Chapter 5 Electric Fields in Material Space

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1 Chapter 5 Electric Fields in Material Space Islamic University of Gaza Electrical Engineering Department Dr. Talal Skaik

2 Introduction In chapter 4, Electrostatic fields in free space were considered. This chapter covers electric fields in materials. Materials are generally classified as conductors and nonconductors. Non-conducting materials are unusually referred as insulators or dielectrics. Materials may be classified in terms of their conductivity σ, in mhos per meter ( / m ), or siemens per meter (s/m). 2

3 Properties of Materials A material with high conductivity (σ>>1) is referred to as metals. A material with small conductivity (σ<<1) is referred to as insulator (or dielectric). A material with conductivity lies between those of metals and insulators are called semiconductors. Table B.1 in Appendix B (Values of conductivity of common materials). Copper and aluminium are metals. Silicon and germanium are semiconductors. Glass and rubber are insulators. 3

4 Properties of Materials Some conductors exhibit infinite conductivity at temperature near absolute zero (T=0 k), and they are called superconductors. Dielectric materials have few electrons available for conduction of current. Metals have abundance of free electrons. 4

5 5.3 Convection and Conduction Currents Electric current is generally caused by motion of electric charges. The current in (Amperes) through a given area is the electric charge passing through the area per unit time. I = dq dt In a current of one ampere, charge is being transferred at a rate of one coulomb per second. 5

6 Convection and Conduction Currents If current ΔI flows through a planar surface ΔS, the current density is: I J = S If current is normal to the surface: I = J S If the current is not normal to the surface: I = J S The total current flowing through a surface S is: I = J S S 6

7 Convection Current Doesn t involve conductors and consequently doesn t satisfy ohms law. It occurs when current flows through an insulating medium such as liquid, vacuum. Ex) beam of electrons in vacuum tube is a convection current. 7

8 Convection Current If there is a flow of charge of density ρ V at velocity u=u y a y the current is: Q ρv V ρv S y I= = = = ρv Su t t t I J y = = ρv uy S In General: J= ρ u V Convection Current Density J= ρ V y u 8

9 Conduction Current Requires a conductor. A conductor has a large number of free electrons that provide conduction current due to an impressed electric field. When electric field is applied, the force on an electron with charge e is : F = ee In free space, the electron would accelerate. In material, the electron suffers continual collisions. The electron will move with different velocities between collisions. 9

10 Conduction Current The average velocity is called the drift velocity u. eτ u = E m τ : average time between collisions. m: mass of electron. Drift velocity is directly proportional to the applied field. If there are n electrons per unit volume, ρ V =-ne 2 The conduction current density is: J= ne τ ρv u = E = σe m J= σ E 10

11 5.4 Conductors A conductor has abundance of charge that is free to move. A perfect conductor (σ= ) can not contain an electrostatic field within it. Isolated Conductor 11

12 Conductors Inside conductor E = 0 and V = 0, which implies V=0. Thus, a conductor is an equipotential body. According to gauss law, E ds= ρv dv If E=0, then the charge density ρ V = 0. ε S V Inside an isolated perfect conductor, E=0, ρ =0, =0 V V ab (V ab is potential difference between two points a and b in the conductor) 12

Conductors When a potential difference is applied at the ends of the conductor, E 0 inside the conductor. ( conductor not isolated, wired to a source).

13 Conductors When a potential difference is applied at the ends of the conductor, E 0 inside the conductor. ( conductor not isolated, wired to a source). An electric field must exist inside the conductor to sustain the flow of current. The direction of the electric field E produced is the same as the direction of the flow of positive charges or current I. This direction is opposite to the direction of the flow of electrons. The opposition to the flow is called Resistance. 13

14 To derive Resistance: V The magnitude of electric field is given by E = l Assuming conductor has uniform cross section of area S, I σv = σ E= S l l V l R = R = = I σ S σ S l ρcl 1 R = =, ρc = Resistivity σs S σ J = I S If the cross section of the conductor is not uniform: R V = = I E dl σ E ds The power P (in watts): 2 P = E J dv or P = I R v 14

15 If, calculate the current passing through (a) a hemispherical shell of radius 20 cm, 0<Ѳ<π/2, 0<φ<2π (b) a spherical shell of radius 10 cm. Solution: I = J ds ds = r θdφdθa 2, sin π/2 2 π π/ θ θ φ θ π θ θ θ 3 r 0.2 r 0.2 r = r = θ= 0φ= 0 θ= 0 ( a) I = ( 2cos ) r sin d d = 2 cos sin d Let u=sin θ, du = cosθdθ I J 1 = 3 2cos a + sin a A/m r 3 ( θ r θ φ ) π /2 2 4π sin θ = = 10π = 31.4 A ( b) 0 < θ < π, r = π sin θ I = = π 0 Example 5.1 r Alternatively for this case: I= J ds = J dv = 0, since J = 0 S v 15

16 A wire of diameter 1 mm and conductivity 5 x 10 7 S/m has free electrons per cubic meter when an electric field of 10 mv/m is applied. Determine (a) the charge density of free electrons. (b) the current density (c) the current in the wire (d) the drift velocity of the electrons (take e=-1.6 x C) Solution ( a) ρ = ne = (10 )( ) = C / m V Example ( b) J = σ E = (5 10 )(10 10 ) = 500 ka / m d 5π -6 5 ( c) I = JS = (5 10 )( π ) = = 0.393A 2 4 J 5 10 d ce J u u m s 5 ( )sin 5 = ρv, = = = ρv / 16

A lead (σ=5 x 10 7 S/m ) bar of square cross section has a hole bored along its length of 4 m so that its cross section becomes that of figure 5.5. find the resistance between the square ends.

17 A lead (σ=5 x 10 7 S/m ) bar of square cross section has a hole bored along its length of 4 m so that its cross section becomes that of figure 5.5. find the resistance between the square ends. Solution R l = σ S = π = 3 π where S d r 2 π 2 = 9 cm 4 R= 4 Hence / 4 10 = 974µ Ω ( π ) 6 4 Example

5.5 Polarization in Dielectrics In dielectric materials, charges are not able to move about freely, they are bound by finite forces. (displacement will take place when external force is applied).

18 5.5 Polarization in Dielectrics In dielectric materials, charges are not able to move about freely, they are bound by finite forces. (displacement will take place when external force is applied). Atoms or molecules are electrically neutral since positive and negative charges have equal amounts. When Electric field is applied, positive charges move in the direction of E, and negative charges move in the opposite direction. The molecules are deformed from their original shape, and each molecule gets some dipole moment. +Q (nucleus) -Q (electron) 18

19 The dipole moment is Where d is the distance vector from Q to +Q of the dipole. If there are N dipoles, the total dipole moment due to the electric field is: Polarization in Dielectrics P=Qd Q1d 1+Q2d 2+ +Q d = Q d N N k k k= 1 As a measure of intensity of polarization, define Polarization P (in coulombs per meter squared) as the dipole moment per unit volume of the dielectric: N P = lim v 0 N k= 1 Qd k v k 19

20 Polarization in Dielectrics Two groups of dielectrics: Nonpolar: nonpolar dielectric molecules do not posses dipoles until the application of electric field. Examples: hydrogen, oxygen, nitrogen, Polar: molecules have built-in permanent dipoles that are randomly oriented. When external E is applied, dipole moments are aligned parallel with E. Polarization of a polar molecule: (a) permanent dipole (E = 0), (b) alignment of permanent dipole (E 0). Examples: water, sulfer dioxide 20

21 Polarization in Dielectrics When polarization occurs, two charge densities are formed: ρ ps (1) An equivalent surface charge density is formed over the surface of the dielectric. ρ ps = Pa n where a n is unit normal to the surface. (2) An equivalent volume charge density ρ pv is formed throughout the dielectric. ρ pv = P Notes: ρ ps and ρ pv are called bound (or polarization) surface and volume charge densities, respectively, as a distinct from free surface and volume charge densities ρs and ρv. Bound charges are those that are not free to move within the dielectric material; they are caused by the displacement that occurs on a molecular scale during polarization. 21

22 If the entire dielectric were electrically neutral prior to application of E and if we have not added free charge, the dielectric will remain electrically neutral. Thus the total charge of the dielectric material remains zero. Q Q bs bv : total bound surface charge : total bound volume charge Q = ρ ds = P. ds bs ps Q = Q = ρ dv =.Pdv bv bs pv Polarization in Dielectrics v total charge= ρ ds + ρ dv = Q Q = 0 S v ps pv bs bs v ρ ps = Pa n ρ pv = P 22

23 If the dielectric region contains free charge with volume charge density of ρ V, the total volume charge density is: ρ = ρ + ρ t v pv 1 1 E= + = ε ε ( ε E+P) 0 ( ρ ) v ρpv ( ρv P) 0 0 = ρ = D 0 v D= ε E + P The application of E to the dielectric material causes the flux density to be greater than it would be in free space. Polarization is proportional to the applied electric field: χ e Polarization in Dielectrics P= χε e 0E Where is the susceptibility of the material. Susceptibility of a material describes its response to an applied field. 23

24 5.6 Dielectric Constant and Strength D= εe + P = εe + χεe 0 0 e 0 D= ε 1+ χ E= εεe ( ) 0 e 0 ε is called the permittivity of the dielectric. ε o is the permittivity of free space. ε 0 =10-9 /36π F/m. ε r is the relative permittivity. r ( ) D= εe, ε= εε, ε = 1+ χ = 0 r r e The dielectric constant (or relative permittivity) the permittivity of the dielectric to that of free space. ε ε ε r 0 is the ratio of Table B.2 in appendix B, values of dielectric constants of some materials. 24

25 No dielectric is ideal. When the electric field in a dielectric is sufficiently large, it begins to pull electrons completely out of the molecules, and dielectric becomes conducting. Dielectric breakdown occurs when a dielectric becomes conducting. The dielectric strength is the maximum electric field that a dielectric can tolerate or withstand without electrical breakdown. 25

26 5.7 Linear, Isotropic, and homogeneous Dielectrics A material is linear if D varies linearly with E. Materials for which ε (or σ) does not vary in the region being considered (the same at all points, i.e. independent of (x,y,z) are said to be homogeneous. Materials are inhomogeneous (or monohomogeneous) when ε is dependent on the space coordinates. Materials for which D and E are in the same direction are said to be isotropic. For anisotropic (or nonisotropic) materials, D, E, and P are not parallel. ε or χ e has nine components. 26

27 Linear, Isotropic, and homogeneous Dielectrics Dx εxx εxy εxz Ex D = ε ε ε E y yx yy yz y Dz εzx εzy ε zz Ez A dielectric material is linear if D= εe and ε does not change with the applied E field. A dielectric material is homogeneous if D= εe and ε does not change from point to point. A dielectric material is isotropic if D= εe and ε does not change with direction. 27

28 A dielectric cube of side L and center at the origin has polarization P=ar where a is a constant and r=xa x +ya y +za z. Find all bound charge densities and show that the total bound charge vanishes. Solution: for each of the six faces, there is a surface charge ρ. for the face located at x=l/2: ρ = P a = ax = ps x x= L/2 x= L/2 Example 5.5 al 2 L/2 L/2 6aL totalbound surface charge is : Qs= ρpsds = 6 ρpsdydz = L = 3aL 2 S L/2 L/2 bound volume charge density is ρ = P = ( a+ a+ a) = 3a total bound volume charge is Q = ρ dv = 3a dv = 3aL 3 3 Hence total charge is Qt = Qs + Qv = 3aL 3aL = 0 v pv pv ps

29 The electric field intensity in polystyrene (εr =2.55) filling the space between the plates of parallel-plate capacitor is 10kV/m. The distance between the plates is 1.5 mm. Calculate: (a) D (b) P (c) the surface charge density of free charge on the plates. (d) the surface density of polarization charge. (e) the potential difference between the plates. Solution (a) D= εε 0 re= (2.55) 10 = nc/m 36π 9 10 (b) P= χε e 0E = ( εr 1) ε0e = (1.55) 10 = 137 nc/m 36π 2 (c) ρ = D a = D = nc/m s n (d) ρ = P a = P = 137 nc/m ps n ( ) 4 3 (e) V V = Ed = = n n Example

30 A dielectric sphere (ε r =5.7) of radius 10 cm has a point charge of 2pC placed at its centre. Calculate: (a) The surface density of polarization charge on the surface of the sphere. (b) The force exerted by the charge on a -4pC point charge placed on the sphere. Solution (a) Assuming point charge is located at the origin, E Q 4πε ε r 0 2 a χ Q P= χεe = a ρ = ps (b) F r e e 0 2 4πε rr ( ε ) 12 r 1 Q (4.7)2 10 = P a r = = = pc/m 2 4 4πε r 4 π (5.7) QQ a 1 2 = 2 r = 4πε 0ε rr r r r Example ( 4)(2) π (5.7) π 4 a r 2 = a pn r 30

31 Find the force with which the plates of a parallel-plate capacitor attract each other. Also, determine the pressure on the surface of the plate due to the field. Solution ρs E = a n where a n is a unit normal to the plate 2ε and ρ is the surface charge density. s ρ ρ S F=QE = ρ. a = a or 2 s s ss n n 2ε 2εε 0 r Example ρs S Q F = = 2ε 2εS 2 ρs The pressure is force per area = 2 εε 0 r 31

32 5.8 Continuity Equation and Relaxation Time From principle of charge conservation, the time rate of decrease of charge within a given volume must be equal to the net outward current flow through the surface of the volume. I OUT dqin = J ds= dt S where Qin is the total charge enclosed by the closed surface. dqin d v vdv dv dt = ρ ρ dt = t V V Using divergence theorem, J ds = J dv ρv J dv = dv t V V S ρv J = t This equation is called continuity of current equation or continuity equation. 32 V J ρv = t

33 Continuity Equation and Relaxation Time Consider introducing a charge at some interior point of a given material (conductor or dielectric). From Ohm's law J= σ E ρv From Gauss's law D= ρv and hence E= ε substituting in the continuity equation, ρv J = t ρv σe= t σρv ρ ρ v v σ = + ρ v = 0 ε t t ε ρv σ + ρ v = 0 t ε 33

34 Continuity Equation and Relaxation Time ρv σ + ρ v = 0 t ε ρv σ Separating the variables: = t ρ ε σt and integrating both sides: ln ρv = + ln ρ ε where ln ρ is a constant of integration. v0 tt / ε r ρv = ρv0 e where Tr = σ T is the time constant in seconds. r v v0 ρ = ρ e v v tt / 0 r 34

35 Continuity Equation and Relaxation Time When we introduce a volume charge density at an interior point in a material, it decays resulting in a charge movement from the interior point at which it was introduced to the surface of the material. The time constant Tr of this decay is called relaxation time or rearrangement time. Relaxation time is the time it takes for a charge placed in the interior of a material to drop to e -1 or 36.8 percent of its initial value. It is very short for good conductors and very long for good dielectrics. For a good conductor the relaxation time is so short that most of the charge will vanish from the interior point and appear at the surface within a short time. For a good dielectric the relaxation time is very long that the introduced charge remains wherever placed for times up to days. 35

36 5.9 Boundary conditions When the field exists in a medium consisting of two different media, the conditions the field must satisfy are called boundary conditions. For the electrostatic field the following boundary conditions are important: Dielectric dielectric interface. Conductor dielectric. Conductor free space. We will use Maxwell Equations: E dl=0, D ds=q enc 36

37 Dielectric-dielectric Boundary conditions Consider the boundary between two dielectrics with permittivities 37

38 Dielectric-dielectric Boundary conditions The fields in the two media can be expressed as: E = E + E E = E + E 1 1t 1n 2 2t 2n Apply the equation E dl = 0 to the path abcda in the figure abcda E L h h h h E dl= E1 t w E1n E2n E2t w+ E2n + E1 n = w E w= 0 1t 2t E = E E1 t = E 1t 2 2t Tangential components of E are equal at the boundary. E t undergoes no change on the boundary and it is continuous across the boundary. t 38

39 Dielectric-dielectric Boundary conditions E = D t D t E = ε D = ε D 1 2 1t 2t 2 1t 1 2t ε1 ε2 The tangential component of D under goes some change across the boundary. So D is said to be discontinuous across the boundary. The boundary conditions for the normal components are obtained by applying Gauss s law on a small pill box shaped volume as in the next figure. S D ds= Q D S D S = Q= ρ S Assuming h 0 D 1n 2n S D = ρ 1n 2n S D D = ρ 1n 2n S 39

40 Dielectric-dielectric Boundary conditions 40

41 D Dielectric-dielectric Boundary conditions 1n 2n S If no free charge exists at the boundary, ρ =0 D D = ρ = D 1n 2n D = D 1n 2n Normal components of D are equal at the boundary. D n undergoes no change on the boundary and it is continuous across the boundary. E1 n ε 2 D1 n = D2n ε1e1 n = ε2e2n = E ε The normal component of E undergoes some change across the boundary. So E n is said to be discontinuous across the boundary. S 2n 1 41

42 Dielectric-dielectric Boundary conditions If the field on one side is known, we can find the field on the other side. θ1 : angle between E1 and normal. θ2 : angle between E2 and normal. 42

43 Dielectric-dielectric Boundary conditions This is called law of refraction 43

44 Conductor-dielectric Boundary conditions Apply the equation E dl=0 to the path abcda in the figure abcda h h h h E dl= Et w En 0 0 w En = E w= 0 E = 0 D = 0 D = εεe = 0 t t t t 0 r t D t = εε 0 ret = 0 44

45 Conductor-dielectric Boundary conditions To find boundary conditions for normal components, apply gauss law: S D ds = Q D S 0 S = Q Assuming h 0 D n Q S n ρ S S S = = = ρ S D = εεe = ρ n 0 r n S 45

46 Conductor-dielectric Boundary conditions Notes No electric field exists inside a conductor. Since E = V two points in the conductor., there is no potential difference between any The electric field must be external to the conductor and must be normal to its surface. 46

47 Conductor-Free space Boundary conditions Special case of Conductor-Dielectric conditions. Replacing ε = 1: D D r = εεe = ρ n 0 r n S = ε E = ρ n 0 n S As in the earlier case D = ε E = 0 47 t 0 t D D = ε E = ρ n 0 n S t = ε 0Et = 0

48 Example 5.9 Two extensive homogeneous isotropic dielectrics meet on plane z=0. for z>0, εr1 =4 and for z<0, εr2 =3. a uniform Electric field E 1 =5a x -2a y +3a z kv/m exists for z 0. Find: (a) E 2 for z 0. (b) The angles E 1 and E 2 make with the interface. (c) the energy densities (in J/m 3 ) in both dielectrics. (d) the energy with a cube of side 2 m centred at (3,4,-5). 48

49 E =E +E Example solution 1 1n 1t E = E.a = E.a = 3 E = 3a 1n 1 n 1 z 1n E = E -E = 5a -2a 1t 1 1n x y z E E 2t 1t x y 2n 1n 0 r2 2n 0 r1 1n 2n = E = 5a -2a D = D εε E = εε E ε 4 4 E E 3a 4a r1 = 1n = 1n = z = z ε r E =E +E = 5a -2a + 4a kv/m 2 2t 2n x (b) α =90- θ, α =90-θ y E1n 3 3 tanα1 = = = α1 = 29.1, θ1 = 60.9 E t z E 4 tanθ ε tanα = = α = 36.6, θ = (Note: = is satisfied) 2n 1 r E2t 29 tanθ2 εr 2 49

50 3 (c) The energy densities in (J/m ) in both dielectrics: ( 9 10 μj/m ) π E1 2 ε1 1 = = w = E 4 (25 4 9) ( 9 10 μj/m) π E 2 2 ε 2 2 = = w = E 3 ( ) (d) The energy with a { x y z } ( ) E E2 E2 E2 x= 2 y= 3z= 6 ( ) cube of side 2 m centred at 3,4, , 3 5, -6 4, z=-5 (Region 2) W = w dv = w dzdydx = w (2)(2)(2) = mj Example solution 50

51 Region y<0 consists of a perfect conductor while region y>0 is a dielectric medium (εr1 =2). If there is a surface charge 2 nc/m 2 on the conductor, determine E and D at: (a) A(3,-2,2) (b) B(-4,1,5) Solution (a) Point A(3,-2,2) is in the conductor since y=-2<0 at A. Hence: E=0=D (b) B(-4,1,5) is in the dielectric medium since y=1>0 at B. D = ρ = 2 nc/m n s -9 D 2 10 E= = π a a = εε (2) 0 r 36π 2 y Example 5.10 y =113.1 a y V/m 51

Lecture 16 Dielectrics: Dipole, Polarization

Lecture 16 Dielectrics: Dipole, olarization Sections: 4.7, 5.7 Homework: See homework file Electric Dipole and its Dipole Moment electric dipole: two point charges of equal charge but opposite polarity