9.8b). This is known as orientation polarisation, which depends upon temperature. Absence of field (a) Fig. 9.7 Presence of field (b) In view of all these polarisations, the total polarisation is the sum of the electronic, ionic and orientation polarisations. This is given by P = P + P + P e i o Fig GAUSS S LAW IN DILCTRICS Gauss s law states that the surface integral of electric field vector over a closed surface is equal to 1/e times the net charge enclosed by the surface, i.e., d S = 1 q ε or ε d S = q or ε S = d q Let us consider a parallel plate capacitor without dielectric, as shown in Fig. 9.9a. Then Gauss s law is written as ε S = d q or e S = q

6 9.6 ngineering Physics or = q ε S where is the electric filed between the plates. (i) Fig. 9.9 or e S = q q or = q q εs ε S As we know that the relative permittivity is K = or = K With the help of q. (i), we get q = = K Kε S In the presence of dielectric material between capacitor plates, it is clear from Fig. 9.9b that the total charge enclosed by Gaussian surface is (q q ), where q is the induced charge in the dielectric due to polarisation. Then Gauss s law says ε d S = ( q q ) (ii) (iii) (iv) (v) Using q. (v) in q. (iii) we obtain q q q = KεS εs εs q or q q K = (vi) The induced charge is therefore = 1 q q 1 K From the above relation it is clear that the induced charge q is less than the free charge q. q By substituting the value of ( q q ) = in q. (ii), we get for Gauss s law K q q q ε ds = or K ds = = or εk ε e ds = q i.e., D d S = q This is the Gauss s law in the presence of dielectric. 9.7 NRGY STORD IN AN LCTROSTATIC FILD Consider a dielectric slab of length l and cross-sectional area S is subject to an external electric field (Fig. 9.1). Under the effect of external field it becomes polarised.

7 Dielectrics 9.7 If q is the magnitude of charge and dl is the small change in displacement, then the change in dipole moment is given by dp = qdl (i) The work done for displacement dl is given as dw = F dl (dl is the displacement) dw = q dl = ( qdl ) or dw = dl (ii) Fig. 9.1 This work done is stored as electrostatic energy in dielectric material. As we know D = P + ε and D = ε = Kε [ ε = Kε ] P = D ε = Kε ε or P = (K 1) ε Now by definition, the polarisation P is induced dipole moment per unit volume. Since V = Sl p P = or p = PV V or d p = VdP = V( K 1) ε d where we have used q. (iii) or dp = (K 1) ε Vd (iv) With the help of q. (iv), q. (ii) reads dw = ( K 1) ε V d (v) By integrating q. (v), we get W = ( K 1) εv d = ( K 1) εv d 1 = ( K 1) εv (vi) Therefore, the work done per unit volume W 1 = ( K 1)ε (vi) V The above equation gives the energy density due to the polarisation. The net energy density is equal to the energy density in free space plus energy density due to the polarisation, given by 1 ε u = ε + ( K 1) 1 u = ε K (viii) In free space K=1 u = 1 ε (ix) (iii)

8 9.8 ngineering Physics 9.8 DILCTRIC LOSS When a dielectric material is placed in an alternating electric field (Fig. 9.11a), a part of the energy is wasted, which is known as dielectric loss. This is because of the fact that the reversing nature of the field causes the direction of the dipoles to reverse. The dielectric loss depends on the frequency and the mechanism by which the polarisation is produced in the material. An ideal dielectric does not absorb electrical energy. However, in a real dielectric, there is always a loss of some electrical energy. Consider a parallel plate capacitor of capacity C, whose plates having area S are separated by a distance d. The space between the plates of the capacitor is filled with dielectric material having permittivity e. The sinusoidal voltage V of angular frequency ω is applied to the capacitor. Then the current through the capacitor is Q V I = CV V t + R = t + R Since ω is inversely proportional to the time period, the current I may be written as I = ωcv + V R or I = Ic + Id (i) where I c and I d are conduction and displacement currents, respectively. From the above relation, it is clear that there are two kinds of currents that flow through the dielectric. The Current I c, I d and I are plotted in Fig. 9.11b, from which it is clear that the resultant current 1 I = ( I + I ) / c d lags behind the displacement current by an angle δ. In an ideal dielectric V R =, which means Ic = becomes zero. In R this situation the resultant current would be I = I d = ωcv Now C = ε S d free space and e e r S for the d capacitor filled with dielectric material. Therefore, ε εrsωv I = I d = (ii) d Ic The angle δ is known as loss angle, which can be calculated from Fig b. In OAD, tanδ = I qs. (ii) and (iii) yield so Ic = Id tan δ (iii) I c Fig rsv = ( ωε ε ) tan δ d d

9 Dielectrics 9.9 Thus, the real power loss in the dielectric materials is Pl = VIc ωεεrsv = tan δ d ωεεr ( Sd) V = tan δ d = f V V π ε εr d tan δ = ( πε ) εr f V tan δ 11 P = V ε f tanδ l r The above expression shows that the power loss depends on the volume V of the dielectric, its dielectric constant e r, frequency f of the alternating field together with its amplitude. 9.9 CLAUSIUS-MOSOTTI QUATION The interaction between the atoms of gases can be neglected as the atoms are separated by sufficiently large distance. So atoms of gases feel the same field as applied to them. However, in solids and liquids, the atoms are closely surrounded (on all the sides) by other atoms which may be polarised under the action of an external field. Hence, the internal intensity of the electric field at a given point of the dielectric is generally not equal to the intensity of the applied field. The internal field is actually the electric field acting at the location of a given atom and is equal to the sum of the electric field created by neighbouring atoms and the applied field. In case of crystal possessing cubic symmetry, the internal field is given by i = + P / 3ε This field is called Lorentz field and also some time referred to as local field. In linear and isotropic dielectric the molecular dipole moment p is directly proportional to the internal field i, that is p = α i p = α( + P / 3ε ) (i) As mentioned earlier, the proportionality constant α here is known as molecular polarisability. In terms of number of molecules per unit volume N, the polarisation density P is given by P = Np = Nα( + P / ε ) (ii) The above equation for P gives Nα P = 1 Nα / 3ε (iii) In terms of the electric susceptibility χ, the above equation is written as P = ε χ (iv) A comparison of qs. (iii) and (iv) yields

10 9.1 ngineering Physics q. (v) gives Nα = εχ 1 Nα / 3ε or 3ε ( K 1) α = N(K + ) K 1 α + = N K 3ε The above equation is known as Clausius-Mosotti equation. This equation is valid for non-polar dielectrics having cubic crystal structure. The Clausius-Mosotti equation is also known as Lorentz-Lorentz equation in view of its application in optics. 9.1 SUMMARY The topics covered in this chapter are summarized below. (1) A dielectric is an insulating material in which all the electrons are tightly bound to the nuclei of the atoms and there are no free electrons available for the conduction of current. Therefore, the electrical conductivity of a dielectric is very low. The forbidden gap ( g ) is very large in dielectrics. Materials such as glass, polymers, mica, oil and paper are a few examples of dielectrics. () A non-polar molecule is the one in which the centre of gravity of the positive charge (protons) and negative charge (electrons) coincide. So such molecules do not have any permanent dipole moment. Nitrogen (N ) and hydrogen (H ) are the examples of non-polar molecules. (3) A polar molecule is the one in which the centre of gravity of the positive charges is separated by finite distance from that of the negative charges. Unbalanced electric charges, usually valency electrons, of such molecules result in a dipole moment and orientation. Therefore, these molecules possess permanent electric dipole. xamples of polar molecules are N O, H O and HCl. (4) An external electric field, when applied to a dielectric material, exerts a force on each charged particle and pushes the positive charge in its own direction while the negative charge is displaced in opposite direction. Consequently, the centres of positive and negative charges of each atom are displaced from their equilibrium positions. Such a molecule (or atom) is then called as induced electric dipole and this process is known as dielectric polarisation. (5) The induced dipole moment developed per unit volume in a dielectric on placing it inside an electric field is known as polarisation density P. If N be the number of atoms in a unit volume and α the atomic polarisability, then polarisation density is P = Nαε. (6) lectric susceptibility χ and the dielectric constant K are related as K = 1+ χ. (7) Polarisation is of three types, namely electronic polarisation, ionic polarisation and orientation polarisation. (8) Gauss s law states that the surface integral of electric field vector over a closed surface is equal to 1 e times the net charge enclosed by the surface, i.e., ds = q ε (9) The energy stored in an electrostatic field is u = 1 ε K which takes the form u = 1 ε in the free space. (v)

11 Dielectrics 9.11 (1) The internal intensity of the electric field at a given point of the dielectric is generally not equal to the intensity of the applied field. The internal field is actually the electric field acting at the location of a given atom and is equal to the sum of the electric field created by neighbouring atoms and the applied field. In case of crystal possessing cubic symmetry, a relation K 1 N K + = α exists 3ε between the dielectric constant K, atomic polarisability a and permittivity e. This equation is known as Clausius-Mosotti equation, which is also known as Lorentz-Lorentz equation in view of its application in optics. SOLVD XAMPLS xample 1 Two parallel plates having equal and opposite charges are separated by a cm thick slab that has dielectric constant 3. If the electric filed inside is 1 6 V/m. Calculate the polarisation and displacement vector. Solution Given = 1 6 V/m = 1 6 N/C, K = 3, = C N m Formula used is D = + P Also D = K or 1 6 D = = C/m P = D 5 = = C / m xample Two parallel plates have equal and opposite charges. When the space between them is evacuated the electric intensity is V/m and when the space is filled with dielectric the electric intensity is V/m. What is the included charge density on the surface of the dielectric? 5 5 Solution Given = 3 1 V/m, and = 1 1 V/m Formula used is P = or P = σ p = ( ) [ P = σ p] s p = [3 1] 1 5 s p = C / m xample 3 Two parallel plates of capacitor having equal and opposite charges are separated by 6. mm thick dielectric material of dielectric constant.8. If the electric field strength inside be 1 5 V/m, determine polarisation vector, displacement vector and energy density in the dielectric.

12 9.1 ngineering Physics Solution Given = 1 5 V/m = 1 5 N/C and K =.8 P = ( K 1), D = K and energy density = 1 K P = (.8 1) 1 5 = C / m = C / m D = = C / m = C / m nergy density = ( 1 ) xample 4 = J / m 3 =.14 J / m An isotropic material of relative permittivity e r is placed normal to a uniform external electric field with an electric displacement vector of magnitude C / m. If the volume of the slab is.5 m 3 and magnitude of polarisation is C / m, find the value of e r and total dipole moment of the slab. Solution Given D = C/m, P = 4 1 C/ m and V =.5 m 3 e r = K =? Formula used is or D = + P ε = (D P)/ ε = 4 ( 5 4) = V/m D 5 1 K = εr = = 4 ε 1 4 = 5 Total dipole moment p P = = Volume V p = PV = =. 1 4 C-m xample 5 Dielectric constant of a gas at N.T.P is Calculate dipole moment of each atom of the gas when it is held in an external field of V/m. Solution Given 4 = 3 V/ m = 3 1 N/ Cand K = = r Formula used is K = 1 + χ or χ = K 1 = =.74 and polarisation density is P = χ = = C / m

13 Dielectrics 9.13 No. of atoms of gas per cubic metre (N) = = Induced dipole moment of each atom (p) = P N = or p = C-m 11 xample 6 Determine the electric susceptibility at C for a gas whose dielectric constant at C is Solution Given K = 1.41 and T = C Formula used is K = 1 + χ or χ = K 1 = = = OBJCTIV TYP QUSTIONS Q.1 Dimension of atomic polarisability in SI units is (a) Cm (b) CV 1 m (c) CVm (d) none of them Q. A non-polar molecule is the one in which the centre of gravity of positive and negative charges (a) coincides (b) gets separated by 1 8 m (c) gets separated by 1Å (d) none of these Q.3 The net charge inside a dielectric before and after polarisation remains (a) negative (b) positive (c) same (d) none of these Q.4 In vacuum, electric susceptibility is (a) less than 1 (b) greater than 1 (c) small but ve (d) zero Q.5 The electric susceptibility and dielectric constant are related as (a) χ = K ~ 1 (b) χ = 1 + K (c) χ = K (d) χ = 1 + K Q.6 Dimension of displacement vector in SI unit is (a) C-m (b) C-m 1 (c) C-m (d) C-m Q.7 The relation between three electric vectors, D and P is (a) D = + P (b) D = + P 1 (c) D = + P (d) D = + P ( )

14 9.14 ngineering Physics Q.8 Gauss s law in dielectrics is (a) ds = 1 q (b) D ds = q (c) ds = q (d) both (a) and (b) Q.9 What changes in the capacitance of a capacitor occur if the dielectric material between the plates of the capacitor is replaced by air or vacuum? (a) increases (b) decreases (c) remains same (d) none of these Q.1 Dielectrics are the substances which are (a) conductors (b) insulators (c) semi-conductor (d) none of these SHORT-ANSWR QUSTIONS Q.1 What are polar and non-polar molecules? Q. Discuss different polarisation mechanisms in dielectrics? Q.3 What is a dielectric? Q.4 How long does polarisation of non-polar molecules last? Q.5 State and prove Gauss law in dielectrics. Q.6 What are dielectric losses? Q.7 Discuss the behaviour of a dielectric in a.c. field. Q.8 Write the Clausius-Mossotti equation. Q.9 Will an atom having spherically symmetric charge distribution be polar or non-polar? xplain. Q.1 What do you understand by polarisation of dielectric and dielectric susceptibility? Find the relation between the two. Q.11 Write note on (i) Dielectrics (ii) Three electric vectors (iii) Dielectric losses. PRACTIC PROBLMS General Questions Q.1 What is a dielectric substance? Give examples. Discuss the importance of dielectrics. Q. What are polar and non-polar molecules? Discuss the effect of electric field on polar dielectrics. What is meant by polarisation of dielectric? Q.3 Discuss different types of polarisations in dielectrics. Q.4 What happens when a non-polar molecule is placed in an electric field? Define atomic dipole moment and atomic polarisability. What are their dimensions? Give their S.I. units. Q.5 What is atomic polarisability? Find a relation between dipole moment and atomic polarisability or show that p = α. Q.6 Show that the electric field inside a polarised dielectric due to induced polarisation charge is P = where P is the polarisation density vector. o

15 Dielectrics 9.15 Q.7 xplain the terms dielectric polarisation, susceptibility, permittivity and dielectric coefficient. Derive their inter-relation equation. Q.8 Define and explain the three electric vectors P, and D. Why electric field inside a dielectric decreases due to polarisation? Show that D = o + P. Also give their units. Q.9 Show that D = o + P, where the symbols have their usual meanings. Q.1 What are three electric vectors in dielectrics? Name and find relation between them. Q.11 What do you understand by polarisation of dielectric and dielectric susceptibility? Find the relation between the two. Q.1 xplain the phenomenon of polarisation of dielectric medium and show that K = 1+ χ e. Here the symbols have their usual meanings. Q.13 Define the terms dielectric constant K and electric susceptibility χ e. Prove the relation K = 1+ χ e. Q.14 Find the relation between induced charge and free charge when a dielectric material of dielectric constant K is placed between the plates of a parallel plate capacitor. 1 Q.15 Prove that induced charge varies with the dielectric as K = σ p 1, where σ σ p and σ free are the free induced and free surface charge densities, respectively. Hence show that for a metal K =. Q.16 The electric field between the plates of a parallel plate capacitor is without dielectric. But if dielectric of relative permittivity ε r is introduced between the plates what will the electric field be? Q.17 What is the effect of temperature on the dielectric constant of a substance containing molecules of permanent dipole moment? Q.18 Derive a relation between electric susceptibility and atomic polarisability on the basis of microscopic description of matters at atomic level. Q.19 Derive Clausius-Mosotti relation for non-polar dielectrics. Q. Discuss the effect of introducing a dielectric between the plates of a capacitor. Show that the capacitance of a charged capacitor when a dielectric material of dielectric constant K is introduced A between the plates is given by K. d Q.1 xplain why the introduction of a dielectric slab between the plates of a capacitor changes its capacitance? Q. State and prove Gauss s law in dielectrics. Q.3 Derive an expression for Gauss s law in the presence of dielectric. Prove that divergence of displacement vector is equal to density of free charge or D = ρ free. Also discuss integral form of Gauss s law. Q.4 Using Gauss s law in dielectric medium, show that D = ρ free, where symbols have their usual meanings. Q.5 xplain the mechanism contributing to dielectric polarisation. Discuss the behaviour of a dielectric in an alternating field. Q.6 Show that the electrostatic energy per unit volume in a dielectric is 1 D, where symbols have their usual meanings. Q.7 Deduce an expression for energy stored in dielectric in electrostatic field.