Jaypee Institute ofJaypee Institute of Information Information Technology University,Technology University,

NoidaNoida

Department of Physics and materials Science and Engineering

Coulombs Law Like charges repel, unlike charges attract.

The electric force acting on a point charge q1 as a result of

the presence of a second point charge q2 is given by

Coulomb's Law:

where 0 = permittivity of space

Gauss's Law • The total of the electric flux out of a closed surface

is equal to the charge enclosed divided by the permittivity.

• Electric Flux over a Closed Surface = Charge enclosed by the Surface divided by o.

o = the permittivity of free space 8.854x10-12 C2/(N m2) = 1/4k)

.• A more intuitive statement: the total number of electric

field line entering or leaving a closed volume of space is directly proportional to the charge enclosed by the volume

• If there is no net charge inside some volume of space then the electric flux over the surface of that volume is always equal to zero.

Electric Flux: The Electric Flux E is the product of Component of the

Electric Field Perpendicular to a Surface times the Surface Area.

Laplace's and Poisson's Equations

• A useful approach to the calculation of electric potentials is to relate that potential to the charge density which gives rise to it. The electric field is related to the charge density by the divergence relationship

and the electric field is related to the electric potential by a gradient relationship

• Therefore the potential is related to the charge density by Poisson's equation

This is Poisson's equation.• In a charge-free region of space, this becomes

Laplace's equation

Potential of a Uniform Sphere of Charge

• Since the potential is a scalar function, this approach has advantages over trying to calculate the electric field directly. Once the potential has been calculated, the electric field can be computed by taking the gradient of the potential.

The use of Poisson's and Laplace's equations will be explored for a uniform sphere of charge. In spherical polar coordinates, Poisson's equation takes the form:

Biot-Savart Law• Currents, i.e. moving electric charges, produce magnetic

fields.  There are no magnetic charges• The Biot-Savart Law relates magnetic fields to the currents

which are their sources. In a similar manner, Coulomb's law relates electric fields to the point charges which are their sources.

• where µ0 is the permeability constant

The Biot-Savart Law is used to calculate the magnetic field at a given position. It is usually only practical to do the calculation in special cases where some symmetry makes the problem simpler.

Ampere's Law• Ampere's law allows us to write down a single

equation that describes all of the ways that electric current can make magnetic field. But, just as with Gauss's law, this single equation is very difficult to solve. Ampere's law says: The path integral of around any (imaginary) closed path is equal to the

current enclosed by the path, multiplied by

When we discussed Gauss's law, we noted that the law was true no matter how distorted the surface or how complicated the electric field. Similarly, Ampere's law is always true,

• magnetic field around an infinitely long wire

But B has the same value a distance a away from the rod and hence

Maxwell's Equations

Gauss' law in Electrostatics (Source of E)

Faraday's law (Source of E)

Ampere-Maxwell’s law (Source of B)

Gauss' law in Magnetostatics

.dl

.dl

Differential form of Maxwell’s Equations

Gauss' law

Faraday's law

Ampere's law

Ampere’s Law (constant currents):

B dl Ienc z 0Ampere’s Law for constant currents.

• What about currents which are not continuous?•

The capacitor holds a charge Q over the two plates. How can there be a current emerging from the capacitor plates?

I

PHYS 102Ampere’s Law (continuous currents?):

B dl Ienc z 0Ampere’s Law for constant currents.

I

+Q charge deposits on the plate

E-field increasing as Q increases!

-Q charge induced by E-field

I

PHYS 102Ampere’s Law (modification?):

E dld

dtB dA

z ze j Faraday’s

Law

A changing magnetic flux creates an electric field!!!

B dl Ienc z 0

Ampere’s Law for constant currents.

Scottish physicist James Clerk Maxwell suggests that a changing electric flux creates a magnetic field!!!

PHYS 102Ampere’s Law (modification?):

B dl Ienc z 0 z 0 0

d

dtE dA Modified Ampere’s Law.

0

d

dtE dA z

Is termed the displacement current.

Id

B dl I Ienc d z 0 ( )

PHYS 102Modified Ampere’s Law

B dl Ienc z 0 z 0 0

d

dtE dA

To see how the displacement current comes about, one has to consider the electric flux through the capacitor’s plate (Gauss’s Law).

E dA

QQ E dAencenc z z

0

0

Q increases on the capacitor, the electric flux also increases at the same rate.

dQ

dt

d

dtE dA Ienc

d z 0

PHYS 102Modified Ampere’s Law:

I I

What does B-field look like in this region?

B dl Ienc z 0 z 0 0

d

dtE dA

PHYS 102Modified Ampere’s Law:

I

What does B-field look like in this region?

• • • •• •• • • •

• • • •• •• • • •• •• • • •• •• • • •• •• • • •

E-field increasing

B-field forms concentric rings with direction given as shown.

B dl Ienc z 0 z 0 0

d

dtE dA

Electromagnetic Waves • Transmission of energy through a vacuum or using no medium is

accomplished by electromagnetic waves, caused by the osscilation of electric and magnetic fields.

• They move at a constant speed of 3x108 m/s. Often, they are called electromagnetic radiation, light, or photons

• Light is not the only example of an electromagnetic wave. Other electromagnetic waves include the microwaves and the radio waves that are broadcast from radio stations.

• An electromagnetic wave can be created by accelerating charges; moving charges back and forth will produce oscillating electric and magnetic fields, and these travel at the speed of light.

• It would really be more accurate to call the speed "the speed of an electromagnetic wave", because light is just one example of an electromagnetic wave.

Reflection and Transmission at Normal incidence

0 1

0 11

ˆ( , ) exp

1 ˆ( , ) exp

I I

I I

E x t E i k x t j

B x t E i k x t kv

��������������

��������������

Suppose yz plane forms the boundary between two linear media. Aplane wave of frequency ω travelling in The x direction (from left) and polarizedalong y direction, approaches the interface from left (see figure)

In medium 1 following reflected wave travels back

0 1

0 11

ˆ( , ) exp

1 ˆ( , ) exp

R R

R R

E x t E i k x t j

B x t E i k x t kv

��������������

�������������� -sign in BR is because Poynting vector mustaim in the direction of propagation

Reflection and Transmission at Normal incidence-2

0 2

0 22

ˆ( , ) exp

1 ˆ( , ) exp

T T

T T

E x t E i k x t j

B x t E i k x t kv

��������������

��������������

In medium 2 we get a transmitted wave:

At x=o the combined fields to the left

EI+ER and BI+BR, must join the fields to the right ET and BT in accordance to the boundary condition. Since there are no components perpendicular to the surface so boundary conditions (i) and (ii) are trivial. However last two yields:

0 0 0

0 0 01 1 1 2 2

1 1 1 1 1

R I T

I R T

E E E

E E Ev v v

Reflection and Transmission at Normal incidence-3

0 0 0

1 1 1 2 1 2

2 2 2 1 2 1

0 0 0 0

1r

2

2 1 20 0 0 0

2 1 2 1

,

where

1 2and

1 1

vIf 1(nonmagnetic media) then =

v

2thus we have, and

I R T

R I T I

R I T I

or E E E

v n

v n

E E E E

v v vE E E E

v v v v

The reflected wave is in phase if v2>v1 and out of phase if v2<v1

2 1 20 0 0 0

2 1 2 1

1 2 10 0 0 0

2 1 2 1

The real amplitudes are related by

2and

cin terms of refractive index n=

v

2and

R I T I

R I T I

v v vE E E E

v v v v

n n nE E E E

n n n n

Reflected wave is 180o out of phase when reflected froma denser medium. This fact was encountered by you during

Last semester optics course. Now you have a proof!!!

Reflection coefficient (R) and Transmission coefficient (T)

• Intensity (average power per unit area is given by):

• If μ1= μ2 = μ0, i.e μr=1 , then the ratio of the reflected intensity to the incident intensity is

20

1

2I vE

2 2

0 1 2

0 1 2

RR

I I

EI n nR

I E n n

Where as the ratio of transmitted intensity to incident intensity is 2 2

02 2 2 1 1 22

1 1 0 1 1 2 1 2

2 4

( )TT

I I

EI v n n n nT

I v E n n n n n

NOTE: R+T=1 => conservation of energy,

Use ε α (n)2

Oblique Incidence

Oblique Incidence