Properties and Boundary Conditions Prepared By Dr. Eng. Sherif Hekal Assistant Professor – Electronics and Communications Engineering

11/30/2016 1 Agenda

• Intended Learning Outcomes

• and

• Dielectric Constant and Susceptibility

• Boundary Conditions for Dielectrics

•

• Multiple-Dielectric

11/30/2016 2 Intended Learning Outcomes

• In this chapter, we consider insulating materials, or dielectrics. • Such materials differ from conductors in that ideally, there is no free charge that can be transported within them to produce conduction current. • An applied electric has the effect of displacing the charges slightly, leading to the formation of ensembles of electric ; this process is called Polarization . The extent to which this occurs is measured by the relative , or dielectric constant. • Boundary conditions for the fields at interfaces between dielectrics are developed to evaluate these differences. • In the end, the methods for calculating capacitance for a number of cases are presented, including geometries, and to be able to make judgments on how capacitance will be altered by changes in materials or their configuration.

11/30/2016 3 Dielectrics and Polarization

 Dielectrics are materials which have no free charges; all are bound and associated with the nearest (Fig. 7.1(a)). They behave as electrically neutral when they are not in an .  An external applied electric field causes microscopic separations of the centers of positive and negative charges as shown in Fig 7.1 (b).  These separations behave like electric dipoles, and this phenomenon is known as dielectric polarization.

11/30/2016 4 Dielectrics and Polarization

Electric moment

11/30/2016 5 Dielectrics and Polarization Dielectrics may be subdivided into two groups : • Non-Polar: dielectrics that do not possess permanent . Electric dipole moments can be induced by placing the materials in an externally applied electric field. • Polar: dielectrics that possess permanent dipole moments which are ordinarily randomly oriented, but which become more or less oriented by the application of an external electric field. An example of this type of dielectric is water.

(a) (b) (a) (b)

11/30/2016Orientations of non-polar when(a) E0 = Orientations of polar molecules when(a) E0 = 0,6 0, and (b) E0 ≠ 0. and (b) E0 ≠ 0. Dielectrics and Polarization

• However, the alignment is not complete due to random thermal motion. • The aligned molecules (induced dipoles) then generate an electric field that is opposite to the applied field but smaller in magnitude.

퐸푃

11/30/2016 7 Dielectrics and Polarization

Suppose we have a piece of material in the form of a cylinder with area A and height h, and that it consists of N electric dipoles, each with electric dipole moment 푝 spread uniformly throughout the of the cylinder. what is the average electric field just due to the presence of the aligned dipoles? To answer this question, let us define the polarization vector 푃 to be the net electric dipole moment vector per unit volume:

(1)

11/30/2016 8 Dielectrics and Polarization

In the case of our cylinder, where all the dipoles are perfectly aligned, the magnitude of 푃 is equal to C/m2

From the equivalence between (Fig. (a) and (b)), we have two net charges ±QP which produce net dipole moment of QPh; hence we note that the equivalent charge distribution resembles that of a parallel-plate , with an (a) (b) equivalent surface σ that is equal to the P (a) A cylinder with uniform dipole distribution. magnitude of the polarization: (b) Equivalent charge distribution.

11/30/2016 9 Dielectrics and Polarization

Thus, our equivalent charge system will produce an average electric field of magnitude 퐸푃 = 푃/휖0. Since the direction of this electric field is opposite to the direction of 푃, in vector notation, we have (2) The total electric field 퐸 is the sum of these two fields:

(3) In most cases, the polarization 푃 is not only in the same direction as 퐸 but also linearly proportional to 퐸0(and hence 퐸.) This is reasonable because without the external field there would be no alignment of dipoles and no polarization. We write the linear relation between and as

11/30/2016 (4) 10 Dielectrics and Polarization

• where χe is called the . Materials that obey this relation are linear dielectrics. Combing Eqs. (2) and (3) gives

(5)

where 휀푟 = 휅푒 = 1 + 휒푒

is the dielectric constant. The dielectric constant 휅푒is always greater than one since 휒푒 > 0.

Thus, we see that the effect of dielectric materials is always to decrease the electric field below what it would otherwise be.

11/30/2016 11 Dielectric Constant and Susceptibility

Electric Density (D): density is defined as charge per unit area and it has same units of dielectric polarization. Electric flux density D at a point in a free space or air in terms of Electric field strength is D0   0E At the same point in a medium is given by D  E As the polarization measures the additional flux density arising from the presence of material as compared to free space

D   0E  P   0 (1 e )E

11/30/2016 12 D   0 rE (6) Definitions

• Polarization: the process of creating or inducing dipoles in a dielectric medium by an external field. • : the ability of dielectric to form instantaneous dipoles. It is a property of matter • Polarization vector: it is defined as the dipole moment per unit volume.

C/m2

• Electric susceptibility (χe) is a dimensionless proportionality constant that indicates the degree of polarization of a dielectric material in response to an applied electric field. The greater the electric susceptibility, the greater the ability of a material to polarize in response to the field, and thereby reduce the total electric field inside the material (and store ). 11/30/2016 13 Definitions

• permittivity (ε): is the measure of resistance that is encountered when forming an electric field in a medium. In other words, permittivity is a measure of how an electric field affects, and is affected by, a dielectric medium.

permittivity ε is measured in per meter, and εr is the of the material. • Dielectric Constant (relative permittivity) gives a measure of the polarizability of a material relative to free space and is defined as the ratio between the permittivity of the medium to the permittivity of free space.   r   0 11/30/2016 14 Boundary conditions for Dielectric materials

• Let us first consider the interface between two dielectrics having 휀1 and 휀2 and occupying regions 1 and 2, as shown in the figure below. • We first examine the tangential components by using

• around the small closed path, obtaining

The small contribution by the normal component of E along the sections of ∆ℎ becomes negligible as ∆ℎ decreases and the closed path crowds the surface. Immediately, then, (7) 11/30/2016 15 Boundary conditions for Dielectric materials

• The boundary conditions on the normal components are found by applying Gauss’s law to the small “pillbox” shown in the figure below. • The sides are again very short, and the flux leaving the top and bottom surfaces is the difference

(8) This charge may be placed there deliberately, thus unbalancing the total charge in and on this dielectric body. Except for this special case, ρS is zero on the interface; Hence (9)

11/30/2016 or 16 Boundary conditions for Dielectric materials

• Let D1 (and E1) make an angle θ1 with a normal to the surface as shown in the figure. Because the normal components of D are continuous, (10)

But

Thus and the division of this equation by (10) gives

(11)

11/30/2016 17 Boundary conditions for Dielectric materials

Example: Find the fields within the Teflon slab (εr = 2.1), given the uniform external field Eout = E0 ax in free space. as shown in Figure, with free space on both sides of the slab and an external field

We also have and

11/30/2016 18 Boundary conditions for Dielectric materials

Inside, the continuity of DN at the boundary allows us to find that Din = Dout = ε0E0 ax. This gives us

To get the polarization field in the dielectric, we use D = ε0E + P and obtain

11/30/2016 19 Boundary conditions for Dielectric materials

X

Y Z

11/30/2016 Region1 Region2 20 Boundary conditions for Dielectric materials X The boundary at plane Z = 0, hence

Y Z Dt1 DN1 Region1 Region2 2 2 Dt1  Dx  Dy

2 2 2 D1  Dx  Dy  Dz

1 DN1 1  cos ( ) D1

P1  D1  0 E1 1  E  D /  11/30/2016 1 1 0 r P1  D1(1 ) 21  r Boundary conditions for Dielectric materials

11/30/2016 22 Capacitance

• The characteristic that all dielectric materials have in common, whether they are solid, liquid, or gas, and whether or not they are crystalline in nature, is their ability to store electric energy. • This storage takes place by means of a shift in the relative positions of the internal, bound positive and negative charges against the normal molecular and atomic . • A capacitor is a device that stores energy; energy thus stored can either be associated with accumulated charge or it can be related to the stored electric field.

11/30/2016 23 Capacitance • A capacitor is a device for storing electrical charge. • Capacitors consist of a pair of conducting plates separated by an insulating material (oil, paper, air). • The measure of the extent to which a capacitor can store charge is called Capacitance. • Capacitance is measured in farads F, or more usually microfarads µF or Pico farads pF.

11/30/2016 24 Capacitance

We define the capacitance of a two-conductor system as “the ratio of the magnitude of the total charge on either conductor to the magnitude of the potential difference between conductors”.

Farad

1 F = 1 = 1coulomb/

11/30/2016 25 Capacitance

• In general terms, we determine Q by a surface over the positive conductors, and we find V0 by carrying a unit positive charge from the negative to the positive surface,

• The capacitance is independent of the potential and total charge, for their ratio is constant. • The capacitance is a function only of the physical dimensions of the system of conductors and of the permittivity of the homogeneous dielectric.   A 2  L C  0 r C  0 r d ln(b / a)

11/30/2016 For parallel plate capacitor For 26 Capacitance

Case 1: Dielectrics without Battery

εr

• As shown in Figure, a battery with a potential difference |∆푉0| is first connected to a capacitor C0, which holds a charge 푄0 = 퐶0|∆푉0| . We then disconnect the battery, leaving 푄0 = const. • insert a dielectric between the plates, while keeping the charge constant, experimentally it is found that | V | | V | 0  r • This implies that the capacitance is changed to

Q0 Q0 C     rC0 | V | | V0 | / r 11/30/2016 27 Capacitance

Case 2: Dielectrics with Battery

εr

• Consider a second case where a battery supplying a potential difference |∆푉0| remains connected as the dielectric is inserted. Experimentally, it is found

Q   rQ0

• where Q0 is the charge on the plates in the absence of any dielectric. • This implies that the capacitance is changed to

Q  rQ0 C     rC0 | V0 | | V0 | 11/30/2016 28 Capacitance

• Parallel-plate Capacitor Consider two metallic plates of equal area A separated by a distance d, as shown in the figure below. a uniform sheet of ±ρS on each conductor leads to the uniform field

and

11/30/2016 29 Capacitance

• Parallel-plate Capacitor The potential difference between lower and upper planes is lower 0 S S V0   E dL   dz  d upper d ε ε

Q  S A  V  S d 0 ε Q εA C   V0 d

11/30/2016 30 Capacitance

• Parallel-plate Capacitor

11/30/2016 31 Capacitance

• Cylindrical Capacitor To calculate the capacitance, we first compute the electric field. Due to the cylindrical of the system, we choose our to be a coaxial cylinder with length ℓ < L and radius r where a < r < b. Using Gauss’s law, we have

E dA  Q /  encolsed S 휌 ℓ 퐸퐴 = 퐸 2휋푟ℓ = 푙 휀 휌 ∴ 퐸 = 푙 2휋휀푟 푄 ℓ Where ρ is the line charge density, and ρ = l l 퐿 11/30/2016 32 Capacitance

• Cylindrical Capacitor The potential difference is given by a V  Va Vb   Er dr b 휌 푎 푑푟 휌 푏 = − 푙 = 푙 ln 2휋휀 푏 푟 2휋휀 푎 Q  L C   l V l ln(b a) / 2 2L C  ℓ ln(b a) 11/30/2016 33 Capacitance

• Spherical Capacitor let’s consider a spherical capacitor which consists of two concentric spherical shells of radii a and b, as shown in Figure

(a) spherical capacitor with two concentric (b) Gaussian surface for calculating the spherical shells of radii a and b. electric field. 11/30/2016 34 Capacitance

• Spherical Capacitor The electric field is non-vanishing only in the region a < r < b. Using Gauss’s law, we obtain

Therefore, the potential difference between the two conducting shells is: a Q a dr Q  1 1  Q  b  a  V  Va Vb   Er dr          b b 2 4 0 r 4 0  a b  4 0  ab  which gives

11/30/2016 35 Capacitance

Find the capacitance shown in the figure, where the region between the plates is filled with a dielectric having a dielectric - - -- - constant εr - aφ E Solution: Assuming the applied to the electrodes is V 0 α  + + + + + + + + + + + + h V0   E dl  + + + + + + aZ Using the cylindrical coordinates system (r, φ, z), it is clear that the + + + + + + field is in the aφ direction ra 0 rb V0   Ea dl Where dl  rd  0   E rd E r 

11/30/2016 E  V0 / r , D  E  V0 / r 36 Capacitance

From the boundary conditions for conductors ------aφ S  Dn  V0 / r E

∴ The total charge on the plate α + + + + + + + + + + + + h Q  S ds , ds  drdz + + + + + +  + + + + + + aZ rb h r Q  (V0 / r)drdz a r 0 a rb

rb V h dr V h  r   ( 0 )  0 ln b  r   a  r   ra  Q h  r    h  r  C   ln b  C  0 r ln b  11/30/2016     37 V0   ra    ra  Multiple-Dielectric Capacitors

Let’s consider the two parallel-plate capacitor having two dielectric materials ε1 and ε2 , as shown in the figure below, and their thickness are d1 and d2, respectively.

parallel-plate capacitor containing two dielectrics with the dielectric interface parallel to the conducting plates.

11/30/2016 38 Multiple-Dielectric Capacitors

Suppose we assume a potential difference V0 between the plates. The electric field intensities in the two regions, E2 and E1, are both uniform,

V0  E1d1  E2d2 At the dielectric interface, E is normal, and from the boundary conditions

DN1  DN2  1E1   2 E2

V0  E1d1  E2d2

 E1d1  1  2 E1 d2

V0 E1  11/30/2016 d1  1  2 d2 39 Multiple-Dielectric Capacitors

From the boundary conditions for conductors, the surface charge density on the lower plate has the magnitude

V0 S1  D1  1E1  (d1 ε1)  (d2 ε2 )

Because D1 = D2, the magnitude of the surface charge is the same on each plate. The capacitance is then Q  S 1 C   S  V0 V0 (d1 Sε1)  (d2 Sε2 ) 1 C  (1 C1)  (1 C2 )

11/30/2016 40 Multiple-Dielectric Capacitors

If the dielectric boundary were placed normal to the S1 S2 two conducting plates and the dielectrics occupied areas of S and S , as shown in the figure 1 2 d ε ε then an assumed potential difference V0 would 1 2 produce field strengths Q  S   S ε ES  ε ES C   S1 1 S 2 2  1 1 2 2 V0 Ed Ed ε S  ε S C  1 1 2 2 d

C  C1  C2 11/30/2016 41 Multiple-Dielectric Capacitors

11/30/2016 42 11/30/2016 43