ELECTRIC FIELD INTNSITY AND POTENTIALbssbcollege.ac.in/Physics-V.pdf · III B.Sc Physics Paper –V...

III B.Sc Physics Paper –V Semister -V

Department of Physics , B.S.S.B Degree College , Tadikonda

UNIT – I

ELECTRIC FIELD INTNSITY AND POTENTIAL

Introduction

Characteristics of charge :

1. Charge is of two types. One is positive charge and other is negative charge

2. Like charges repel to each other and unlike charges attracted to each other

3. Charge is quantized

4. Charge is conservative

Coulimb’s law :- The force of attraction or repulsion between the two point charges is

directly proportional to product of two charges and inversely proportional to square of the

distance between them.

. q1 q2

. q1 and q2 be the point charges are separated by a distance r .The force acting between the

charges F is F ∝ q1 q2 F ∝ 1

𝑟2 => F ∝

𝑄1 𝑄2

𝑟2

F = 1

4𝜋𝜖 𝑄1 𝑄2

𝑟2

Electric field :- The region surrounding an electric charge or a group of charges in which

another charge experience a force is called electric field.

Intensity of electric field E :- The intensity of electric field is defined as the force

experienced by a unit positive charge placed at that point.

E = 𝐹

𝑞 N/C

Electric Lines of Force : It is the imaginary smooth curve drawn in an electric field along

which a free isolated positive charge will move.



Properties :

1. The Electric lines of force starting from Positive charge and ending from negative

charge

2. The Tangent drawn at any point drawn on the line of force gives the direction of

electric field at that point.

3. The electric lines of force start from positive charge and end on a negative charge

4. No two electric line of force intersect each other.

Electric field intensity due to a point charge

Consider a point charge +q coulomb is placed at a point O in air as shown below. The

electric field intensity can be determine at p

According to columbs law The electric force acting on q0 is given by F = 1

4𝜋𝜖 𝑞1 𝑞0

𝑟2

q .P

r

But the electric field intensity E = 𝐹

𝑞

E=

1

4𝜋𝜖 𝑞1 𝑞0

𝑟2

𝑞0 => E =

1

4𝜋𝜖 𝑞

𝑟2

Electric Flux : - The electric flus through a surface placed represented the total number of

electric lines of force crossing the surface in the direction normal to the surface.

∅E = ∫ 𝐸. 𝑑𝑠



Gauss’s Law

Statement :- Gauss’s law states that total normal electric flux ∅E over a closed surface is

1/𝜖0 times the total charge Q enclosed within the surface .

Mathematically it can be expressed as ∅E = ∫ 𝐸. 𝑑𝑠 = ∫ 𝐸 𝑑𝑠 cos 𝜃 = (1/𝜖0 ) Q

𝜖0 is the permittivity of the free space .

Proof : - i) When the charge is with in the surface

Let a charge +Q is placed at O with in a closed surface of irregular shape as

shown below. Consider a point P on the surface at a distance r from O. Consider a small

area dS around P. the normal surface dS is represented by a vector dS makes an angle 𝜃

with the direction of electric field E along OP.

The electric flux through the area dS is

∅E = 𝐸. 𝑑𝑠 = 𝐸 𝑑𝑠 Cos 𝜃 ------1

From columbs law the electric intensity E at a point P is

=> E = 1

4𝜋𝜖 𝑄

𝑟2 ………… 2

From equations 1 and 2 we get d∅E =𝐸. 𝑑𝑠 = 𝐸 𝑑𝑠 Cos 𝜃

=1

4𝜋𝜖

𝑄

𝑟2 . dS Cos 𝜃

= 𝑄

4𝜋𝜖 (

dS Cos 𝜃

𝑟2 )

( dS Cos 𝜃

𝑟2 ) is called solid angle d𝜔 subtended by dS at O.

= > d∅E = 𝑄

4𝜋𝜖 d𝜔

The total flux ∅E over the entire whole surface is given by ∅E = 𝑄

4𝜋𝜖0 ∫ d𝜔

Where ∫ d𝜔 is the solid angle subtended by the whole surface at O. this is equal to 4𝜋

=> ∅E = 𝑄

4𝜋𝜖0 4𝜋 = Q/𝜖0



Let the closed surface enclosed several charges any +Q1 , +Q2 , +Q3 , …… , +Q11 , +Q2

1 ,

+Q31, ……. Thus , the total flux is given by

∅E = ( 1/𝜖0 ) [+Q1 +Q2 +Q3 …… +Q11 +Q2

1 +Q31 …….]

= (1/𝜖0 ) ∑Q

ii) When the charge is outside the surface

Let a point charge +Q be submitted at a point O outside the closed surface as

shown below. now a cone of solid angle d𝜔 from O cuts the surface areas dS1 , dS2 , dS3 ,

dS4 at points P,Q,R and S respectively. The electric flux for an outward normal is positive

and inward normal is Negative. There fore the flux through areas dS2 , dS4 are positive and

dS1 , dS3 are negative.

There fore the electric flux at p through area dS1 = - 𝑄

4𝜋𝜖0 d𝜔

the electric flux at p through area dS2 = + 𝑄

4𝜋𝜖0 d𝜔

the electric flux at p through area dS3 = - 𝑄

4𝜋𝜖0 d𝜔

the electric flux at p through area dS4 = + 𝑄

4𝜋𝜖0 d𝜔

Total Electric flux = - 𝑄

4𝜋𝜖0 d𝜔+

𝑄

4𝜋𝜖0 d𝜔-

𝑄

4𝜋𝜖0 d𝜔+

𝑄

4𝜋𝜖0 d𝜔= 0

So the total electric flux over the whole surface due to an external charge is zero.

This verifies Gauss’s Law.

Differential form of Gauss’s Law

According to Gauss’s law as ∅E = ∫ 𝐸. 𝑑𝑠 = (1/𝜖0 ) Q

Let the Q be distributed over the volume V and ρ be the density of charge. Then

Q = ∭ 𝜌 𝑑𝑉 => ∫ 𝐸. 𝑑𝑠 = (1/𝜖0 ) ∭ 𝜌 𝑑𝑉 -------i

From divergence theorem ∫ 𝐸. 𝑑𝑠 = ∭ ∇. 𝐸 𝑑𝑉 ------- ii

From i and ii ∭ ∇. 𝐸 𝑑𝑉 = (1/𝜖0 ) ∭ 𝜌 𝑑𝑉 =>∇. 𝐸 = 𝜌/𝜖0



Electric field due to a Uniformly charged sphere

Case(i) : At a point outside the charged sphere

Consider a sphere A of radius R with centre O as shown in figure. Let a charge q be

uniformly distributed over it. Suppose P be an external point at a distance r from the centre

O. Construct a Gaussian surface of radius r . In Gaussian surface the electric field intensity

is same. Both E and dS are in the same direction at P then 𝜃 = 0

The flux in that surface is 𝐸. 𝑑𝑠 = 𝐸 𝑑𝑠 Cos 𝜃 = 𝐸 𝑑𝑠 Cos00 =E dS

Electric flux though the entire Gaussian surface is ∅E = ∫ 𝐸. 𝑑𝑠 = ∫ 𝐸 𝑑𝑠 = 𝐸 ∫ 𝑑𝑠

According to Gauss’s law the total electric flux over a closed surface is 1//𝜖0 times the

charge enclosed in the in the surface

∅E = ∫ 𝐸. 𝑑𝑠 = 𝑞

𝜖0

∫ 𝐸 𝑑𝑠 = 𝑞

𝜖0

𝐸 ∫ 𝑑𝑠 = 𝑞

𝜖0

E(4πr2) = 𝑞

𝜖0

=> E = 1

4𝜋𝜖0 𝑞

𝑟2 newton / coulomb -----------I

Case(ii) : At a point on the surface

When the point P lies on the surface of the sphere , then r=R, in this case the field

intensity is

E = 1

4𝜋𝜖0 𝑞

𝑅2 newton / coulomb

Case (iii) At a point inside the charged sphere

The electric field E at a point P which is inside the charged sphere at a distance r1

from the centre. The Gaussian surface at p1 is shown below

𝐸. 𝑑𝑆 = 𝐸 𝑑𝑆 Cos 𝜃 = 𝐸 𝑑𝑆 Cos00 =E dS



The electric flux through the entire surface is ∅E = ∫ 𝐸. 𝑑𝑠 = 𝑞

𝜖0

∫ 𝐸 𝑑𝑠 = 𝑞1

𝜖0

𝐸 ∫ 𝑑𝑠 = 𝑞1

𝜖0

E(4πr1 2) = 𝑞1

𝜖0 -----(a)

The total charge enclosed by the Gaussian surface

= Volume enclosed x charge per unit volume

= 4

3 πr3 x ρ

Charge q is distributed over a sphere of radius R . Hence ρ = 𝑇𝑜𝑡𝑎𝑙 𝑐ℎ𝑎𝑟𝑔𝑒

𝑉𝑜𝑙𝑢𝑚𝑒 =

𝑞4

3 πr3

.; Charge enclosed in Gaussian surface q1 = 4

3 πr3 x

𝑞4

3 πr3

= q [ 𝑟1

𝑅 ]3 ----(b)

From gauss’s law equation (a) and (b) is

E(4πr1) 2 = 𝑞1

𝜖0

E(4πr1) 2 = q [

𝑟1

𝑅 ]3

𝜖0

E= 1

(4πr1) 2 q [

𝑟1

𝑅 ]3

𝜖0 => E =

1

4𝜋𝜖0 𝑞 𝑟1

𝑅3 ---- II

This gives the electric field intensity is directly proportional to the distance r1

The variation of electric field with distance is shown below

The field outside the sphere is inversely proportional to the distance . the electric field

intensity is maximum at the surface of the sphere . At the centre of the sphere the electric

filed intensity is 0.



Electric field due to infinite conducting sheet of charge

Consider a charged conducting surface of charge density . Now we determine the field

due to a Point P near the surface and outside the conductor. Construct a Gaussian

cylindrical surface.

The direction of the electric field near the surface is perpendicular to the surface.

From the figure the electric flux through the

Cylindrical Gaussian surface results from the

two ends and curved surface of the cylinder.

At right end the electric field E,

is parallel to dS.

then the electric flux is EdS

and

At left end there is no electric field. There fore the flux through this end is zero

The electric flux through the curved surface is zero because Both E and dS are

perpendicular.

∅ = ∮ 𝐸. 𝑑𝑆 + ∮ 𝐸. 𝑑𝑆 + ∮ 𝐸. 𝑑𝑆

Righ left centre End end end

∅ = ES + 0 + 0 = ES

According to gauss’s law ES = 𝑞

𝜖0 =

𝜎𝑆

𝜖0 => E =

𝜎

𝜖0

This is the electric field intensity in a infinite conducting sheet of charge.



ELECTRIC POTENTIAL

Electric Potential Difference :- The work done experience by a unit test charge placed

from one point to other point in the electric field is called electric potential difference

between these points.

If W be the work done in moving the test charge q0 from q0

point B to point A the potential difference VA-VB = 𝑊

𝑞0 +q *A *B * F

Electric Potential : The work done by an external agent in carrying a unit positive test

charge from infinity to that point against the electric force of the field.

The electric potential V = 𝑊

𝑞0 J/C or Volts

Volt :- one volt is defined as the difference in the potential between the points so that one

joul work is done in carrying one coulomb of positive charge from one point to other .

1 volt = 1 joul / 1 coulomb

Equipotential surfaces

The equipotential surfaces in an electric field in a surface on which the potential is

same at every point.

The locus of all points which have the same electric potential is called equipotential

surface.

The potential difference in any two points on the equipotential is zero

No work is done in taking a charge from one point to other.

In equipotential surfaces lines of force at every point is perpendicular to the

surface.



In case of uniform electric field the lines of force are straight and parallel the the

equipotential surfaces are perpendicular to the lines of force as shown in figure(a).

In case of point charge or sphere of charge the equipotential surfaces are concentric

spheres as shown in figure(b).

When the charge is infinite , the equipotential surfaces is plane.

The equipotential surfaces acts as wave –fronts in optics.

POTENTIAL DUE TO A PINT CHARGE

Consider a point charge +q and its outward electric field is E. we determine the

electric potential at a point B situated at a distance rb from the charge +q.

Consider the points A and B in the radial line as shown above. Let the test charge q0

moving towards from A to B.

The force exerted by the field of charge q on the test charge is q0 is Eq0 .

The work done by external agent to move the charge q0 through a small distance dr is

given by dw= Eq0 . dr = Eq0 dr Cos 1800 = - Eq0 dr

We know that E = 1

4𝜋𝜖0 𝑞

𝑟2

dw = - Eq0 dr = − (1

4𝜋𝜖0 𝑞

𝑟2 ) q0 dr

The total work done in moving the test charge from A to B

WAB = − ∫1

4𝜋𝜖0 𝑞𝑞0

𝑟2

𝐵

𝐴 . dr

Where rA is the distance of point A from q

WAB = −𝑞𝑞0

4𝜋𝜖0 (−

1

r )ra

rb = 𝑞𝑞0

4𝜋𝜖0 (

1

rB−

1

rA )

The potential difference between two points will be VA-VB = 𝑊

𝑞0 =

𝑞

4𝜋𝜖0 (

1

rB−

1

rA )



When point A is taken at infinity so that VA= 0

VB = 1

4𝜋𝜖0 𝑞

𝑟𝑏

The potential at point is V = 1

4𝜋𝜖0 𝑞

𝑟

Electric Potential due to Charged Spherical shell Conductor

i) When point P lies outside the shell

Consider a conducting charged spherical conductor with centre O and radius R as

shown below. Let q is the charge and 𝜎 be the charge density. When whole charge is

distributed on the surface of the sphere and will be charged inside the sphere. Consider

P be a point outside the sphere at a distance r from the centre . the sphere is divided in

to number of rings with centre on O. Consider a ring ABCD between two planes AB

and CD.

From the figure CP = x , < 𝐶𝑂𝑃 = 𝜃 𝑎𝑛𝑑 < 𝐴𝑂𝐶 = 𝑑𝜃 .

From Right angle triangle OEC Sin 𝜃 = 𝐶𝐸

𝑂𝐶 => CE = OC Sin 𝜃 = R Sin 𝜃

The radius of the ring = R Sin 𝜃

From the sector AOC , AC = R 𝑑𝜃 ( arc length = radius x angle )

The thick of the shell = R 𝑑𝜃



Circumference of the ring = 2𝜋 x radius of the shell = 2𝜋 x R Sin 𝜃 = 2𝜋 R Sin 𝜃

Area of the ring = Circumference x thickness of the ring

= 2𝜋 R Sin 𝜃 x R 𝑑𝜃

= 2𝜋 R2 Sin 𝜃 𝑑𝜃

Charge on the ring = area of the ring x surface density

= 2𝜋 R2 Sin 𝜃 𝑑𝜃 x 𝜎

𝜎 = 𝑡𝑜𝑡𝑎𝑙 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑠ℎ𝑒𝑙𝑙

𝑡𝑜𝑡𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 =

𝑞

4𝜋𝑅2

Charge on the ring dq = 2𝜋 R2 Sin 𝜃 𝑑𝜃 x 𝑞

4𝜋𝑅2

= 𝑞

2Sin 𝜃 𝑑𝜃 ------------------1

The Potential at P due to the charge on the ring dv = 1

4𝜋𝜖0 𝑑𝑞

𝑥 -----------2

dv = q Sin 𝜃 𝑑𝜃

4𝜋𝜖0 𝑥 ------------ 3

from figure 𝑥 2 = R2 + r2 – 2Rr Cos 𝜃

Differentiating the above equation we get 2𝑥 d𝑥 = - 2 Rr (- Sin 𝜃 𝑑𝜃 )

Sin 𝜃 𝑑𝜃 = 𝑥 𝑑𝑥

𝑅𝑟 ---------------- 4

Substituting the value of Sin 𝜃 𝑑𝜃 we get

dv = q 𝑥 𝑑𝑥

8𝜋𝜖0 𝑅𝑟𝑥 =

q 𝑑𝑥

8𝜋𝜖0 𝑅𝑟 ------------------- 5

to obtained potential due to whole spherical shell we integrate the above equation with

the limits 𝑥 = r-R to 𝑥 = r+R

V = ∫q 𝑑𝑥

8𝜋𝜖0 𝑅𝑟

𝑟+𝑅

𝑟−𝑅 =

q

8𝜋𝜖0 𝑅𝑟 𝑥 |𝑟+𝑅

𝑟−𝑅 = q

8𝜋𝜖0 𝑅𝑟 ( 𝑟 − 𝑅 − 𝑟 + 𝑅 )

= q

8𝜋𝜖0 𝑅𝑟 ( 2𝑅 )



V = 1

4𝜋𝜖0 𝑞

𝑟 ---------------- 6

This gives the potential at a point on the out side the spherical shell. Potential is

inversely proportional to the distance

ii) When P lies on the surface

In case r = R

Potential at the surface V = 1

4𝜋𝜖0 𝑞

𝑅 -------------- 7

iii ) When P lies inside the Shell

In this the limits of the above is changed to 𝑥 = R-r to 𝑥 = R+r and integrate

the equation we get

V = ∫q 𝑑𝑥

8𝜋𝜖0 𝑅𝑟

𝑅+𝑟

𝑅−𝑟 =

q

8𝜋𝜖0 𝑅𝑟 𝑥 |𝑅+𝑟

𝑅−𝑟 = q

8𝜋𝜖0 𝑅𝑟 ( 𝑅 − 𝑟 − 𝑅 + 𝑟 )

= q

8𝜋𝜖0 𝑅𝑟 ( 2𝑟 )

V = 1

4𝜋𝜖0 𝑞

𝑅

The potential at an internal point is same on the surface.

The variation of potential with distance is shown below



UNIT - I

2. Dielctrics

Dielectrics are the substances which are do not contain free electrons. The electrons are

tightly bound to the nucleus of the atom.

Ex: Mica , Glass , wood ,Paper , Plastic etc.

Difference Between Dielctrics and conductors

1. Dielectrics are the substances which are do not contain free electrons. Conductors

contain large number of free electrons

2. The dielectric does not conduct electricity. Conductors conduct electricity.

3. The charge given to a dielectric remains localized whereas it resides on the other

surface in case of conductor.

Uses

1. For large dielectric constant and for large dielectric strength glass and paper are used.

2. For making high capacity condensers , paper and mica are used.

3. Quartz , mica and glass are used for high insulation

Dielectric Polarization and charge Density

Consider a dielectric slab is paced between the two metallic plates which can be charged to

equal and opposite charge. Below figure shows the electronic structure of an atom when

the two plates are not charged. When the plates are charged the electrons are slightly

displaced towards the positively charged plate and positively charged nucleus towards the

negatively charged plate. In this the dielectric is acted upon by the forces and said to be

polarized. This distorted atom is called Electric dipole.it has an electric dipole moment.



The electric dipole moment per unit volume is called as dielectric polarization P.

Explanation:-

Consider a dielectric slab of area of cross section A and length l is placed in an electric

field as shown below. The induced charged on faces ABCD and EFGH be –q1 and +q1

respectively respectively. The dipole moment P is q1l .

The volume of the slab is Al.

The electric dipole moment per unit volume is dielectric polarization P = 𝑞1

𝐴𝑙𝑙 =

𝑞1

𝐴

The dielectric polarization is numerically equal to the induced surface charge density.

Definitions :-

Dielctric constant (K) : It is the ratio of the capacitance of the condenser with dielectric to

the capacitance of the same condenser without dielectric.

K = 𝐶 ( 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑛𝑑𝑒𝑛𝑠𝑒𝑟 𝑤𝑖𝑡ℎ 𝑑𝑖𝑒𝑙𝑐𝑡𝑟𝑖𝑐 )

𝐶0 (𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑛𝑑𝑒𝑛𝑠𝑒𝑟 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑑𝑖𝑒𝑙𝑐𝑡𝑟𝑖𝑐 ) (or)

K = 𝑉0 ( 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑑𝑖𝑒𝑙𝑐𝑡𝑟𝑖𝑐 )

𝑉(𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑤𝑖𝑡ℎ 𝑑𝑖𝑒𝑙𝑐𝑡𝑟𝑖𝑐 ) =

𝐹0 ( 𝑓𝑜𝑟𝑐𝑒 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑑𝑖𝑒𝑙𝑐𝑡𝑟𝑖𝑐 )

𝐹( 𝑓𝑜𝑟𝑐𝑒 𝑤𝑖𝑡ℎ 𝑑𝑖𝑒𝑙𝑐𝑡𝑟𝑖𝑐 ) =

𝜖 ( 𝑃𝑒𝑟𝑚𝑖𝑡𝑡𝑖𝑣𝑖𝑡𝑦 𝑤𝑖𝑡ℎ 𝑑𝑖𝑒𝑙𝑐𝑡𝑟𝑖𝑐 )

𝜖0 (𝑃𝑒𝑟𝑚𝑖𝑡𝑖𝑣𝑖𝑡𝑦 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑑𝑖𝑒𝑙𝑐𝑡𝑟𝑖𝑐 )

Electric Susceptibility ( χ ):-

The electric susceptibility (χ ) may be defined as the ratio of polarization vector to the

electric intensity in the dielectric .

P ∝ E = > P = χ E

It is also defined as the ratio of induced surface charge density produced in the dielectric

to the resultant electric field in the dielectric .



Relation between Dielectric constant and susceptibility

The Polarization P is proportional to the electric field E within the dielectric

P ∝ E = > P = χ E

Where χ is called electric susceptibility of the medium.

We know that D = ε0 E +P

ε E = ε0 E + χ E

ε = ε0 + χ

Dividing ε0 on both sides we get

𝜖

𝜖0 = 1 +

χ

𝜖0 => K = 1 +

χ

𝜖0 or χ = 𝜖0 ( K – 1 )

Three electric vectors and their relations

1. Electric field Intensity (E) :- The force experienced by a unit positive charge placed

in the electric field at any point. The direction of the field is same as that of the field.

2. Dielctric polarization (P) :- The electric dipole moment per unit volume is called as

dielectric polarization P.

3. Electric displacement D :- It is the vector quantity whose surface integral over any

charged surface is equal to the free charge only within the surface.

Relation

According to gauss law the electric flux ∅𝐸 = ∫ 𝐸. 𝑑𝑆 = 𝑞

∈0

E is the electric field vector and dS is the surface area.

Consider a parallel plate condenser without dielectric as shown below. A is the area

of the plate. The charge q on the plates is the same. The Gaussian surface is PQRS as

shown below. The from gausses law

∮ 𝐸0. 𝑑𝑆 = 𝑞

∈0

𝐸0 ∮ 𝑑𝑆 = 𝑞

∈0



E0 A = 𝑞

∈0

E0 = 𝑞

𝐴∈0 ----------- 1

When the dielectric slabe is placed between the plates of the condenser , the charge is

negative on the surface of nearer the positive plate and positive charge is nearer the

negative plate. Let q1 be the induced surface charge. The net charge within the Gaussian

surface is P1Q1R1S1 is (q-q1). Let E be the electric field intensity the by gauss’s law

∮ 𝐸. 𝑑𝑆 = 𝑞−𝑞1

∈0 = > 𝐸 ∮ 𝑑𝑆 =

𝑞−𝑞1

∈0

E . A = 𝑞−𝑞1

∈0

E = 𝑞−𝑞1

𝐴∈0 ------------ 2

By the definition of the dielectric constant K , 𝐸 = 𝐸0

𝐾

Then from equation 2 we get E = 𝑞−𝑞1

𝐴∈0

𝐸0

𝐾 =

𝑞

𝐴∈0 −

𝑞1

𝐴∈0

From equation 1 we get E0 = 𝑞

𝐴∈0 =>

𝑞

𝐾𝐴∈0=

𝑞

𝐴∈0 −

𝑞1

𝐴∈0

𝑞

𝐴 ∈0=

𝑞

𝐾𝐴 ∈0+

𝑞1

𝐴 ∈0

𝑞

𝐴= ∈0 (

𝑞

𝐾𝐴 ∈0 ) +

𝑞1

𝐴

Where 𝑞

𝐾𝐴∈0 = 𝐸 and

𝑞1

𝐴 = 𝑃

𝑞

𝐴= ∈0 𝐸 + 𝑃

q/A is the electric displacement D. Hence 𝑫 = ∈𝟎 𝑬 + 𝑷

This is the relation between D,E,P



The vectors are

1. The displacement D is connected with free charge only. The lines of D begins and

end on free charge.

2. The dielectric polarization P connected with polarization charge only. The lines

begin and end on polarization charges.

3. The electric intensity E is connected with all charges that are actually present where

free or polarization .E is reduced inside the dielectric

Boundary condition at the di-elctric surface

The rules governing the behavior of E and D at the boundary between two dielctrics are

known as boundary conditions.

Statements and Proof :

1. First Boundary Condition :-

The normal component of electric displacement D is the same on both sides of the

boundary of two media of different dielectrics

Proof :

Let AB represent a small portion of the boundary between two media of absolute

permitivities ε1 and ε2 as shown below. The media are the homogeneous and

isotropic. Consider a small surface area dS on the boundary, so that curvature may

be neglected. Suppose D1 and D2 be the electric induction vectors in the media on

either side of dS respectively. 𝜃1 and 𝜃2 are the angles with D1 and D2 makes with

the normal to dS.

To find the boundary condition for D, construct a small pill box shaped

surface wich intersects the boundary with its end faces parallel to it. The height h of



the pill box is assumed to be negligibly small in comparission with the diameter of

the box.

From gausses law ∮ 𝐸. 𝑑𝑆 = 𝑞

∈0 => ∈0 ∮ 𝐸. 𝑑𝑆 = 𝑞

∮ 𝐷. 𝑑𝑆 = 𝑞 -------------1

Let D1n be the average normal component of displacement vector D1 in medium 1

D2n be the average normal component of displacement vector D2 in medium 2

D2n is along inward normal.

By gausses law D1n dS - D2n dS = q ----------- 2

Q is the total charge enclosed by the surface. D2n and dS are negative.

D1n - D2n = 𝑞

dS = 𝜎 ------------ 3

Where 𝜎 is the charge density on the boundary of two dielectrics.

If the boundary is free from charge , then 𝜎 = 0

From equation 3 The displacement vector D depend of 𝜎

D1n - D2n = 0 => D1n = D2n --------- 4

The normal component of displacement vector is continuous across the charge free

boundary between two dielectrics.



2. Second Boundary Condition :-

The tangent component of the electrical intensities are the same on both sides of the

dialectics

Proof :

Consider a rectangle PQRS of small width and its length parallel to the boundary

separation two dielectrics as shown below. The rectangle lies with its longest sides

parallel to the boundary . Let E1 and E2 be the electric intensities in two media. The

direction of the electric field vector makes angle 𝜃1 and 𝜃2 with the normal to the

boundary.

The workdone in taking a unit charge around the rectangle PQRS must vanish.

= > ∮ 𝐸. 𝑑𝑙 = 0

E1 sin 𝜃1 𝑑𝑙 - E2 Sin 𝜃1 𝑑𝑙 = 0 ------1

We neglect the contribution of the short sides QR and SP are small

From equation 1 is modified as E1t - E2t = 0 => E1t = E2t

Thus , the tangential component of the field is continuous at the boundary of two dielctrics.



UNIT – III

1. Alternating Currents

Introduction :

An alternating current or a.c is defined as one which passes through a cycle of changes

at regular intervals. The wave front of a.c voltage or current is shown below.

In mathematically it can be represented as current is i=i0 Sin ωt

Voltage E=E0 Sin ωt

The current or voltage maximum value is i0 or E0

The variation of current and voltage in different components as follows

1) A.C through pure resistance

The circuit containing pure resistance R.

Let the applied voltage is E=E0 Sin ωt --1

By ohm’s law E = i R

E0 Sin ωt = i R

i= 𝐸0

𝑅 Sin ωt -------------2

i will be large when Sin ωt is unity i0 = 𝐸0

𝑅

i = i0 Sin ωt ------------3



comparing equation 1 and 3 we get the voltage and current are in phase with each

other as shown by the above figure (b) and (c)

2) A.C through pure Inductance only

Consider a circuit containing pure inductance only as shown below. Let apply an a.c

voltage E=E0 Sin ωt . then by the self inductance of the coil an emf is generated in

opposes the rise or fall of current through it. Then self inductance is

E = L 𝑑𝑖

𝑑𝑡

E0 Sin ωt = L 𝑑𝑖

𝑑𝑡 --------------1

di= 𝐸0

𝐿 Sin ωt dt

integrate on both side we get

i=∫𝐸0

𝐿 Sin ωt dt

i=𝐸0

𝐿ω(−Cos ωt)

= − 𝐸0

𝐿ω Cos ωt = −

𝐸0

𝐿ω Sin (ωt −

π

2 ) ------- 2

Maximum value of i is i0 = 𝐸0

𝐿ω

i=i0 Sin (ωt −π

2 ) ----------------- 3

E=E0 Sin ωt --------------- 4

comparing equation 3 and 4 we get current lags behind the voltage by π

2 or 900

as

shown above figure in (b) and (c)

in equation 2 ωL = XL , where XL is called inductive reactance.

Where ω = 2πf



3) A.C through pure Capacitance only

Consider a circuit consists of pure Capacitance as shown below. q is the charge on the

plates at any instant then ,

q=CE -----1

putting E = E0 Sinωt we get

q=CE0 Sinωt ----- 2

differentiating we get 𝑑𝑞

𝑑𝑡 =

d

dt(CE0 Sinωt ) = > i=ωCE0 Cos ωt

i=𝐸01

ωC

Cos ωt = 𝐸01

ωC

Sin (ωt +π

2 )

Maximum current will flow when Sin (ωt +π

2 ) is unity.

i0 = 𝐸01

ωc

= 𝐸0

𝑋𝐶

where Xc = 1

ωc = Capacitive reactance

Applied voltage E=E0 Sin ωt

i= i0 Sin (ωt +π

2 )

Comparing these two equations we get Current leads voltage by π

2 or 900 as shown in

figure (b) and (c).



A.C Circuit containing Inductance and resistance ( LR – Circuit )

Consider a circuit consists of a resistance R and inductance L in series with an

alternating e.m.f as shown below. Let I be the instantaneous value of current in the

circuit. By the changing the current an induced emf is set up and oppeses the applied

emf and is given by -L𝑑𝑖

𝑑𝑡 . the effective emf driving the current in the circuit is E-L

𝑑𝑖

𝑑𝑡.

= E0 Sin ωt -L𝑑𝑖

𝑑𝑡

According to ohms law this driving current is equal to Ri

E0 Sin ωt -L𝑑𝑖

𝑑𝑡 = Ri

L𝑑𝑖

𝑑𝑡+ Ri = E0 Sin ωt -------------1

This equation is a.c current with simple harmonically with the frequency same as applied

emf.

The solution of this equation is in the form i=i0 Sin (ωt- ∅ ) ------------ 2

Where i0 and ∅ are constants. These can be determined by substituting the value of i in

equation 1 we get

𝑑𝑖

𝑑𝑡 = i0 Cos (ωt- ∅ ). ω = >

𝑑𝑖

𝑑𝑡 = i0 ω Cos (ωt- ∅ ) ----2

Substituting these values in equation 1 we get

L𝑑𝑖

𝑑𝑡+ Ri = E0 Sin ωt ----- 3

Li0 ω Cos (ωt- ∅ ) + R i0 Sin (ωt- ∅ ) = E0 Sin ωt

Li0 ω Cos (ωt- ∅ ) + R i0 Sin (ωt- ∅ ) = E0 Sin (ωt- ∅ + ∅)



=>Li0 ω Cos (ωt- ∅ ) + R i0 Sin (ωt- ∅ )= E0 [ Sin (ωt- ∅) Cos∅ + Cos(ωt- ∅) Sin ∅ ]

= E0 Sin (ωt- ∅) Cos∅ + E0 Cos(ωt- ∅) Sin ∅

This equation is true for all values of t.

Equating the coefficient of Cos (ωt- ∅ ) on both sides we get

Li0 ω = E0 Sin ∅ ------------ 4

Equating the coefficient of Sin (ωt- ∅ ) on both sides we get

R i0 = E0 Cos∅ ----------- 5

Squring and adding the equation 4 and 5 we get

( Li0 ω )2 = ( E0 Sin ∅ ) 2

( R i0 ) 2

= ( E0 Cos∅ ) 2

( Li0 ω )2 + ( R i0 ) 2 = ( E0 Sin ∅ ) 2 + ( E0 Cos∅ ) 2

L2 i02 ω2 + R2 i0

2 = E02 Sin2 ∅ + E0

2 Cos2 ∅

i02 (L2 ω2 + R2 ) = E0

2 (Sin2 ∅ + Cos2 ∅ )

i02 (L2 ω2 + R2 ) = E0

2

i02 =

E02

(L2 ω2 + R2 ) => i0 =

𝐸0

√L2 ω2 + R2

i0 = 𝐸0

√L2 ω2 + R2

i0 = 𝐸0

√XL2 + R2

-----------6

where XL = ωL it is the inductive reactance of the Circuit

dividing equation 4 / 5 we get Tan ∅ = Li0 ω

𝑅 𝑖𝑜 =

L ω

𝑅

∅ = Tan -1 ( L ω

𝑅 ) -------------- 7



Substituting the value of i0 in equation 2 we get

i = 𝐸0

√L2 ω2 + R2 Sin (ωt- ∅ ) --------------- 8

The amplitude of current i0 = 𝐸0

√L2 ω2 + R2 and

The current lag behind the voltage by an angle ∅ = Tan -1 ( L ω

𝑅 ) = Tan -1 (

xL

𝑅 )

The impedance Z of the circuit = 𝐸0

𝑖0 = √ ω2L2 + R2

Vector Diagram :-

The vector diagram is plot between the voltage across resistance is remains in phase

with the current and voltage across inductance is lead by 90 0 in phase with current.

Let EL and ER be the magnitude of the voltage across the inductor L and resistor R

respectively. As current i is same in R and L

ER may be represent along the current line while EL at 900 a head to ER

as shown

below. Both ER and EL are right angle to each other. The vector OA represents the

magnitude and direction of voltage across R, while vector OB represents the

magnitude and direction of voltage across L. By parallegrom law the diagonal OR

represents the voltage across inductance and resistance in series.

E2 = EL2 + ER

2

i2Z2 = L2 i2 ω2 + R2 i2

Z2 = L2 ω2 + R2

Z = √L2 ω2 + R2

And Tan ∅ = 𝐸𝐿

𝐸𝑅 =

L ω

𝑅



A.C Circuit containing Capacitance and resistance ( CR – Circuit )

Consider a circuit consists of a resistance R and Capacitor C in series with an

alternating e.m.f as shown below. Let q be the Charge on the capacitor at any instant

t and I be the current in the circuit as shown below.

The potential difference across the capacitor at this instant is 𝑞

𝑐. This opposes the

original applied e.m.f .

The effective e.m.f in the circuit is E0 Sin ωt - 𝑞

𝑐

According to ohms law this is equal to R i

E0 Sin ωt - 𝑞

𝑐 = Ri

Ri +𝑞

𝑐 = E0 Sin ωt -------------------1

Differentiation this equation we get R 𝑑𝑖

𝑑𝑡 +

1

𝑐 𝑑𝑞

𝑑𝑡 = E0ω Cos ωt

R 𝑑𝑖

𝑑𝑡 +

𝑖

𝑐 = E0ω Cos ωt -----------------2 ( i=

𝑑𝑞

𝑑𝑡 )

The solution of the equation is i=i0 Sin (ωt- ∅ )

Where i0 and ∅ are constants and are determined

𝑑𝑖

𝑑𝑡 = i0 Cos (ωt- ∅ ). ω = >

𝑑𝑖

𝑑𝑡 = i0 ω Cos (ωt- ∅ ) ----3

The equation 2 is modified as

R i0 ω Cos (ωt- ∅ ) +

𝑖0

𝐶 Sin (ωt- ∅ ) = E0ω Cos ωt



R i0 ω Cos (ωt- ∅ ) +

𝑖0

𝐶 Sin (ωt- ∅ ) = E0ω

Cos (ωt- ∅ + ∅)

= E0ω [Cos (ωt- ∅) Cos ∅ - Sin (ωt- ∅) Sin ∅ ]

= E0ω Cos (ωt- ∅) Cos ∅ - E0ω Sin (ωt- ∅) Sin ∅

Equating coefficient of Cos (ωt- ∅ ) Terms on both sides we get

R i0 ω = E0ω Cos ∅ -----------4

Equating coefficient of Sin (ωt- ∅ ) Terms on both sides we get

𝑖0

𝐶 = - E0ω Sin ∅ -----------5

Squaring and adding equation 4 and 5 we get

(R i0 ω )2 + (

𝑖0

𝐶 )2 = E0

2 ω

2 Cos2 ∅ + E02 ω

2 Sin2 ∅

i02 [ R2 ω 2 +

1

𝑐2 ] = E0

2 ω

2 --------------- 6

i02 ω 2 [ R2 +

1

ω2 𝑐2 ] = E0

2 ω

2

i02 [ R2 +

1

ω2 𝑐2 ] = E0

2

i02 =

E02

(1

c2 ω2 + R2 )

i0 = E0

√(1

c2 ω2 + R2 ) => i0 =

𝐸0

√Xc2 + R2

--------------7

dividing equation 5 / 4 we get Tan ∅ = i0 ω

𝐶𝑅 𝑖𝑜 ω =

1

𝑅ωC

∅ = Tan -1 ( 1

𝑅ωC ) = Tan -1 (

Xc

𝑅 ) -------------- 7

Substituting the value of i0 in the above equation we get

i = E0

√(1

c2 ω2 + R2 ) Sin (ωt- ∅ ) --------------- 8



The amplitude of current i0 = E0

√(1

c2 ω2 + R2 ) and

The current lag behind the voltage by an angle ∅ = Tan -1 ( 1

𝑅ωC ) = Tan -1 (

Xc

𝑅 )

The impedance Z of the circuit = 𝐸0

𝑖0 = √

1

c2 ω2 + R2

Vector Diagram

The vector diagram can be plotted by considering the voltage across the resistance always

remains constant in phase with the current but the voltage across condenser lags behind by

900 . Let ER and EC be the magnitude of the voltage across the resistance R and condenser

C respectively . Then

ER = R I and Ec = i Xc = 𝑖

ω𝑐

This is shown in OA and OB. The resultant vector OR represents the voltage E.

E2 = ER2 + EC

2 ----------9

Where E = iZ , where Z is the impedance of the circuit

(Iz)2 = ( R i )2 + ( 𝑖

ω𝑐 )2

Z2 = R2 + ( 1

ω𝑐 ) 2 = R2 + (

1

ω2𝑐2 )

Z =√R2 + ( 1

ω2𝑐2 )

Tan ∅ = 𝐸𝐶

𝐸𝑅 =

1

𝑅 ωC => ∅ = Tan -1 (

Xc

𝑅 )



LCR Series circuit

Consider a circuit consisting an Inductor , resister and Capacitor are connecting in series as

shown below. An alternating e.m.f is applied to the circuit E = E0 Sin ωt . Let I be the

current in the circuit at any instant t and q is the charge on the capacitor at that instant.

The potential across the capacitor is 𝑞

𝑐

The emf due to the inductance is -L 𝑑𝑖

𝑑𝑡

According to ohms law E0 Sin ωt -L 𝑑𝑖

𝑑𝑡 -

𝑞

𝑐 = Ri

L 𝑑𝑖

𝑑𝑡 +

𝑞

𝑐 + Ri = E0 Sin ωt

Differentiating the above equation with respect to ‘t ‘ we get

L 𝑑2𝑖

𝑑𝑡2 +

1

𝑐 𝑑𝑞

𝑑𝑡 + R

𝑑𝑖

𝑑𝑡 = E0 ω Cos ωt

L 𝑑2𝑖

𝑑𝑡2 + R

𝑑𝑖

𝑑𝑡 +

1

𝑐 𝑑𝑞

𝑑𝑡 = E0 ω Cos ωt

L 𝑑2𝑖

𝑑𝑡2 + R

𝑑𝑖

𝑑𝑡 +

𝑖

𝐶 = E0 ω Cos ωt --------------------1

The solution of the above equation is is i=i0 Sin (ωt- ∅ )

Where i0 and ∅ are constants and are determined

𝑑𝑖

𝑑𝑡 = i0 Cos (ωt- ∅ ). ω = >

𝑑𝑖

𝑑𝑡 = i0 ω Cos (ωt- ∅ )

𝑑2𝑖

𝑑𝑡2 = - i0 ω

2 Sin (ωt- ∅ )



Substituting these values in equation 1 we get

L 𝑑2𝑖

𝑑𝑡2 + R

𝑑𝑖

𝑑𝑡 +

𝑖

𝐶 = E0 ω Cos ωt

-L i0 ω2 Sin (ωt- ∅ ) + R i0 ω Cos (ωt- ∅ ) +

1

𝐶 ( i0 Sin (ωt- ∅ ) ) = E0 ω Cos (ωt- ∅ + ∅ )

= E0ω [Cos (ωt- ∅) Cos ∅ - Sin (ωt- ∅) Sin ∅ ]

= E0ω Cos (ωt- ∅) Cos ∅ - E0ω Sin (ωt- ∅) Sin ∅ -----------2

Equating the coefficient of Cos (ωt- ∅) terms on both sides we get

R i0 ω = E0ω Cos ∅ ---------------3

Equating the coefficient of Sin (ωt- ∅) terms on both sides we get

-L i0 ω2 +

𝑖0

𝐶 = - E0ω Sin ∅ ------------4

Squaring and adding the equation 3 and 4 we get

( -L i0 ω2 +

𝑖0

𝐶 ) 2 + R2 i0

2 ω2 = E0 2 ω2

i02 [ ( -L ω2 +

1

𝐶 )2 + R2 ω2 ] = E0

2 ω2

i02 ω2 [ ( L ω -

1

ω𝐶 )2 + R2 ] = E0

2 ω2

i02 [ ( L ω -

1

ω𝐶 )2 + R2 ] = E0

2

i02 =

E02

[ ( L ω − 1

ω𝐶 )2 + R2 ]

=> i0 = 𝐸0

√( L ω − 1

ω𝐶 )2 + R2

i0 = 𝐸0

√R2 + ( L ω − 1

ω𝐶 )2

------------------ 5

Dividing the equation 4 by equation 5

Tan ∅ = - −L i0 ω2 +

𝑖0𝐶

R i0 ω

Tan ∅ = - −L ω2 +

1

𝐶

R ω =

L ω2− 1

𝐶

R ω = ω ( L ω −

1

ω𝐶 ) =

ω ( L ω− 1

ω𝐶 )

Rω



Tan ∅ = ( L ω−

1

ω𝐶 )

R =

𝑋𝐿− 𝑋𝐶

𝑅

The current in the circuit is i=i0 Sin (ωt- ∅ )

i= 𝐸0

√R2 + ( ω L − 1

ω𝐶 )2

Sin (ωt- ∅ )

this equation shows

1) The maximum current i0 = 𝐸0

√R2 + ( L ω − 1

ω𝐶 )2

2) The impedance of the of the circuit Z = √R2 + ( ωL − 1

ω𝐶 )2

3) The current lags in phase from the emf by an angle ∅

Tan ∅ = ( L ω−

1

ω𝐶 )

R =


𝑅

When ωL > 1/ωc then ∅ is positive i.e current lags behind the applied voltage

When ωL = 1/ωc then ∅ is 0 i.e current is in phase with applied voltage

When ωL < 1/ωc then ∅ is Negative i.e current leads the applied voltage

Vector Diagram

The vector diagram of LCR circuit is shown below.

The potential across the resistance ER is represented by the vector OA = Ri0 .

The potential across the Inductance EL = ωLi0 . it is 900 lags the voltage and is

antiphase to the Ec .

The resultant E = EL - Ec

This represents the OR

E02 = (OR)2 = i0

2 [ R2 + ( ω L - 1

ω𝐶 )2 ]

(OR) = i0 √R2 + ( ω L − 1

ω𝐶 )2

Impedence Z = E0

𝑖0 = √R2 + ( ω L −

1

ω𝐶 )2

From figure Tan ∅ = ( L ω−

1

ω𝐶 )

R =


𝑅



The value of ∅ is lead or lag depending on the value of ω L and ω c

Series resonance

In the LCR series circuit the maximum current is i0 = 𝐸0

√R2 + ( L ω − 1

ω𝐶 )2

The impedence of the circuit is Z = √R2 + ( ω L − 1

ω𝐶 )2

In this very large capacitive reactance at low frequencies and a very large inductive

reactance at high frequencies. At a particular frequency the reactance is zero. This

frequency is called resonance frequency. At resonance frequency the impedance is

minimum.

At resonance frequency ω L − 1

ω𝐶 = 0 => ω L =

1

ω𝐶

ω2 = 1

𝐿𝐶 => (2𝜋𝑓0)2 =

1

𝐿𝐶 => 𝑓0 =

1

2𝜋√𝐿𝐶

Where 𝑓0 is the resonance frequency. This equation gives the resonance frequency depends

on the values of L and C and does not depend on R

The variation of Current with frequency

is shown below.

This graph drawn with different resistance values.

In this the current increases slowly , then goes to maximum at resonant frequency f and

finally decreases.

When R is low the peak is high . this is the sharpness of resonance.



Quality factor

Definition :- The quality factor is defined as 2π times the ratio of the energy stored to the

average energy loss per period.

Q = 2π 𝐸𝑛𝑒𝑟𝑔𝑦 𝑠𝑡𝑜𝑟𝑒𝑑

𝐸𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑠 𝑝𝑒𝑟 𝑝𝑒𝑟𝑖𝑜𝑑

Explanation :- the energy stored between the plates of the capacitor with voltageV0 is

1

2CV0

2. The energy is stored in magnetic field around the inductor with current i0 is

1

2Li0

2. The energy is loss through resistance is

1

2 𝑅i0

2

The quality factor is Q = 2π 𝐸𝑛𝑒𝑟𝑔𝑦 𝑠𝑡𝑜𝑟𝑒𝑑

𝐸𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑠 𝑝𝑒𝑟 𝑝𝑒𝑟𝑖𝑜𝑑 = 2π 𝑓

𝐸𝑛𝑒𝑟𝑔𝑦 𝑠𝑡𝑜𝑟𝑒𝑑

𝐸𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑠 𝑝𝑒𝑟 𝑆𝑒𝑐𝑜𝑛𝑑

Q = 2π 𝑓 1

2Li0

2

1

2 𝑅i0

2 = 2π 𝑓

𝐿

𝑅 =

ω L

𝑅

Q = ω L

𝑅 for inductance

Q = 1

ωC𝑅 for Capacitance is used.

The quality factor is defined as the ratio of reactance of either inductance of capacitance at

the resonant frequency to the circuit.

When XL = Xc the value of Q is same.

Sharpness of resonance : is is defined as the rapidly with which the current falls from

its resonant value with the change in applied frequency.

LCR - PARALLEL RESONANT CIRCUIT

A LCR parallel resonant circuit is shown below. An inductance L and resistance r are

connected in series in one branch and a capacitor is another bronch. An alternating e.m.f is

connected in the circuit. The current from the generator is i0 . the distribution of current in

two branches is shown below.



From kirchoff’s first law i0= i1 + i2

Z is the impedance of the circuit .

The impedance in the resistor and inductance branch is Z1 = R + jωL

The impedance in the capacitor branch Z2 = jωC

The two branches are in parallel . so the resultant impedance is

1

𝑍 =

1

𝑍1 +

1

𝑍2

1

𝑍 =

1

R + jωL +

11

jωC

1

𝑍 =

1

R + jωL + jωC

The admittance Y = 1

𝑍 =

1

R + jωL + jωC

Y = 1

R + jωL∗

R − jωL

R − jωL+ jωC

Y = R − jωL

(R + jωL ) ( R – jωL)+ jωC

Y = R − jωL

( R2 –(jωL) 2)+ jωC

Y = R − jωL

R2+ω2L2+ jωC

Y = R

R2+ω2L2−

jωL

R2+ω2L2+ jωC



Y = R

R2+ω2L2− j(

ωL

R2+ω2L2+ ωC )

The equation consists of two terms . one is real and another is imaginary.

The magnitude of admittance is

| Y | = ((R

R2+ω2L2 )2 + ( ωC −

ωL

R2+ω2L2)2 )

1

2

The impedance is maximum or the admittance is minimum at particular frequency.

ωC = ωL

R2 + ω2L2

C = L

R2 + ω2L2

C (R2 + ω2L2 ) = L => R2 + ω2L2 = 𝐿

𝐶 => ω2L2 =

𝐿

𝐶 - R2

ω2 = 𝐿

𝐶L2 -

R2

L2 => ω2 =

1

𝐿𝐶 -

R2

L2

ω = √1

𝐿𝐶 −

R2

L2 => 2πf = √

1

𝐿𝐶 −

R2

L2

f = 1

2𝜋 √

1

𝐿𝐶 −

R2

L2 . this frequency is called resonance frequency

the variation of current with frequency is shown below.



At resonance frequency the value of admittance is of the circuit is Y = 𝑅

R2+ω2L2

Z = R2+ω2L2

𝑅

Where R 0 , the impedence Z ∞. The variation of impedence with frequency is ω is

shown above.

SERIES RESONANCE VERSUS PARALLEL RESONANCE

Series Resonance Parallel Resonance

Series resonance frequency is

𝑓0 = 1

2𝜋√𝐿𝐶

Parallel Resonance frequency is

f = 1

2𝜋 √

1

𝐿𝐶 −

R2

L2

At resonance the power factor is

unity and impedance is purely

resistive Z = R

Power factor is also unity but impedance

is given by Z = L/CR

At resonance the current is

maximum

At resonance the current is maximum

At resonance the impedance of the

circuit is minimum

At resonance the impedance of the circuit

is minimum

The circuit is called as acceptor The circuit is called as rejector

Power in Alternating Currents

The power in an electric circuit is the rate at which electrical energy is consumed in the

circuit . Let an e.m.f E= E0 sinωt be applied to a circuit. The current in the circuit

i=i0 Sin (ωt- ∅ ) where ∅ is the phase difference between A.C voltage and Current.

The instantaneous power is given by

E i= E0 i0 sinωt Sin (ωt- ∅ )

Average Power P = ∫ 𝐸 𝑖 𝑑𝑡

𝑇

0

∫ 𝑑𝑡𝑇

0

= 1

𝑇 ∫ 𝐸0

𝑇

0𝑖0 Sin ωt Sin (ωt − ∅ ) dt



= 1

𝑇 ∫

E0 i0

2

𝑇

0 [ Cos (ωt − ωt + ∅ ) − Cos (ωt + ωt − ∅ ) ] dt

= 1

𝑇 E0 i0

2 [ ∫ cos ∅ 𝑑𝑡

𝑇

0 - ∫ cos( 2ωt − ∅ ) 𝑑𝑡

𝑇

0

P = 1

𝑇 E0 i0

2 [ t cos ∅ ] 𝑇

0- [

𝑆𝑖𝑛(2ωt−∅ )

2ωt ]

𝑇0

P = 1

𝑇 E0 i0

2 [ T cos ∅ ] 𝑇

0- [

𝑆𝑖𝑛(2ωt−∅ )

2ω ]

𝑇0

P = 1

𝑇 E0 i0

2 [ T cos ∅ ]-

1

2ω[ 𝑆𝑖𝑛 (2

2𝜋

𝑇 T − ∅ ) − 𝑆𝑖𝑛 ∅ ]

P = 1

𝑇 E0 i0

2 [ T cos ∅ ] - 0

P = E0 i0

2 cos ∅ =

𝐸0

√2

𝑖0

√2 cos ∅ = Erms x irms cos ∅

P = Erms x irms cos ∅

True power = Apparent power x Power factor

cos ∅ is known as Power factor.

When the power consumed in the circuit is Zero , the current is said to be Wattless. This is

possible when the power factor is zero

𝐶𝑜𝑠 ∅ = 0 or ∅ = 𝜋

2 When the current and voltage are differ in phase by

𝜋

2.



UNIT – III

2 . MAXWELL’S EQUATIONS IN ELECTROMAGNETIC WAVES

The basic laws of electricity and magnetism

The basic equations in electricity and magnetism are

1. Gauss’s law in electrostatics are ∮ 𝐸. 𝑑𝑠 = 𝑞

∈0

2. Gauss’s law in Magnetism are ∮ 𝐵. 𝑑𝑠 = 0

3. Faraday’s law of electromagnetic induction ∮ 𝐸. 𝑑𝑙 = −𝑑∅𝑏

𝑑𝑡

4. Amphere’s law for magnetic field is ∮ 𝐵. 𝑑𝑙 = 𝜇0 𝑖

Maxwell equations in differential form

1. ∇. 𝐸 = 𝜌

𝜖0

2. ∇. 𝐵 = 0

3. ∇ x 𝐸 = −𝜕𝐵

𝜕𝑡

4. ∇ x 𝐵 = 𝜇0 ( 𝐽 + 𝜖0 𝜕𝐸

𝜕𝑡 )

Maxwell’s wave equation of Electromagnetic equation

Consider a maxwell’s electromagnetic wave equation to a homogeneous , isotropic

dielectric medium. In isotropic medium is one which infinite resistance to the current and

then conductivity is Zero ( σ = 0 ). In homogeneous medium there is no volume

distribution of charge. i.e charge density ρ = 0. There fore J =0 , ρ = 0

1. We that mexwells equations are ∇. 𝐸 = 𝜌

𝜖0

2. ∇. 𝐵 = 0

3. ∇ x 𝐸 = −𝜕𝐵

𝜕𝑡

4. ∇ x 𝐵 = 𝜇0 ( 𝐽 + 𝜖0 𝜕𝐸

𝜕𝑡 )

Maxwells equations for Dielectric medium is

∇. 𝐸 = 0 ----------1

∇. 𝐵 = 0 -----------------2



∇ x 𝐸 = −𝜕𝐵

𝜕𝑡 -------------3

∇ x 𝐵 = 𝜇 ∈ 𝜕𝐸

𝜕𝑡 -----------------4

The equation of propagation in dielectric medium is

Elimination equation B from equations 3 and 4 we get

From equation 3 is ∇ x 𝐸 = −𝜕𝐵

𝜕𝑡

Apply curl on both sides we get

∇ x ∇ x 𝐸 = ∇ x −𝜕𝐵

𝜕𝑡

∇ ( ∇ . E ) − ∇2 E = –𝜕( ∇ x 𝐵 )

𝜕𝑡

From equation 1 , ∇ . E = 0 and from equation 4 is ∇ x 𝐵 subtituted

0 − ∇2 E = –𝜕

𝜕𝑡 (𝜇 ∈

𝜕𝐸

𝜕𝑡 )

- ∇2 E = - 𝜇 ∈ 𝜕2𝐸

𝜕𝑡2 )

∇2 E = 𝜇 ∈ 𝜕2𝐸

𝜕𝑡2 ) --------------------5

From equation 4 is ∇ x 𝐵 = 𝜇 ∈ 𝜕𝐸

𝜕𝑡

Curl on both sides we get ∇ x ∇ x 𝐵 = ∇ x 𝜇 ∈ 𝜕𝐸

𝜕𝑡

∇ ( ∇ . B ) − ∇2 B = 𝜇 ∈ 𝜕 (∇ x𝐸 )

𝜕𝑡

From equation 2 ∇ . B = 0 and from equation 3 substitute ∇ x𝐸 value above we get

0 − ∇2 B = 𝜇 ∈ 𝜕

𝜕𝑡 (−

𝜕𝐵

𝜕𝑡 )

− ∇2 B = - 𝜇 ∈ 𝜕2𝐵

𝜕𝑡2

∇2 B = 𝜇 ∈ 𝜕2𝐵

𝜕𝑡2 ------------------- 6

The equations 5 and 6 are the electromagnetic wave equations.



We know the general wave equation ∇2𝜑 = 1

𝑣2 𝜕2𝜑

𝜕𝑥2 -------7

Whare v is the velocity of the wave

Comparing equations 6 and 7 we get

1

𝑣2 = 𝜇 ∈ => v2 =

1

𝜇∈ => v =

1

√𝜇∈

Where 𝜇 and ∈ are permeability and permittivity of the medium

The electric wave and magnetic wave travel with the velocity v = 1

√𝜇∈

These waves are electromagnetic waves.

For free space v = 1

√𝜇0∈0

𝜇0 = 4π x 10 -7 and ε0 = 8.9 x 10-12

Then v = 1

√4π x 10 −7 x 8.9 x 10−12 = 3 x 108 m/s

The velocity of E and B are traveling in velocity of light.

Transverse nature of Electromagnetic waves

Consider an electromagnetic wave which the Component of vectors E and B vary

with one coordinate i.e in ‘ x ‘ and also in time t.

E( x , t ) and B( x , t ) are in homogeneous and isotropic medium

From maxwell’s First equations in that medium ∇. 𝐸 = 0

𝜕𝐸𝑥

𝜕𝑥+

𝜕𝐸𝑦

𝜕𝑦 +

𝜕𝐸𝑧

𝜕𝑧 = 0

𝜕𝐸𝑥

𝜕𝑥 = 0

𝐸𝑥 = Constant -----------------------1

From maxwell’s Second equations in that medium ∇. 𝐵 = 0

𝜕𝐵𝑥

𝜕𝑥+

𝜕𝐵𝑦

𝜕𝑦 +

𝜕𝐵𝑧

𝜕𝑧 = 0



𝜕𝐵𝑥

𝜕𝑥 = 0

𝐵𝑥 = Constant -----------------------2

From equations 1 and 2 the derivation of E and B with respect to Y and Z are Zero.

From Maxwell’s Third equation ∇ x E = −𝜕𝐵

𝜕𝑡

[

𝑖 𝑗 𝑘𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝐸𝑥 𝐸𝑦 𝐸𝑧

] = −𝜕

𝜕𝑡 [ 𝑖 𝐵𝑥 + 𝑗 𝐵𝑥 + 𝑘 𝐵𝑥 ]

i [ 𝜕𝐸𝑧

𝜕𝑦−

𝜕𝐸𝑦

𝜕𝑧 ] = - i

𝜕𝐵𝑥

𝜕𝑡 ---------- 3

i [ 𝜕𝐸𝑧

𝜕𝑦−

𝜕𝐸𝑦

𝜕𝑧 ] = 0

.𝜕𝐸𝑧

𝜕𝑦−

𝜕𝐸𝑦

𝜕𝑧 = 0

From Equation 3 𝜕𝐵𝑥

𝜕𝑡 = 0 => 𝐵𝑥 = Constant

From Maxwells Fourth relation ∇ x 𝐵 = 𝜇 ∈ 𝜕𝐸

𝜕𝑡

[

𝑖 𝑗 𝑘𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝐵𝑥 𝐵𝑦 𝐵𝑧

] = 𝜇 ∈𝜕

𝜕𝑡 [ 𝑖 𝐸𝑥 + 𝑗 𝐸𝑥 + 𝑘 𝐸𝑥 ]

i [ 𝜕𝐵𝑧

𝜕𝑦−

𝜕𝐵𝑦

𝜕𝑧 ] = 𝜇 ∈

𝜕𝐸𝑥

𝜕𝑡 ----------4

i [ 𝜕𝐵𝑧

𝜕𝑦−

𝜕𝐵𝑦

𝜕𝑧 ] = 0

.𝜕𝐵𝑧

𝜕𝑦−

𝜕𝐵𝑦

𝜕𝑧 = 0

From equation 4 𝜕𝐸𝑥

𝜕𝑡 = 0

𝐸𝑥 = Constant --------5



From the above equations we get Ex and Bx are constants as regarding to time and

space.

E = j Ex + k Ez

B = j Bx + k Bz

As E and B do not contain x-component and X-direction being the direction of the

propagating wave . so both these vectors are perpendicular to the direction of the

propagation of the wave.

So the Electromagnetic waves are purely in transverse nature.

POYNTING THEOREM

Statement :- The amount of field energy passing through unit area of the surface

perpendicular to the direction of propagation of energy is called as pointing vector. It is

denoted by P.

The plane electromagnetic wave E and B are perpendicular to each other then the pointing

vector P = 1

𝜇0 ( E x B ) or E X H

Derivation

Consider an elementary volume in the form of rectangular parallelepiped of

sides dx , dy and dz as shown below. The volume of the parallelepiped of sides dx dy dz.

Suppose the electromagnetic energy is propagated along the X-axis. The area

perpendicular to the direction of propagation of energy is dx dy dz. The electromagnetic

energy in this volume is U. the rate of change of energy is 𝜕𝑈

𝜕𝑡 .

𝜕𝑈

𝜕𝑡 = − ∮ 𝑃 . 𝑑𝑆 ----------1

The Negative sign in the above equation is energy entering in the volume. So

∮ 𝑃 . 𝑑𝑆 = - 𝜕𝑈

𝜕𝑡 ---------- 2

The energy density per unit volume in Electric field E is given by

𝑈𝐸 = 1

2 ∈0 𝐸2 ------------ 3



The energy density per unit volume in Magnetic field E is given by

𝑈𝐵 = 1

2 𝜇0𝐻2 ------------- 4

The total energy U = 𝑈𝐸 + 𝑈𝐵 U = 1

2 ∈0 𝐸2 +

1

2 𝜇0𝐻2

The rate of decrease of energy in volume dV is given by

− 𝜕

𝜕𝑥 (

1

2 ∈0 𝐸2 +

1

2 𝜇0𝐻2 ) dV

Rate of decrease of energy for volume V

− 𝜕𝑈

𝜕𝑡 = -

𝜕

𝜕𝑡 ∫(

1

2 ∈0 𝐸2 +

1

2 𝜇0𝐻2 ) dV

= - 𝜕

𝜕𝑡 ∫(

1

2 ∈0 𝐸 . 𝐸 +

1

2 𝜇0 𝐻 . 𝐻) dV

= ∫ − [ ∈0 𝐸 ( 𝜕𝐸

𝜕𝑡 ) + 𝜇0𝐻 (

𝜕𝐻

𝜕𝑡 ) ] dV ----------------- 5

From maxwell’s 4th equations for non conducting medium

∇ x 𝐻 = ∈0 𝜕𝐸

𝜕𝑡 ( B = 𝜇0H )

. 𝜕𝐸

𝜕𝑡 =

∇ x 𝐻

∈0 -----------------6

From maxwell’s 3rd equations ∇ x E = − 𝜕𝐵

𝜕𝑡 = - 𝜇0

𝜕𝐻

𝜕𝑡

𝜕𝐻

𝜕𝑡 =

∇ x E

𝜇0 --------------- 7

Substituting the equation (6) and (7) values in equation (5) we get

− 𝜕𝑈

𝜕𝑡 = ∫ − [ ∈0 𝐸 (

∇ x 𝐻∈0

) − 𝜇0𝐻 ( ∇ x E

𝜇0 ) ] dV

Equation (6) can be written as

− 𝜕𝑈

𝜕𝑡 = ∫ [ 𝐻. ( ∇ x 𝐸 ) – 𝐸 . ( ∇ x 𝐻 ) ] dV

− 𝜕𝑈

𝜕𝑡 = ∫ ∇ . ( 𝐄 X 𝐇) 𝑑𝑉



Using Gauss theorem of divergence the Volume integral is converted into surface integral

Then ∫ ∇ . ( 𝐄 X 𝐇) 𝑑𝑉 = ∫(𝐄 X 𝐇). 𝑑𝑆 --------- 8

− 𝜕𝑈

𝜕𝑡 = ∫(𝐄 X 𝐇). 𝑑𝑆 ----------9

Comparing equation (2) and (9) we get

∮ 𝑃 . 𝑑𝑆 = ∫(𝐄 X 𝐇). 𝑑𝑆

P = 𝐄 X 𝐇

This vector gives the energy flows takes place in a direction perpendicular to the plane

containing E and H .

Production and Detection of Electromagnetic waves

(Hertz Experiment )

Production :

Consider Two metal square plates A and B which are placed at a certain distance.

These plates acts as a capacitor plates. The opposite faces of A and B are connected to

highly polished spheres C and D through thick wire.



An induction coil is connected across the wire attached to the plates of the capacitor .

So the capacitor plates may be charged to a high potential .In this A and B acts as

Capacitor and thick wire acts as inductor. This circuit acts as oscillatory LC circuit and

produce electromagnetic waves.

When the D.C Voltage supplied by induction coil becomes high , the air gap

between C and D gets ionized and the gap become conducting. The electrical discharge

takes place between the two plates A and B. Due to the discharge , the voltage across the

capacitor plate A and B falls down and conduction process in air gap stop. The process is

repeated again and again every time and continuous series E.M waves is produced.

Detection

Hertz gives a thick wire ring connected to polished brass sphere E and F with a small gap

as shown above in figure (b). The brass spheres serve as the plates of a capacitor and

connecting wires as inductor. This is an oscillatory LC circuit and the frequency is

f= 1

2𝜋 √𝐿𝐶

When the electromagnetic waves passing through this circuit , an alternating e.m.f is

induced around the wire. When the natural frequency of the detector is equal to the

frequency of E.M waves produced spark between C and D, resonance occur. The

amplitude of the oscillations of the detector circuit increases. Then a spark is produced in

the air gap, the E.M Waves are detected.



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