Loading in 5 sec....

Course Name: Physics-II Course Code: 10B11PH211 Course Credits: 4 (3 1 0) PowerPoint Presentation

Course Name: Physics-II Course Code: 10B11PH211 Course Credits: 4 (3 1 0)

- By
**xuan** - Follow User

- 71 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' Course Name: Physics-II Course Code: 10B11PH211 Course Credits: 4 (3 1 0) ' - xuan

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Magnetostatics: Maxwell’s equations in free space and dielectric media. Propagation of EM waves through boundary-

Course Code: 10B11PH211

Course Credits: 4 (3 1 0)

Total number of Lectures: 40

- COURSE DESCRIPTION
- Broad Area: Electromagnetism
- Fiber Optics
- Thermodynamics
- Quantum mechanism
- Solid state physics

- Electromagnetism
- Electrostatics:
- Coulomb’s law,
- Gauss law and its applications,
- Treatment of electrostatic problems by solution
- of Laplace and Poisson’s equations.

- Electrostatics:

- Biot-Savart law,
- Faraday’s Law,
- Ampere’s law,

- Reflection
- Refraction
- Absorption and
- Total Internal Reflection.

- Books:
- Introduction to Electrodynamics
- by D.J. Griffith
- Principles of Electromagnetics
- by Matthew N. O. Sadiku
- Electromagnetics
- by Edminister (Schuam series)
- Engineering Electromagnetic
- by W H Hayt & J A Buck

What is Electromagnetics?

What is Electromagnetics?

What is Electromagnetics?

What is Electromagnetics?

What is Electromagnetics?

What is Electromagnetics?

What is Electromagnetics?

What is Electromagnetics?

Choice is based on symmetry of problem

To understand the Electromagnetic, we must know basic vector algebra and coordinate systems. So let us start the coordinate systems.

COORDINATE SYSTEMS

- RECTANGULAR or Cartesian

Sheets - RECTANGULAR

- CYLINDRICAL

Examples:

Wires/Cables - CYLINDRICAL

- SPHERICAL

Spheres - SPHERICAL

Orthogonal Coordinate Systems:

1. Cartesian Coordinates

z

P(x,y,z)

Or

y

Rectangular Coordinates

x

P (x, y, z)

z

z

P(r, , z)

2. Cylindrical Coordinates

P (r, , z)

y

r

x

Φ

z

3. Spherical Coordinates

P(r, θ,)

θ

r

P (r, θ, )

y

x

Φ

z

Cartesian Coordinates

P(x, y, z)

P(x,y,z)

P(r, θ, Φ)

θ

r

y

x

y

x

Φ

Cylindrical Coordinates

P(r, Φ, z)

Spherical Coordinates

P(r, θ, Φ)

z

z

P(r, Φ, z)

y

r

x

Φ

Differential quantities:

dx

y

6

2

3

7

x

AREA INTEGRALS

- integration over 2 “delta” distances

Example:

AREA =

= 16

Note that: z = constant

Cylindrical Coordinates: Visualization of Volume element

Differential quantities:

Limits of integration of r, φ,z are 0<r<∞ , o<φ <2π ,0<z <∞

Spherical coordinate system (r,,φ)

Radius=r

0<r<∞

- -Zenith angle
- 0<θ <

( starts from +Z reaches up to –Z) ,

-Azimuthal Angle

0<φ <2

(starts from +X direction and lies in x-y plane only)

Spherical coordinate system (r,,φ)

P(r, θ, φ)

Z

r sin θ

dr

P

r cosθ

r dθ

θ

r

dθ

Y

dφ

φ

r sinθ dφ

r sinθ

X

Spherical coordinate system (r,,φ)

P(r, θ, φ)

Z

r sin θ

dr

P

r cos θ

r dθ

θ

r

dθ

Y

dφ

φ

r sinθ dφ

r sinθ

X

Try Yourself: Surface area of the sphere= 4πR2 .

Basics of fields, Gradient, Divergence and Curl.

Coulomb’s law, Electric Flux & Gauss’s law

Scalar and Vector Fields

- A scalar field is a function that gives us a single value of some variable for every point in space.
voltage, current, energy, temperature

- A vector is a quantity which has both a magnitude and a direction in space.
velocity, momentum, acceleration and force

Gradient, Divergence and Curl

The Del Operator

- Gradient of a scalar function is a vector quantity.
- Divergenceof a vector is a scalar quantity.
- Curl of a vector is a vector quantity.

Operator in Cartesian Coordinate System

gradT: points the direction of maximum increase of the function T.

Divergence:

Curl:

where

Fundamental theorem for divergence and curl

- Gauss divergence theorem:
- Stokes curl theorem

Conversion of volume integral to surface integral and vice verse.

Open S

Closed L

Conversion of surface integral to line integral and vice verse.

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‘sLaw:

where e0 = permittivity of space

The number of electric field lines through a surface

A

E

E=EA,

- Conclusion:
- The total flux depends on
- strength of the field,
- the size of the surface area it passes through,
- and on how the area is oriented with respect to the field.

- The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity (eo.).
- eo = the permittivity of free space 8.854x10-12 C2/(N m2)

da

E

+q

Integral Form

Differential Form

where

+ve Flux

The defining conditions of a Guassian surface:

- The surface is closed.
- At each point of the surface E is either normal or tangential to the surface.
- E is sectionally constant over that part of the surface where E is normal.

Electric lines of flux and Derivation of Gauss’ Law using Coulombs law

- Consider a sphere drawn around a positive point charge. Evaluate the net flux through the closed surface.

Net Flux =

For a Point charge

dA

Gauss’ Law

Differential form of Gauss Law:

Proof:

Gauss Law

Gauss divergence theorem:

or

Note: Gauss law is also known as Maxwell’s first equation.

C

A

da

E

da

D

B

Where dΩ is solid angle

Asmnt 2:

Proof of the Gauss’s law for the charge inside

da

E

+q

and outside the Gaussian surface

Applications of Gauss law(Spherical distribution systems)

- Conducting Sphere of charge ‘q’ and radius ‘R’:
- E at an external point: Eo
- E at the surface: Es
- E at an internal point: Ei

- Nonconducting Sphere
- E at an external point: Eo
- E at the surface: Es
- E at an internal point: Ei

(Spherical systems: Conducting Sphere)

Gaussian surface

- Conducting Sphere of charge ‘q’ and radius ‘R’:
- E at an external point: Eo r>R
- E at the surface: Es r=R
- E at an internal point: Ei r<R

Case-I: E at an external point;

Net electric fux through ‘P’:

R

P

r

S1

The Electric field strength at any point outside a spherical charge distribution is the same as through the whole charge were concentrated at the centre.

r=R

Gaussian surface

R

r

(Spherical systems: Conducting Sphere)Case-II: E at the Surface;

Case-III: E at an internal point;

(Spherical systems: Nonconducting Sphere)

Nonconducting sphere (Volume charge density)

- E at an external point: Eo
- E at the surface: Es
- E at an internal point: Ei

R

P

r

S1

(Spherical systems: Nonconducting Sphere)

- Nonconducting Sphere of charge ‘q’ and radius ‘R’:
- E at an external point: Eo r>R
- E at the surface: Es r=R
- E at an internal point: Ei r<R

Case-I: E at an external point;

Net electric flux through ‘P’:

r=R

Gaussian surface

R

r

(Spherical systems: Nonconducting Sphere)

Case-II: E at the Surface;

Case-II: E at an internal point;

R

P

P

r

r

E

E

r=0

r=R

r=R

r

r

(Spherical systems: Conducting Sphere)

(Spherical systems: Nonconducting Sphere)

r=0

Problems: Spherical Symmetry

2. Non conducting spherical shell of inner radius r1, outer radius r2 and charge density ρ= k/r2 , where k is a constant. Also determine Max E at any value of r .

1. Non conducting solid sphere of radius R and charge density ρ=k/r2, Where k is a constant.

Determine Electric field everywhere by using Gauss Law for the following;

E3

R

E1

E5

Ei

Eo

E2

E4

Es

Spherical shell

Solid sphere

Applications of Gauss law(Cylindrical distribution systems)

- Conducting long Cylinder of charge ‘q’ and radius ‘R’:
- E at an external point: Eo
- E at the surface: Es
- E at an internal point: Ei

- Nonconducting long Cylinder
- E at an external point: Eo
- E at the surface: Es
- E at an internal point: Ei

Cylindrical distribution systems: Conducting Cylinder

- Conducting long Cylinder of charge ‘q’ and radius ‘R’ :
- E at an external point: Eor>R
- E at the surface: Es r=R
- E at an internal point: Eir<R

Gaussian

surface

Case-I: E at an external point;

Net electric flux through ‘P’:

E

l

R

O

P

r

l

R

O

P

r

Cylindrical distribution systems: Nonconducting Cylinder- Nonconducting Cylinder of radius ‘R’, height ‘l’ and charge density ‘ρ’:
- E at an external point: Eo r>R
- E at the surface: Es r=R
- E at an internal point: Ei r<R

Gaussian

surface

Case-I: E at an external point;

Net electric flux through ‘P’:

Numerical: Non conducting Cylindrical shell (r1, r2 and height h) having volume charge density ρ=k/r. Determine E everywhere.

Case-I: E at an external point r0; E0

Gaussian

surface

E

l

O

P

r1

r2

r0

Applications of Gauss law(Infinitely long sheet of Charge)

The plane is infinitely large, any point can be treated as the center point of the plane, so E at that point must be normal to the surface and must have the same magnitude at all points equidistant from the plane.

A cylindrical Gaussian surface is used to find the electric field of an infinite plane sheet of charge.

Assume the surface charge density is σ.

‘E’ does not depend on the distance ‘r’. Therefore, the field is uniform everywhere.

Electric field just outside the surface of a charged conductor

Assume the surface charge density is σ.

σ

- conductorσ

-σ

-σ

+σ

(2)

(3)

Find the field in each of three regions

(i) to the left of both

(ii) between them

(iii) to the right of both

+σ

+σ

E- E- E- E-

E+ E+ E+ E+

E+ E+ E+ E+

E+ E+ E+ E+

(1)

0

0

0

E- E- E- E-

E- E- E- E-

0

Force on the surface of conductor conductor

Electric Field outside a conductor

Note1: Force on a charge q (or surface charge density σ) placed in an external field E:

Note 2: “ But E is discontinuous across a surface charge distribution” Therefore

On the surface, force per unit area :

In our case: Eabove=σ/ε0 and Ebelow is zero. Hence Force (per unit area) on the conductor surface:

or

Where P is outward electrostatic Pressure on the conductor surface.

Prob. 2.37: conductor

Two large metal plates (each of area A) are held a distance d apart. Suppose we put a charge Q on each plate, what is the electrostatic pressure on the plates ?

Line integral of Electric field: Electric Potential conductor

Electric field at a field point r, due to a point charge at origin:

Area integral of E over the surface (Flux q/ε0 )

What about Line integral of E from some point a to some other point b?

The electric potential at a distance r from the point charge is the work done per unit charge in bringing a test charge from infinity to that point.

Line integral of Electric field: Electric Potential conductor

Electric field at a field point r, due to a point charge at origin:

Area integral of E over the surface (Flux q/ε0 )

What about Line integral of E from some point a to some other point b?

Curl of E ?

The integral around a closed path is zero

Using Stokes’ theorem =>

In electrostatics only means no moving charge or current

Hence curl of E

Numerical conductor

Find the potential inside and outside a uniformly charged solid sphere of radius R and total charge q. Use infinity as your reference point. Sketch V(r) .

V(r)

Electric field at r > R

r < R

Therefore, Electric potential

at r > R

R

r < R

R conductor

Find the potential inside and outside a spherical shell of radius R, which carries a uniform surface charge. Set the reference point at infinity.

r< R

r

Notice that the potential is not zero inside the shell , even though the field is. V is constant in this region, so that V=0

R conductor

Find the potential inside and outside a spherical shell of radius R, which carries a uniform surface charge. Set the reference point at infinity.

r< R

r

Notice that the potential is not zero inside the shell , even though the field is. V is constant in this region, so that V=0

Poisson's and Laplace's Equation conductor

- The electric field is related to the charge density by the divergence relationship
- 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
- 2 : Laplacian operator.
- In a charge-free region of space,
- Laplace's equation

Note: Del Operator conductor

Note: Laplace’s Operator conductor

- Show that potential function V( conductorx,y,z) or V(r) satisfies the Laplace’s equation.
- Given
- In Cartesian Coordinates
- In Spherical Polar coordinates

In Cartesian coordinates conductor

S conductorpherical polar coordinates

Calculate the numerical value for V and conductorρv at point P in free space if (D 7.1 (page 175, 7th Ed. Hayt)

Ans: (a) 12V, - 106.2 pC/m3

(b) -22.5V, 0

(c) 4V, -141.7pC/m3

Numerical conductor

- Does potential function 2(x2-y2+z) satisfies Laplace’s equation?
- Determine potential outside a charged conducting sphere of radius R, using Laplace’s equation.
Given V=V0 at r=R

= 0 at r=infinite.

Ans: Yes

Solution: conductor

Applications: Laplace’s and Poisson’s equation conductor

In Cartesian or Rectangular coordinates

One dimensional solution of Laplace’s Equation in rectangular coordinate system

Let V be a function of z only. Then Laplace’s Equation reduces to

Equation (3) represents a family of equi – potential surfaces with z taking up constant values.

Applications: Laplace’s and Poisson’s equation conductor

Consider two such equi – potential surfaces one at z = z1 and the other at z = z2.

Let V= V1 at z = z1 and V = V2 at z = z2

Z

This is the case with a parallel plate capacitor with a plate separation of z1 ~ z2 = d and a potential difference V1 ~ V2.

Applying the above two conditions, called boundary conditions, we get,

z=z2

V = V2 = Va

Let , V1 = 0 at z1 = 0 and V2 = Va at z2=z

d

z1=0

V = V1 = 0

Fig 1 Parallel plate capacitor

We find that V is a linear function of z

Similarly, V as a function of x or y the solution of the Laplace’s equation can be solved.

Applications: Laplace’s and Poisson’s equation conductor

In Cylindrical coordinates: r dependent only

Now, in cylindrical coordinates, the Laplace’s equation becomes

Consider V is a function of r only . Thus the Laplace’s equation reduces to

From this equation, we observe that equipotential surfaces are given by r = constant and are cylinders. Example of the problem is that of a coaxial capacitor or coaxial cable.

Applications: Laplace’s and Poisson’s equation conductor

In Cylindrical coordinates: r dependent only

Let us create the boundary conditions by choosing

V = Va at r = a and V = Vb at r = b, b > a.

Applications: Laplace’s and Poisson’s equation conductor

In Cylindrical coordinates: dependent only

Consider V as a function of Φ only.

From this equation, we observe that equipotential surfaces

are given by Ф = constant planes,

Choose two such equipotential surfaces, V = Va at Ф =α and V = 0 at Ф =0. Example: corner reflector antenna ,Used in communication systems.

Applications: Laplace’s and Poisson’s equation conductor

In Spherical coordinates: r dependent only

Consider that V is a function of r only. Laplace’s equation reduces to

where A and B are arbitrary constants to be evaluated. This equation represents a family of equi – potential surfaces for r = constant.

Let us choose two such equi-potential surfaces at r = a and r = b, b > a , such that at

r = a, V = Va and at r = b, V = Vb

Applications: Laplace’s and Poisson’s equation conductor

In Spherical coordinates: r dependent only

This is an example of concentric spheres or Spherical capacitor

Applications: Laplace’s and Poisson’s equation conductor

In Spherical coordinates: dependent only

Finally let us consider V as a function of θ only . In this case the Laplace’s equation reduces to

Integrating once again, we get,

This equation represents a family of equi-potential surfaces for constant θ. Let us consider two such equi-potential surfaces at θ =π/2, V = 0and at θ = α, V = Va.

Applications: Laplace’s and Poisson’s equation conductor

In Spherical coordinates: dependent only

The equi-potential surfaces are cones as shown in figure below.

Such a system is called a conical antenna

Applying these two boundary conditions to the equation (i) and solving it for A and B, substituting these values in (i),we get,

Download Presentation

Connecting to Server..