chapter 03.ces

8/7/2019 chapter 03.ces

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CHAPTER 3 ELECTRIC FIELD IN DIELECTRICS

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3.0 INTRODUCTION

In previous chapter, you have learned about electric field due tocharges. In this chapter, you will study the electric field in dielectricsor the effect on electric field due to dielectric. It is very importanttopic as dielectric plays very vital role in capacitors.

________________________________________________________3.1 OBJECTIVES

At the end of this chapter, you will be able to

yknow different kinds of materials and moleculesyderive an expression for the Gausss law in dielectrics, andyobtain an expression for the electric field in dielectrics.

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3.2 MATTER AND MOLECULES

Matter is composed of small particles called molecules containingatoms. An atom has a positive charged nucleus and negativelycharged electrons moving around the nucleus in circular orbits havingradii of the order 10-10 m. This means that these electrons are confinedin a very small region and hence these electrons are known as boundelectrons. The nucleus has positively charged particles, protons andneutrons that have no charge. The number of protons and electronsare always equal in an atom, meaning that an atom is electricallyneutral. The nucleus has the radius of the order much less than 10-10m.

In case of solids, the electrons are closely packed. The electrons closeto the nucleus are more strongly bound than the electrons in outer


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orbits. The electrons in outer orbits are known as free electrons asthey may be detach by giving some external energy to the atoms.

________________________________________________________3.4 Conductors, insulators and semiconductors

3.4.1 Conductors are materials in which electric charges (electrons)can move quite freely as shown in Fig 1.5.1. Examples aremetals, copper, and iron. In solids the number of atoms percubic meter is of the order of 1028.

Metal rod

Charged Neutral

Fig. 1.5.1: Electric charge flows in conductors easily

3.4.2 Insulators (or bad conductors) are materials through whichcharges (electrons) can flow with great difficulty as shown inFig 1.5.2. Examples are glass and nylon. Such substances arealso known as dielectrics. Gases are mostly insulators so asthe liquids with the exception of mercury.

Wooden rod

Charged Neutral

Fig. 1.5.2: Electric charge does not flow in insulators

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3.4.3 Semi-conductors are the materials whose electrical propertiesare somewhere between those of conductors and insulators(fig. 1.5.3). Their conductivity is about 10-10 times theconductivity of good conductors. Examples are silicon andgermanium.

Semiconductor material

Charged NeutralFig. 1.5.3: Electric charge flows in semiconductors partially

3.4.4 Dielectrics are the materials which does not conductelectricity. That is, electrons are hardly free to move.Examples are glass, mica, ebonite, paraffin, air, etc.

3.5 CONDUCTORS IN ELECTROSTATIC EQUILIBRIUM

A conductor in electrostatic equilibrium has the following properties:

1. The electric field is zero everywhere inside the conductor.2. Any excess charge on an isolated conductor must reside entirely on

its surface.

3. The electric field just outside a charged conductor is perpendicularto the conductors surface and has a magnitude W/Io where W is the

charge per unit area at that point.

4. On an irregularly shaped conductor, charge tends to accumulate atplaces where the radius of curvature of the surface is the smallest,

that is, at sharp points.

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3.6 POLAR AND NON-POLAR MOLECULES

The radius of the electrons orbit is of the order of 10-10 m. This is sosmall that in the ground state of an atom, the centre of gravity of

negative charges lies exactly at the centre of the nucleus of the atom.The nucleus of the atom consists of positive charge. The dipolemoment is equal to the product of one of the charges and theseparation between them. Hence the dipole moment of an atom is zerobecause the separation between the centers of positive and negativecharge is zero. The molecules of a dielectric can be categorized intotwo types: (1) non-polar and (2) polar molecules.

3.6.1 Non-polar moleculesThese are the molecules in which the centre of gravity of positivecharges coincides exactly with the centre of gravity of negative

charges. The net dipole moment of such molecules is zero and do nothave any permanent dipole moment because of the dipole length iszero. H2, O2, N2, and CO2 are a few examples of non-polar molecules.

3.6.2 Polar molecules

In polar molecules the centre of gravity of positive charges does notcoincide with the centre of gravity of negative charges. Thesemolecules have permanent dipole moment. H2O, HCl, and NH3 aregood examples of polar molecules. The polar and non-polarmolecules are shown in figure 1.

(a) (b)Figure 1: (1) Non-polar molecules (b) polar molecules

s+ -


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The dipole moment of polar molecules is of the order 10-29 Coulombmeter which gives the separation between central positive change andnegative charge of the order of about 10-10 m. In the absence ofelectric field, the polar molecules are randomly oriented and thus theirdipole moments point in all possible directions, cancelling the effect

of each other as shown in figure 2.

Figure 2: Polar molecules in the absence of electric field

Macroscopically, the resultant dipole moment per unit volume in amaterial is zero. It means that individual molecules do have the dipolemoment and the net dipole moment of all molecules is zero.

3.6.3 Polar molecules in electric field

In the presence of an electric field, the dipole moments of polarmolecules are oriented (or tend to orient) in the direction of electricfield and thus their dipole moments point in a particular direction asshown in figure 3. IfE is the electric field and p is the moleculardipole moment then the torque, X, experienced by each molecule is

X = p

E (1)

+

E

Figure 3: Polar molecules in the presence of electric field


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The torque tends to align the molecules in the direction of the electricfield. However, this tendency is opposed by the molecules and hencethe degree of alignment will depend upon the strength of the electricfield and the temperature of the material. When the molecules of a

material (dielectric) are aligned in the direction of electric field due toits presence, the dielectric is said to be polarized as shown in figure 3.If the strength of the electric field increases, it increases the separationbetween the centers of positive and negative charges, increasing thelength of the dipole and hence increasing the dipole moment per unitvolume. Thus the alignment tendency of molecules is increased as theelectric field increases. At lower temperature, the thermal vibrationsof the molecules will be less and hence dipole molecules will alignthemselves easily at low temperature.

3.6.4 Non-polar molecules in electric field

When a non-polar molecule is placed in an electric field the centers ofpositive charges move in the direction of electric field and the centersof negative charges move in the opposite direction of the electricfield. The separation between the centers of positive and negativecharges keep on increasing until the force on them due to the electricfield is balanced by the internal force due to their relative separation.Thus the molecule develops a kind of induced dipole moment and themolecule is said to be polarized. This induced dipole momentbecomes zero when the electric field is removed.

________________________________________________________3.7 Atomic dipole moment and polarization

The atomic dipole moment is the product of the positive or negativecharge of the molecule (or atom) and the atomic separation betweenthe centers of the positive and negative charge in the direction of theelectric field. Thus if dl is the separation between the centers of apositive and a negative charge and q represent either positive ornegative charge on the molecule then the atomic dipole moment p isgiven byp = q dl (2)


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We have already mentioned that the induced dipole moment of amolecule due to the electric field E is proportional to the strength ofthe electric field. Therefore,pwE orp = E (3)

Where, is a constant and known as the atomic polarizability. It canbe defined as the induced dipole moment of the atom divided by thestrength of the electric field. The direction of the dipole moment is inthe direction of the electric field.

Example 1Find the unit and dimension of the polarizability.

Solution

Unit of: = p/E = (coulomb meter)/(Newton/coulomb) = C2 N-1 m

Or = p/E = (coulomb meter)/(volt/meter) = C V

-1

m

2

= F m

2

Because coulomb/volt = Farad (F)

Dimension of:Dimensions of are C V-1 m2However, if we represent as = p/IoE, then its dimensions are m

3.

Example 2Show that the induced atomic dipole moment is given byp = Eo = 4 Io R

3Eo where R is the radius of the atom and Eo is theexternal applied electric field.

Solution

Let us consider that an atom has atomic number Z and e is the chargeon a proton or on an electron. The charge on the nucleus will be q = +Ze and the net negative charge on all the electrons will be q = Ze.Now the electric field Eo is applied and due to the polarization, r isthe separation between the centers of positively and negativelycharges then the magnitude of the induced dipole moment of the atomisP = q r = Ze r (i)


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If we take the volume charge density of electrons as then = Ze/(4 R3/3) and therefore, the net negative charge within thesphere of radius r isqr= (4 r

3/3) = [ Ze/(4 R3/3)] (4 r3/3) = Ze r3/R3

The electric field due to the negative charge qr within the sphere ofradius r is given by

E = k qr/r2 = k Ze r/R3 (ii)

Where, denotes the unit vector ofr.

Therefore, the net force on the nucleus due to negative charge isF = k Ze qr/r

2 = Ze E = k Z2 e2 r/R3 (iii)

The force on the nucleus due to the applied electric field Eo is

F = Ze Eo = Ze

Eo (iv)Note that E and Eo are different, E is the electric field produced bythe net negative charge and Eo is the external applied electric field.Since the system is in equilibrium, the net force on the atom will bezero and henceZe Eo k Z

2 e2 r/R3 = 0 givingZe r = Eo R

3/k = 4 I R3 Eo (v)

Comparing equations (i) and (v), we getp = 4 I R3 Eo orp = 4 I R

3Eo (vi)

If N is the number of molecule per unit volume then the polarizationdensity vectorPd is given by

Pd = N p = 4 N I R3Eo (vii)

Pd is a vector quantity (as it is the vector sum of dipole vectors andgives the resultant vector dipole moment per unit volume) and P is theinduced atomic dipole moment.


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3.8 Electric field inside a polarized dielectric

Let us find an expression for the electric field inside a polarizeddielectric due to the induced polarization charge.

L + + + + + Wp = + PdAWp + Wp Ep

Wp = Pd(a) (b)

Figure 4: (b) Dielectric material (b) Surface charge density andelectric field

Consider a rectangular block of length L and uniform cross sectionarea of A of a dielectric material polarized by the electric field E asshown in figure 4a. The surface charge density Wis given by the

charge per unit area and hence the polarized surface charge density atthe end faces of the block are - Wp and +Wp as shown in figure 4b.Therefore, the induced charge on respective face will be Wp Aand +Wp A. The magnitude of the net polarized induced dipole moment isp = Wp A L

The vector sum of all the induced dipoles in a unit volume(polarization density) is also given by

Pd = p/V = N p, where, N is the number of atoms per unit volume.This equation gives

p = Pd V = Pd A LComparing both equations, we get Pd = Wp (4)Thus the polarization vector per unit volume (or polarization density)equals the induced (bound) surface charge density. This can bewritten in vector form asWp = Pd . n or simply P . n (5)

Where, n is unit vector normal to the surface.

E


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The electric field inside the polarized dielectric material due toinduced polarized charge is (see figure 4b)

Ep = Wp/Io orEp=Pd/Io (6)

________________________________________________________3.9 Gausss law in dielectrics

Let us consider a parallel-plate capacitor with free surface chargedensity Wfree as shown in Fig. 5.5.1a. When there is no dielectricpresent between the plates, the electric field between the plates is Eo.When a dielectric is introduced in between the plates, a charge isinduced on the dielectric as shown in Fig. 5.5.1b. The direction of theelectric field inside the dielectric is opposite to the direction of Eo.We may call it as polarized induced electric field Ep.

+ Wfree WfreeVacuum

Eo

(a)

+ W - Wi + Wp Wfree

DielectricGaussian surface

(b)Fig. 5.5.1: (a) Parallel-plate capacitor filled in air, (b) parallel-plate capacitor filled with a dielectric of dielectric constant k

If Ed is the resultant polarized electric field in the presence of adielectric, then

+ Wfree Wp

Eo

+

+

+

+

+

+

- +Ep

- +


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Ed = Eo Ep

Since Eo = Wfree/Io and Ep = Wp/Io, then

Ed = Wfree/Io Wp/Io = (WfreeWp)/Io (5.5.1)

Therefore,

(Wfree Wp) = EdIo (5.5.2)

or

Wp = Wfree EdIo (5.5.)

Since Ed = Eo/k, then

Wp = Wfree Io (Eo/k) (5.5.4)

Also Eo = Wfree/Io, therefore,

Wp = Wfree Wfree/k = Wfree (1 1/k) (5.5.5)

or

Wfree Wp = Wfree/k (5.5.6)

Equation 5.5.6 can also be obtained directly from equation 5.5.1. Thedielectric constant, k, can be defined as the ratio of free chargedensity to the difference between the free and polarized chargedensity. In case of metals, Wfree = Wp and therefore k is infinity and forvacuum, Wp = 0 so that k is unity. Since k is always greater than one,(1 1/k) will always be a positive quantity. This means that theinduced polarized charge Wp will always be less than Wfree and hence kis always greater than unity for most of the dielectrics.

Now applying Gausss law to the gaussian surface as shown in figure5.5.1b, we get


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Eo . dA = Qin/Io (5.5.7)

or

Eo A = Qin/Io = Wfree A/Io

This gives

Eo = Wfree/Io in the absence of dielectrics (5.5.8)

In the presence of the dielectric, equation 5.5.7 becomes

Eo A = (Wfree Wp) A/Io (5.5.9)

Using equation 5.5.6 for (Wfree Wp), we have

Eo A = Wfree A/kIo

k Eo A = Wfree A/ Io

k Eo A = Qin/Io (5.5.10)

In integral from, equation 5.5.10 is

kEo . dA = Qin/Io (5.5.11)

Equation 5.5.11 is the mathematical integral form of Gausss lawin dielectrics. We will derive its differential form later in thischapter.

Relation between three electric vectors E, Pd = P, and D

Let us consider a plane parallel plate capacitor whose plates have Wfreecharge density. When there is no dielectric between the plates of thecapacitor, then the magnitude of the electric field in between theplates of the capacitor (free space of vacuum), Eo is given by


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Eo = Wfree/Io

Now the capacitor is filled with a dielectric having dielectric constantk. IfP is the polarization per unit volume and parallel to Eo. Thepolarized electric field Ep is opposite to Eo, then,

Ep = P/Io

The net electric field in the dielectric is given by

E = Eo + Ep = Eo P/Io and therefore,

Eo = E + P/Io orIo Eo = Io E + P orD = Io E + P (1)Where, D = Io Eo in a dielectric is known as electric displacementvector and has the unit of coulomb/meter2. In free space (or vacuum),there is no polarization, hence

D = Io E (in space) (2)

________________________________________________________3.10 Electric susceptibility

Electric susceptibility is defined as the ratio of polarization per unitvolume to the electric field IoE and is denoted by Ge. Therefore,Ge = P/IoE (3)

Now from equation 1, D = Io E + P, we get

D/Io E = 1 + P/Io E = 1 + Ge (4)But D = Io Eo so thatIo Eo/Io E = Eo/E = 1 + Ge (5)

Since Eo/E = k, we can write equation 5 as

k = 1 + Ge = Ir (6)

Where, Ir= Eo/E = D/Io E = I/Io is the relative permotivity of thedielectric. Thus


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D = IoIrE = IE (7)

Example 3The electric field inside a capacitor filled with a dielectric and withouta dielectric is 1.0 105 N/C and 3 105 N/C respectively. Find the

induced (polarized) charge density on the surface of the dielectric.Given Io = 8.85 10-12 C2N-1m-2

Solution

Io = 8.85 10-12 C2N-1m-2

P = Wp but Io Eo = Io E + P givingP = Io Eo Io E = Io (Eo E) = 8.85 10

-12 (3 105 1 105)= 1.77 10-6 C/m2

Example 4

The relative permitivity (dielectric constant) of a gas is 1.000075 andis kept inside an electric field of 3 105 N/C. Find the dipole moment

of each atom of the gas. Given Io = 8.85 10-12

C2

N-1

m-2

Solution

Avogadro number = the number of gas atoms per gram = 6.06 1023Volume of the gas per gram = 22.4 litre = 22.4 10-3 m3Number of gas atoms per unit volumeN = 6.06 1023/22.4 10-3 = 2.7 1025

Ir = 1.000075 = 1 + Ge, giving Ge = 1 1.000075 = 7.4 10-5

But P/Io E = Ge so thatP = Io E Ge

= 8.85 10-12 3 105 7.4 10-5 = 1.96 10-11 C m

Dipole moment per atom isp = P/N = 1.96 10-11/2.7 1025= 7.26 10-37 C m

SELF MARK TEST 1

1. Two parallel plates each of area A are separated by twoinsulating slabs of thickness t1 and t2 having dielectricconstants k1 and k2 respectively. If the plate charge is Q, find

(i) the electric field within each insulator


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(ii) potential difference across the plates, and(iii) the capacitance of the capacitor

2. Show that the polarization of a dielectric is given by . P = p

3. Derive the differential form of Gausss law for dielectric.4. A thin dielectric rod of unifrom cross section area A extend

along the x-axis from x = 0 to x = L. The polarization of therod along x-axis is given by Px = (3 x

2 + 2) i. Find(i) The bound (polarized) volume charge density, and(ii) The bound surface charge density at x = 0 and x = L

SOLUTIONS OF SELF MARK TEST 1Two parallel plates each of area A are separated by two insulating

slabs of thickness t1 and t2 having dielectric constants k1 and k2respectively. If the plate charge is Q, find(i) the electric field within each insulator

Consider the plates and slabs as shown in the figurePlate P1 A + Q

t1

t2

Plate P2 A - QW+ = Q/A and W- = Q/AThe electric field E1 in the slab of dielectric constant k1 isE1 = W/IoIr1 from plates P1 to P2

The electric field E2 in the slab of dielectric constant k2 isE2 = W/IoIr2 from plates P1 to P2

(ii) potential difference across the platesThe work done by a unit postive charge from plate P2 to P1 is

E1 k1 = Ir1

E2 k2 = Ir2


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W = E1 t1 + E2 t2 = (W/IoIr1)t1 + (W/IoIr2)t2. This is the potential atplate P1 and the potentia at plate P2 is zero. Therefore the potentialdifference across the plates isV = VP1 VP2 = VP1 = (W/IoIr1)t1 + (W/IoIr2)t2

(iii) the capacitance of the capacitorThe capacitance of the capacitor isC = Q/V = WA/[(W/IoIr1)t1 + (W/IoIr2)t2]

Q. Show that the polarization of a dielectric is given by . P = pSolution

Imagine an element of area dA inside a non-polar dielectric as shownin figure below. When the dielectric is polarized, the center ofpositive charge +q of a molecule lies at a distance x from the center ofthe negatively charge q. This x is the same for all the molecules overan infinitesimal region.

E

- + dA

- +

x

- +

Figure 5: Bound and surface charge density

When electric field is applied, n+ positive charges cross the element ofarea dA by moving in the direction ofx, and n- negative charges crossit by moving in the opposite direction. The net charge that crosses dAin the direction ofx is therefore

dq = n+ q n- (q) = (n+ + n-) = N q (1)

Where (n+ + n-) = N is simply the number of molecules within theimaginary parallelepiped of the figure whose volume is x dA. Then

dq = N q xdA = N pdA = P dA (2)


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Where, q x is the dipole moment p of a single molecule and N p = P.If now dA lies on the surface of a dielectric, then dq accumulatesthere and the bound surface charge density

npdAdqb

T

!!W (3)

Where, n

is the unit vector normal to the surface and is pointingoutward.

Thus Wb (also denoted as Wp polarized surface charge density) is equalin magnitude to the normal component ofP and is pointing outward.

The Bound Volume Charge Density Vb or Polarized Volume

Charge Density Vp

The net bound charge (polarized charge) that flows out of a volume vacross an element dA of its surface is P dA is calculated already.

The net bound charge that flows out of the closed surface of area A is

qout = AP. dA (5)

The net charge that remains within v must be qout. IfVb is the volumedensity of the charge remaining within v, then

vVb dv = qout = AP. dA = v . P dv (6)

The integrals are equal at every point in the dielectric and thereforethe bound volume charge density is

Vb = P = Vp (7)

Q. Derive the differential form of Gausss law for dielectric.Solution


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Consider that a given volume v contains various dielectrics and thenet (total) volume charge density in terms of total free and boundcharge density within v is given by

Vt = Vf+ Vb Or qt = qf+ qb

There are no surface charges on the surface of v. Gausss Law relatesthe flux of the electric field intensity E through the closed surface ofarea A to the total net charge qt enclosed within that surface. So

AE. dA = v . E dv = qt/Io = (qf+ qb)/Io (8)

If the volume v lies entirely inside a dielectric then there are nosurface charges and qt must include bound as well as free charges. So

qt = v (Vf+ Vb)dv

Therefore,

AE. dA = (1/Io) v (Vf+ Vb)dv(9)

AE. dA = (1/Io) vVt dv (10)

Where, Vt is the total volume charge density. Applying the divergencetheorem to the surface integral ofE gives the volume integral ofE.That is

AE. dA = vE dv = (1/Io) vVt dv

Equating the integrals, we get

E = Vt/Io (11)This is Gausss Law in differential form. This is one of Maxwellsfourth fundamental equations of electromagnetism.

Q. A thin dielectric rod of unifrom cross section area A extendalong the x-axis from x = 0 to x = L. The polarization of the rodalong x-axis is given by Px = (3 x

2 + 2) i. Find(i) The bound (polarized) volume charge density

P = (3 x2 + 2) i


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P = Vp = (x/xx i+x/xyj+x/xz k) . (3 x2 + 2) i

x/xx (3 x2 + 2) = 6 x and therefore, Vp =6x

(ii) The bound surface charge density at sufrace x = 0 willbe P (x = 0) = 2 i

bound surface charge density at x = 0 will beWp = P. n = 2 i . i = 2Where n is outwarddrawn unit vector.The total bound charge at sufrace x = L will beP (x = L) = (3 L2 + 2) ibound surface charge density at x = L will beWp = P. n = (3 L

2 + 2) i . i = (3 L2 + 2)

Summary

In this chapter you have learntydifferent kinds of materials and moleculesyhow to use Gausss law in dielectrics, andyelectric field in dielectrics.

chapter 03.ces

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