Electromagnetic Induction-hsa

ELECTROMAGNETIC INDUCTION

In the year 1820, it was discovered by Orested that an electric current produces a magnetic field. Soon after this, efforts were made to observe the converse of the magnetic effects of current, i.e. to show that magnetic field may produce electric current. Michel Faraday in England in 1831 demonstrated that electric current can be produced by employing a changing magnetic field. This phenomenon is called as electromagnetic induction. Faraday’s Laws of Electromagnetic Induction Faraday summed up his experimental results in the form of two laws known as Faraday’s Laws of electromagnetic induction .These are stated as follows First law: When the magnetic flux linked with the coil changes, an emf is induced in it which lasts so long as the change of magnetic flux continues. Thus condition for an emf to be induced in a coil is changing magnetic flux. Second law: The magnitude of the induced emf is directly proportional to the rate of change of magnetic flux. Mathematically,

e

d dt

e K

d dt

i.e.

Where K is constant of proportionality and is taken as 1. Induced emf  d e dt The direction or sense of polarity of the induced emf is such that it tends to produce an induced current that will create a magnetic flux to oppose the change in the magnetic flux through the coil. This is known as Lenz’s Law and is stated below. Lenz’s Law Whenever an induced emf is set-up, the direction of the induced current through the loop is such that it opposes the cause which produces it. Thus induced emf in a coil becomes

e 

d dt

The Lenz’s law is the consequence of the law of conservation of energy.

Integral and Differential form of Faraday’s Law of em induction Consider a closed circuit or a coil of any shape and is moving in a stationary magnetic. Let S be the surface enclosed by the coil C. Let magnetic flux r  B density in the neighborhood of the coil C. Then the magnetic flux through a small elementary area is a scalar product . r r r dS B.dS

Total magnetic flux through the entire coil is 

r r  B   B.dS S

According toFaraday’s law of electromagnetic induction the induced emf in a circuit is the –ve time rate of change of magnetic flux linked with the circuit. inducedemf  dB e  dt ----------------(1)

e 

d r r B.dS dt S

Also by definition, the line integral of the electric field over a closed path give the induced emf in the circuit ---------------r r e Ñ  E.dl (2) Where

r E

circuit. From

is the induced electric field at the current element

r dl

of the closed

(1) and (2) n

eq s -----------------(3)

r r d r r E Ñ  .dl   dt S B.dS This

(3) is known as integral form of Faraday’s law of electromagnetic

eq n induction. Differential form : If the circuit (coil C) remains stationary and only magnetic flux density changing then time derivative in

is

(3) may be taken inside the integral sign

eq where it becomes a partial derivative. i.e. r r r B r Ñ  E.dl   S t .dS Now, by Stoke’s Theorem

r B

n

--------------------(4)

r r

r r

Ñ  E.dl   curl E.dS S

(5) becomes 

eq

r r r B r S curl E.dS   S t .dS

n

Since the surface is arbitrary above

is true for any surface

eq 

n

r r B curl E   t

or

---------------(5)

r r B  E   t

This is the differential form of Faraday’s law of electromagnetic induction. Proof :

r r e Ñ  E.dl

Consider a wire loop or frame of any shape which occupies the positions

at C1

time t. It is moving witha velocity

so that it occupies the position

r v

C2

. t  dt

Let elementary length through a distance

r dS

r dl

r v .dt

of theloop is displaced

in the time

swept by the element

, then the area dt

r dl

is given by

r r r dS  v .dt  dl

If

r B

----------------(1)

is the magnetic flux density t any point on this

area, then the magnetic flux the area

at time

r dS

is

r r B.dS

the total magnetic flux crossing the ribbon shaped

.Hence

surfaceS spanned by the boundary of the loop is 

r r

 B.dS S

The integral

r r  B.dS

, therefore represent the change in magnetic flux crossing the

S

wire loop, as it moves from position

in a time C1 to C2

Thus d 

. dt

r r B  .dS S

Substituting the value of

from

r dS

d 

(1), we get

eq r r

n

r

 B.(v .dt  dl ) S

Now

is independent of integration dt

---------------(2) 

Now

r r r d   B.(v  dl ) dt S

r r r r r r B.(v  dl )  ( B  v ).dl

[

sincecross and dot product

are

interchangeable ] r r r   (v  B) . dl -----------------(3) 



d  dt

r

r

r

Ñ  (v  B) . dl

As the integration is now with respect to

r dl

which is a line element

and

the

integration is to be carried out over the boundary of the loop, the surface integral has been changed to the line integral .

 S

Ñ 

If

r E

is the electric field associated with the elementary length

moving with velocity

r v

r dl

when it is

then

r r r E  vB Substituting the value of



r E

d  dt

in

(3)

eq

n

r r E Ñ  . dl

According to Faraday’s law , induced emf is

e e 

r

d dt

r

Ñ  E . dl

Thus induced emf = line integral of

r E

over the circuit.

SELF INDUCTION and COEFFICIENT of SELF INDUCTION The phenomenon due to which a coil opposes any change in the current that flows through it by inducing an opposing emf in itself is called as self induction. The induced emf is called as back emf and obeys the faraday’s law of electromagnetic induction. According to Lenz’s law this induced emf have a direction so as to oppose the cause (changing current ) due to which it is produced . Coefficient of Self Induction or Self Inductance (L) Whenever a current is passed through a coil magnetic field is produced in the surrounding of the coil. The number of lines of induction passing normally through an area near the coil i.e. magnetic flux is found to be directlyproportional to the current passing through the coil.  I or

-----------(1)   LI

Where

is constant of proportionality and is called as coefficient of self induction or self L

inductance of the coil. Its value depends upon the following factors 1. The number of turns of the coil N. 2. Length of the coil 3. Area of cross-section of the coil A 4. Nature of the material of the core on which coil is wound



(1) maybe put as

eq

n

L 

 I

i.e.

when L  

I  1 unit

Thus self inductance of a coil is numerically equal to the magnetic flux linked with the coil when unit current flows through it. The SI unit of L is henry (H). Also according to faraday’s law induced emf in a coil is

e  

L

d dt

 

d (L I ) dt

 L

d I dt

e dI dt

If

then

dI  unity i.e. 1 A / s dt

L  e

Thus self inductance of a coil is numerically equal to the induced emf when the current flowing through it changes at the rate of unity ( ). 1A/ s 1 henry The self inductance of a coil is said to be 1 henry when a current changing at the rate of 1 A/s through it induces an emf of 1 volt in it. NOTE: Inductance in a circuit plays the analogous role as mass in mechanics. Mass opposes the motion of a particle and inductance opposes the change in the current. In other words the effect of inductance in a circuit is same as inertia in mechanics and inductance is therefore called as electrical inertia.

MUTUAL INDUCTANCE and COEFFICIENT of MUTUAL INDUCTANCE The phenomenon by virtue of which an induced emf is produced in a coil due to change in current in a neighboring coil is called as mutual induction. Consider two coils P and S close to each other. Let be the current flowing in the coil P at I1 some instant

and t

be the magnetic flux linked with the coil S at that instant. 2

Now flux linked with the coil S is directly proportional to the current flowing in the coil P. i.e. S  I p

-----------(1) S  M I p Where M is the constant of proportionality and is known as the coefficient of mutual induction or mutual inductance of coil S with respect to coil P. Now According to Faradays law of electromagnetic induction emf induced in the coil S to change in Current in the coil P is i.e.

eS  

dS dt

eS  

d  M Ip  dt

or eS   M

d  Ip  dt

------------------(2)

M  

eS dI p dt

then 

dI p dt

M  eS

 1A / s

the coefficient of mutual induction or mutual inductance of two coil is 

numerically equal to the emf induced in the secondary coil when the current flowing through the primary coil decreases at the rate of 1 A/s. from (1)

eq n i.e.

M

S Ip

when M  S

I p  1 unit

Thus coefficient of mutual inductance is numerically equal to the magnetic flux linked with the secondary coil when a unit current flows through primary coil. Unit of M is henry denoted by H. Its dimensions are .  M 1 L2 T 2 A2  ################################################## ####################

UNIT-II

CAPACITOR Introduction A capacitor is a device which stores electric charge. Capacitors vary in shape and size, but the basic configuration is two conductors carrying equal but opposite charges. Capacitors have many important applications in electronics. Some examples include storing electric potential energy, delaying voltage changes when coupled with resistors, filtering out unwanted frequency signals, forming resonant circuits and making frequency-dependent and independent voltage dividers when combined with resistors.

A capacitor consists of two metal plates separated by a nonconducting medium known as the dielectric medium or simply the dielectric, or by a vacuum. It is represented by the electrical symbol shown in the figure. Capacitance of a capacitor If a potential difference is maintained across the two plates of a capacitor (for example, by connecting the plates across the terminals of a battery) a charge +Q will be stored on one plate and - Q on the other. The ratio of the charge stored on the plates to the potential difference V across them is called the capacitance C of the capacitor. Thus C

Q V

Q = CV If, when the potential difference is one volt, the charge stored is one coulomb, the capacitance is one farad, F. Thus, a farad is a coulomb per volt. It should be mentioned here that, in practical terms, a farad is a very large unit of capacitance, and most capacitors have capacitances of the order of microfarads,

.The dimensions of capacitance are F

Capacitors in Parallel

The potential difference is the same across each, and the total charge is the sum of the charges on the individual capacitor. Therefore: C p  C1  C2  C3

V

Capacitors in Series The charge is the same on each, and the potential difference across the system is the sum of the potential differences across the individual capacitances. Hence V 1 1 1 1    Cs C1 C2 C3

Capacity of a Parallel Plate Capacitor

Consider a parallel plate capacitor consisting of two conducting plates having area A and separated by a small distance

in vacuum or air as shown in the fig. d

Let



and



be the surface charge density on the plates due to charges +Q and –Q respectively.

Neglecting the edge effect (fringing effect near the edges of the plates) the electric field between the

platescan be calculated using Gauss’s law

r r

Q

Ñ  E.dS   S

o

i.e.

or

E. A 

Q o

E. 

Q o A

-------------(1) 

E 

where

 0

is the permittivity of the free space (

0

).

 o  8.85  10

12

Now the work done W required to carry a test charge

C /( N  m ) 2

2

from one plate to other plate is equal to the qo

product of the force

and the distance d (as the electric field between the plates is uniform) qo E

i.e. work done W  ( qo E ).d

thepotential difference V between the plates 

V

W  E.d qo -----------(2)

V 

 .d 0

Now magnitude of charge on either plate is given by Q  .A Now capacity of the parallel plate arrangement is given by

(Using

C



Q V

C

.A  .d 0

(1) and (2))

eq

n

---------------------(3)

C 

0 A d

Thus capacity of a parallel plate capacitor depends on the size and the geometrical arrangement of the plates . C of a parallel plate capacitor is 1.

directly proportional to the area of the plates

2.

Inversely proportional to the separation between the plates.

3.

If the space between the plates is completely filled with a dielectric of dielectric constant K then

capacity is given by

i.e. capacity increases by K times when the space between the

CK

0 A d

capacitor plates completely filled with the dielectric. Example:

A parallel plate capacitor is to be designed with a voltage rating 1 KV using a material of

dielectric constant 3 and dielectric strength about

.For safety we would like, the field never to 7

10 V / m exceed, say 10 % of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pf ? Solution:

Maximum permissible voltage V =

103 V Maximum permissible electric field E= 10%

= 7

10 V / m Now potential difference between the parallel plate capacitor or V  E.d

d

V 103 V  6  103 m E 10 V / m

Now capacity of a parallel plate capacitor with dielectric constant K is given by

106 V / m

or CK



0 A d

A

A

Cd K0

50  1012  103 3  8.85  1012

 18.8  104 m 2

Dielectrics “…As our mental eye penetrates into smaller and smaller distances and shorter and shorter times, we find nature behaving so entirely differently from what we observe in visible and palpable bodies of our surroundings that no model shaped after our large-scale experiences can ever be "true". A completely satisfactory model of this type is not only practically inaccessible, but not even thinkable. Or, to be precise, we can, of course, think of it, but however we think it, it is wrong.” Erwin Schrödinger

A dielectric is a material which does not conduct electricity ie. a dielectric a basically an insulator. In these materials all the electrons are bound to the nuclei of the atoms. Thus there are no free electrons to carry the current. However when an external electric field is applied these charges gets separated through a small distances creating bound charges. An ideal dielectric is a perfect insulator .Real dielectric have

times smaller than that of the good conductors. Glass, plastic, mica,

feeble conductivities nearly 10

20

oil, water, bakelite etc. are some examples of the dielectric constants. According to modern theory neutral atoms in their ground state consist of a central positively nucleus surrounded by a spherically symmetric cloud of equal negative charge of smoothly varying density. Thus for an atom in ground state, the centre of gravity of its –ve charge lies exactly at its nucleolus which is taken to be point positive charge. The dipole moment of an atom is therefore zero. For a molecule the positive charge is supposed to concentrate at the nuclear points and the negative change forming a cloud of smoothly varying density. around constituent nuclei. Depending upon the shape of the cloud and variation of charge density inside it the molecules can be classified into two types non-polar molecules and polar molecules. 1. Non-polar molecules The molecules in which the centre of gravity of positive charges exactly coincides with that of the negative charges and there are no dipoles are called as non-polar molecules. Net dipole moment is zero

etc.

because dipole length is zero. Familiar examples are H 2 , O 2 , N 2 , CO 2 , CH 4 , CCl 4

2. Polar molecules The molecules in which the centre of gravity of positive charges does not coincide with that of the negative charges are known as polar molecules. Such molecules, therefore constitutes a permanent dipole and have a dipole moment. Familiar examples are

etc. These molecules consist H 2 O, CO, HCl, NH3

of dissimilar atoms and their dipole moment is of the order of

which means a 1029 colomb-meter

separation of

between the centers of the positive and negative charges of magnitude o

: 1A (10

10

m)

one electronic charge (

).

1.6  10

19

C

Dielectric Constant or Specific Inductive Capacity The capacity of a parallel plate capacitor in vacuum (free space or air) is given by

. Now if

Co 

0 A d

the space between the plates is completely filled with a dielectric material it is observed that capacity increases by a factor

.i.e. new capacity K = C  Co K

This factor

. K

0 A d

is called as the dielectric constant of the medium. K

K 

Thus dielectric constant of a medium is defined as

K

Capacity of a capacitor with dielecric Capacity of a capacitor with air or vacuum as dielecric

C Co

Also Coulomb’s law for force of attraction or repulsion between two charges of magnitude q1 and q2

separated by a distance r in free space is given by

Fo 

1 q1q2 4 o r 2

But when the same charges are placed in some other medium of permittivity



, the force of attraction

or repulsion is found to be

F 

1 q1q2 4 r 2

It is observed that the force between the charges in dielectric medium is reduced by a factor

. K

i.e.

or

F

Fo K

K 

Fo F

from which it can be shown that

K  r  Where

 o

is called as relative permittivity of the medium.

r Thus dielectric constant of the medium can also be is defined as i.

K

force between two charges in vacuum or air force between two charges in dielectric medium

ii.

K

Permitivity of the medium permittivity of vacuum or air

Dielectric Strength (

) E max

In a dielectric if the applied electric field is increased beyond a certain value, the field will be able to pull out the electrons from the molecules. As a result there will be free electrons and the material becomes a conductor. This called dielectric breakdown. The maximum electric field strength which a dielectric can withstand without breakdown is called the dielectric strength of medium or material .

Dielectric in an Electric Field Consider a dielectric material made up of non-polar molecules. These molecules do not have any electric dipole moment by themselves. However when an external electric field

r Eo

is applied, the centre of

positive charges is pulled in the direction of electric field ,while the centre of negative charges is pulled in the direction opposite to the direction of electric field . As a result these non-polar molecules turns into polar molecules and do acquire a dipole moment due to induction. The induced dipole moment is found to increase with the strength of the applied external electric field and almost independent of temperature. The induced polar molecules are aligned in the direction of external field and cause the surface charges to be induced on the opposite faces of the dielectric. This phenomenon is called as the polarization of dielectric. The induced charges produce an

internal electric field

r Ei

(

r r Ei  Eo

always) inside

dielectric in the direction opposite to the externally applied electric field. So the resultant electric field r E

is less than the applied external electric field

r Eo

.The resultant electric field

r E

is given by

r r r E  Eo  Ei If K is the dielectric constant of the dielectric material then the electric field inside the deictic material is reduced by a factor of dielectric constant K 

r r Eo E K

r r r Eo Eo  Ei  K

or

r r 1 Ei  Eo (1  ) K

In case of polar dielectrics, molecules are permanent dipoles and have net dipole moments. However in absence of external electric field these dipoles are randomly oriented in all possible directions due to thermal agitations. So net dipole moment of these dielectrics in the absence of external electric field is zero .When an external electric field is applied these molecular dipoles experience torque acting on them and tend to align in the direction of electric field. This alignment of dipoles is opposed by the random thermal motion of the molecules. The alignment of the dipoles is directly proportion to the applied external electric field and is inversely proportional to the temperature. If

field and

r Ei

r Eo

is the intensity of externally applied electric

is the internal electric intensity due to polarization of dielectric. Then resultant

electric field intensity

r E

is

r r r E  Eo  Ei

Electric Polarization Vector (

r P

)

When dielectricis placed in an electric field, its molecules become electric dipoles and the dielectric is said to polarized. When a dielectric is polarized the induced charge

appears on the surface area the Q'

dielectric material and therefore electric polarization is defined as ‘the induced surface charge per unit area’. i.e.

P 

Q'   pol A

multiplying and dividing Numerator and denominator with d, the thickness of the dielectric material slab.

P 

Q' d Ad (since

P  Now

) V  A.d

Q 'd V

= induced electric dipole moment Q 'd

Thus

P

Induced dipole moment volume of dielectric

Thus polarization P is also defined as the induced dipole moment per unit volume. Since dipole moment is a vector quantity polarization is a vector quantity. The direction of polarization vector

is same as the direction of resultant electric field

r P

is

r E

inside the dielectric. The polarization

. C

m2

Capacity of a parallel plate capacitor partially filled with a dielectric Consider a parallel plate capacitor having capacity

in free space. Let d be the separation and A be Co

the area of each plate. Capacity of parallel plate capacitor in free space is given as ---------------(1) Co 

Let

o A d

be the charge on each plate. The electric field Q

capacitor can be calculated by Gauss’s law

in the free space between the plates of the Eo

or r r

Q

Ñ  E.dS   S

Eo A 

o

or

Q o

-------------------(2) Eo 

Q o A

Suppose a dielectric slab of thickness t and dielectric constant K is inserted between the plates as shown in the figure. The electric field remains same as

in the free space between the dielectric slab and the Eo

plate. However electric field inside the dielectric is reduced by a factor K i.e. inside the dielectric electric field E is given by

E 

Eo K

----------------------(2) E 

Q K o A

Now the work done required to carry a test charge

from one plate to other is equal to work done qo

in the free space between the dielectric slab and capacitor plates and work done W1

inside the W2

dielectric. totalwork done 

W  W1  W

i.e.

( W  (qo Eo )(d  t )  (qo E ) t

workdone = force×displacement

  Q  Q   qo  (d  t )   qo    o A K o A  

W  qo

)



t  

Q  t  d t    o A  K

Therefore potential difference between the plates is

V 

V 

W qo

Q  t  d t    o A  K

Now, Capacity of a parallel plate capacitor with dielectric slab is given by is given by

C

or

C 

Q V



o A t   d  t  K 

------------------(3) C 

o A 

1    d  t  1  K     

or

----------------(4) 

C

From

K o A

 Kd  t  K  1 

(3) it is clear that effect o introducing a slab of thickness t is the same as decreasing the free

eq

n

space by an amount

. Thus the capacitance of a parallel plate capacitor is increased when the 1  t  1  K  

space between the plates is filled with a dielectric. Special Case: Capacity of a parallel plate capacitor completely filled with a dielectric If the space between the plates is completely filled with a dielectric then td

Substituting

in td

(4) we get

eq

n

or C 



K o A

 Kd  d  K  1 

C K

o A d

C  K Co

Thus the capacity of a parallel plate capacitor is increased by a factor dielectric constant

when the K

space between the plates is completely filled with a dielectric. Note : If the space between the plates is partially filled with a conducting slab then effective separation between the plates is reduced to (d-t) .Hence the capacity of a parallel plates increases.

Example1:Thedistance between the plates of a parallel plate capacitor of a capacitance

is

C dielectric constant

and thickness

.A slab of

d

is inserted between the plates. What is the capacitance of the

K

3d 4

system?

[RTMNU Summer-2007, 2 Marks]

Solution:

We know that capacity of a parallel plate capacitor partially filled with a dielectric is C'



K o A

 Kd  t  K  1 

Here t





3d 4

C'

K o A

 

3d   Kd  4  K  1 

i.e.

C'



4K  o A



 4 Kd  3d  K  1 

C' 

o A d

4K o A

 4 Kd  3dK  3d 



4K o A

 Kd  3d 

4K  K  3

But originalcapacity without dielectric is

C

C' 

o A d

4K C K   3

Electric Displacement (D)

Electric field can be represented by electric lines of force These lines of force originates from the positive charges and terminates at the negative charges. Suppose one lines of force emanates from a unit charge. Then number of lines force coming out from charge

will be equal to the magnitude of the charge i.e. q

.Imagine a hallow sphere of radius r with a charge q

at its centre. q

The total number of lines of force immerging out of the surface of hallow sphere normally will be equal to

. q

Now the surface area of the hollow sphere is

.The 4 r

2

number of lines of force coming out through a unit area is

.Maxwell named this quantity as the electric displacement D

q 4 r 2

D

q 4 r 2

In general electric displacement D is equal to the surface charge density of free charges. i.e.

D

q   free A

Also electric intensity E at a point at a distance r from the charge is given by

E 

1 q 4 r 2

we may write 

D

q  4 r 2 

Thus D E

Where



is the permittivity of the medium.

Electric displacement is a vector quantity and is directed along the same direction as electric intensity. The SI unit of

r D

is C / m2

Note: The electric displacement vector

r D

represents a partial field in the sense that its sources are free

charges (NOT bound charges). It has nothing to do with the induced charges due polarization of the dielectric materials.

Relation between

r D

,

r E

and

r P

Consider polarization of a dielectric slab placed between the plates of a parallel plate capacitor. Let and Q

be the charges on the plates of the capacitor then the surface charge densities on the Q

plates of the capacitor is and

 free 

Q A

 free 

Q A

Now the electric displacement is equal to the surface charge density due to free charges. 

D   free

The electric field intensity in the free space between the capacitor plates is ---------------(1) Eo 

 free o

Due to polarization of the dielectric charges are induced on the opposite faces of the dielectric. Let and Qi

be the induced charges on the faces of the dielectrics. Consequently the induced surface Qi

charge densities is and

 i 

Qi A

 i 

Qi A

The electric field inside the dielectric dueto induced charges is

----------(2)

Ei This electric field



i o

is in the direction opposite to the to electric field

due to free charges. Hence

Ei

Eo

the resultant electric field

inside the dielectric is E

E  Eo  Ei

i.e. E

 free  i  o o

or

 o E   free   i Now

is the induced surface density on the face of a dielectric and is equal to the magnitude of the

i electric polarization

r P

.



oE  D  P

or

----------(3) D  oE  P

In vector form r r r D  oE  P

------------(4)

In the free space where no polarization charges are present

r P

= 0,

(4) may be written as

eq

n

-------------(5)

r r D  oE

Gauss’s Law in term’s of electric displacement vector Consider the situation in which the charge is distributed over a volume V such that



is the volume

charge density. Then the charge enclosed by surface enclosing volume V is -----------(1) Q    .dv V

Now According to Gauss’s law in electrostatics electric flux through closed surface enclosing volume V is

r r Q

Ñ  E.dS   S

Where



is the permittivity of the medium.

Above equation may be written as



r r

Ñ  E.dS

 Q

S

or

r r Ñ   E.dS  Q

--------(2)

S

But

Therefore in terms of

r r D  E (2) may be written as

r D

eq

n

r r Ñ  D.dS  Q S

------------(3)

This is Gauss’s Law for electrostatics in terms of displacement vector

Using

r D

(1)

eq

n

r r Ñ  D.dS  S

-------------(4)

 .dv

V

Applying Gauss’s divergence theorem surface integral in LHS of

(4) may be written as volume

eq

n

integral as r r

Ñ  D.dS S



r   div D.dv V

r

 div D.dv    .dv

V

V

Since the volume is arbitrary this is true for all the volumes therefore integrands must be equal. r div D  

---------------(5)

or

r  div E   or

----------------(6)

r  gE   (5) and (6) are differential forms of the Gauss’s law in electrostatics. n

eq s

Schematic Representation of

r D

,

r E

and

r P

Boundary conditions Satisfied by

(A)

Boundary conditions satisfied by

r D

and

r E

r D

Let’s consider a boundary separating the two medium as shown in the fig Construct a cylindrical Gaussian pillbox of height

and base area h

Let

r D1n

S

Average normal component of displacement vector



r D

to the bottom of the box in

medium 1. It is inward normal. r D2n medium 2

Average normal component of displacement vector



r D

to the face of the box in

By making the height of the cylinder

approaching to zero (

), the contribution to the total h0

h

outward flux from the curved surface can be made zero. Now, by Gauss’s law the net outward flux through the cylindrical Gaussian pillbox is r r r r D1n gS  D2 n gS  Q

Where

is the total charge enclosed by the c Q

cylindrical Gaussian surface. But

Fig

r r D1n gS  D1n S cos 

(

) since  = 180

  D1n S

r r D2 n gS  D2 n S



 D1n S  D2 n S  Q

D2 n  D1n 

Where

Q S



Charge per unit area on the boundary of the two medium. Q  S

Thus the normal component of the displacement vector

r D

changes at the charged boundary

between the two dielectric by an amount equal to the surface charge density



.

However if the boundary is free from charge i.e.

, then  0

D2 n  D1n

i.e. Normal component of the displacement vector is CONTINEOUS across the charge free boundary between the two dielectrics.

(B)

Boundary Conditions Satisfied by

Let’s consider two Electric fields

r E1

and

r E2

r E

in the two media. Here we will make use of

the fact that electrostatic force field is conservative in nature i.e. around any closed path, the potential difference vanishes. In other words the line integral of electric field around any closed path is zero. V (closed path)  

r r

Ñ  E.dl  0

Consider a rectangular path ABCD as shown in the fig. Let AB and CD each have length

and BC and AD x

being negligibly small. Now

r r

Ñ  E.dl

0

ABCD

i.e.

r r

r r

r r

r r

 E.dl   E.dl   E.dl   E.dl

AB

BC

CD

DA

Since paths BC and DA are negligibly small



BC

r r E.dl  0 and



DA

r r E.dl  0

0

Thus

r r E Ñ  .dl 

ABCD

Now





AB

r r r r E.dl   E.dl  0 CD

and

r r E.dl  E1t . x



AB

Where

CD

Tangential components of E1t

 Tangential components of

E2t

r r E.dl  E2t . x



r E1

and

r E2

E1t . x  E2t . x  0

 E1t  E2t  .

x  0

(As 

can not be zero) x

E1t  E2t

Thus tangential components of same on both sides of a boundary between the two dielectrics. Or The tangential components of electric fields

r E

are continuous across the boundary.

Example :Let two isotropic dielectric media be separated by a charge free plane boundary as shown in

.Then show that

the figure .Let the permittivities be

1 and  2

.where tan 1 K1  tan  2 K 2 Solution:

and K1

are dielectric constants of the two medium. K2

As shown in the figure medium 1 and medium 2 are separated by a charge free boundary

i.e. surface charge density on the boundary is zero. The boundary conditions on

r D

and

r E

are

----------(1) D1n 

D2 n

and

----------(2) E1t  E2t

Now from fig. i.e. normal components of D1n

is D1

D1n  D1 cos 1

and D2 n  D2 cos  2

sothat

(1) can be written as --------------(3) D1 cos 1  D2 cos  2

Also E1t  E1 sin 1

and E2t  E2 sin  2

sothat

--------------(4) E1 sin 1  E2 sin  2

Taking ratios of (4) and (3) E1 sin 1 E2 sin  2  D1 cos 1 D2 cos  2

and

But D1  1 .E1

D2   2 .E2



E1 sin 1 E2 sin  2  1.E1 cos 1  2 .E2 cos  2

i.e. tan 1 tan  2  1 2

So that tan 1 1  tan  2  2

If

and K1

are relative permittivities (dielectric constant) then K2

and 1   o K1

 2   o K2

tan 1  o K1  tan  2  o K 2

Or tan 1 K1  tan  2 K 2

Molecular Polarizability When a dielectric is placed in an external electric field its molecules becomes electric dipoles oriented in the direction of the field. Thus the dielectric acquires a net dipole moment and its molecules are polarized. For linear dielectrics induced dipole moment is directly proportional to the electric field intensity causing polarization. i.e. p E

p  αE

where

is constant of proportionality to known as the polarizability.

α



α 

If

p E

then α = p

E 1

Thus polarizability is defined as the induced dipole moment per unit electric field strength. SI unit of polarizability is

. 2

coulomb meter / N

i.e. C  m / N 2

Contribution to the total polarizability is due to following types 1. Electronic Polarizability

 αe  The displacement of the electron cloud relative to the nucleolus in an atom constituting the molecules induces the dipole moment in a molecule. This is called electronic polarizability. 2. Ionic polarizability

 αi  If we have a solid dielectrics whose molecules are made up of ions there is relative motion of positive and negative ions resulting induced dipole moment. This is called ionic polarizability. 3. Orientational or dipolar polarizability

 αo  If there are molecules with permanent dipole moments, randomly oriented, they tend to align in the direction of the applied electric field producing a net dipole moment. This is called orientational or dipolar polarizability. Thus total dielectric polarization may be considered as sum of these three contributions α = αe  αi  αo

Subscripts e, i, or o stands for electronic, ionic and orientational polarizability respectively. 32-

Larentz Local Field (Internal molecular field)

When a dielectric solid is placed in an external electric field a molecule or an atom of solid experiences not only the external field but field produced by the dipoles as well. This resultant field experienced by the atom or molecule is called internal molecular field or Larentz local field or simply local field.

Calculation of Local Field Consider a dielectric, uniformly polarized by placing it in a uniform electric field between two oppositely charged parallel plates. In order to calculate the local field experienced by the atom consider a small sphericalcavity with the atom for which local field is to be calculated at its centre. The radius of the cavity is chosen large enough. The part of the dielectric external to the spherical cavity can be replaced by a system of charges induced at the surface of cavity.

at the point at the centre

The net local field Elocal

of the spherical cavity is given by ----------------(1) Elocal  Eo  E ' E1  E2

where 1.

externally applied electric field due to the Eo 

charges on the capacitor plates.

Eo  2

 o

Electric field at O due to induced charges on



E'

the plane surface of the dielectric E'  3.

 pol o

Field at O due to polarized charges on the



E1

surface of the cavity. 4.

Field at O due to permanent atomic dipoles . But in the present case i.e. for nonpolar E2 

isotropic

dielectric E2  0

Thus local field Elocal  Eo  E ' E1

Now,

effectively produces a net field

.So that E

Eo and E ' E  Eo  E '

----------------(2) Elocal  E  E1

Calculation of E1

Consider a small elementary area ds surrounding a point A on thesurface of the cavity of angular width

and at an d

angle

with the direction of field

 The polarization vector

thecomponent of 

r P r P

r E

.

will be parallel to

r E

normal to area

is dS

PN  P cos 

Since polarization is induced surface charge per unit area, the total charge on area dS is dq  P cos  .dS the electric intensity at O, due to this elementary 

charge

is dq ------------(3)

dE1 

1 P cos  .dS 4 o r 2

This field

is directed along OA. Resolving this field dE1

intensity

along the applied field (parallel ) and dE1

perpendicular to it we have the component along the field



 1 P cos dE1 cos   .dS cos 2  4 o r  i.e.

r E

along the field 1 P cos   .dS 4 o r2 2

The perpendicular component of

r E

at O is zero because they cancel out each other. dE1

If the area

be a ring shaped element as shown in fig. of radius

and width r sin 

dS

surface of sphere, then area of this element is dS  2  r sin   r d

 2 r 2 sin  d Thus parallel component of

due to this ring is dE1

, on the r d



P cos 2   2 r 2 sin  d 2 o r 2

 the field 

P cos 2  sin  d 2 o

at O due to the entire induced charge on the spherical cavity is : E1

E1 



P

 2 0

cos 2  sin  d

o

 2 P  2  cos 2  sin  d 2 o 0

Substitute



cos   x

 sin  .d  dx

when

and

  0;

x 1

when



 ; 2

0

E1 

P 2 x (  dx)  o 1

P  x3     o  3  or

0

1

--------(4)

E1 

P 3 o

(2) becomes 

eq

n

-------(5)

Elocal 

E

P 3 o

x 0

This

(5) is known as Larentz relationgives the actual field acting at the position a molecule at O

eq

n

within the dielectric.

Clausius- Mossotti Equation Claussius and Mossotti tried to correlate the dielectric constant

(macroscopic parameter) to the K

molecular polarizability



(microscopic parameter) of a non-polar dielectric. This relation is known as

Claussius –Mossotti equation. The dipole moment of a single molecule is, p   Elocal

Now, if there are n molecules per unit volume then polarization P is given by Pnp

 n  Elocal

------------(6)

 P  n   E  3 o   But we know that in a dielectric D  o E  P

(In dielectric

 E  oE  P ) D E



P  E    o  P  E  K o   o 

medium of



permittivity

or P   K  1  o E (6) may be written as 

eq

n

 K  1

  K  1  o E  o E  n   E   3 o  

  K  1  n  E  1 3 

 K  1

  K  1 o  n   1 3 

 K  1

o  n   

  

  

 K 2   3  ------------------(7)

K 1 n  K 2 3 o Molecular polarizability



is given by

-----------------(8)

 

3 o  K 1    n  K 2

This equation is called as Claussius-Mossotti equation. The claussius-Mossotti equation holds best for dilute substances such as gases. For liquids and solids, this relation is only approximately correct.

Solved Problems

Hendrik A. Lorentz (1853–1928, Dutch)

Lorentz extended Maxwell’s work in electromagnetic theory and used it to explain the reflection and refraction of light. He developed a simple and useful model for dielectric media. He hypothesized that the atoms were composed of charged particles, and that their movement was the source of light. He won the Nobel prize in 1902 for his contributions to electromagnetic theory.

Michael Faraday (1791–1867, English)

Faraday was one of the greatest experimental physicist in history. He is perhaps best known for his work that established the law of induction (i.e. changing magnetic fields produce electric fields). He also discovered that magnetic fields can interact with light. When a magnetic field is oriented along the direction of travel for light in a dielectric, the polarization of the light will rotate. This effect is used to build optical isolators, which prevent

light from reflecting back into an optical system.

Gustav Kirchhoff (1824–1887, German)

Kirchhoff studied the spectra emitted by various objects. He coined the term “blackbody” radiation. He understood that an excited gas gives off a discrete spectrum, and that an unexcited gas surrounding a blackbody emitter produces dark lines in the blackbody spectrum.

James Clerk Maxwell (1831–1879, Scottish)

Maxwell is best known for his fundamental contributions to electricity and magnetism and the kinetic theory of gases. He studied numerous other subjects, including the human perception of color and color-blindness, and is credited with producing the first color photograph. He originally postulated that electromagnetic waves propagated in a mechanical “luminiferous ether,” but subsequent experiments have found this model untenable. He founded the Cavendish laboratory at Cambridge in 1874, which has produced 28 Nobel prizes to date.