Download Presentation
SPECIAL TECHNIQUES

Loading in 2 Seconds...

1 / 41

# SPECIAL TECHNIQUES - PowerPoint PPT Presentation

SPECIAL TECHNIQUES. To find the electric field of a stationary charge distribution :. Find the potential of the distribution. To Solve : Poisson’s / Laplace’s Equation. To determine V. Poisson's / Laplace’s equation. +. A set of boundary conditions. Proof ?. Uniqueness Theorem.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
Download Presentation

## PowerPoint Slideshow about 'SPECIAL TECHNIQUES' - lamar-mathis

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

SPECIAL TECHNIQUES

To find the electric field of a stationary charge distribution :

Find the potential of the distribution

To Solve :

Poisson’s / Laplace’s Equation

Dr. Champak B. Das (BITS, Pilani)

To determine V

Poisson's / Laplace’s equation

+

A set of boundary conditions

Proof ?

Uniqueness Theorem

Dr. Champak B. Das (BITS, Pilani)

First Uniqueness Theorem :

The solution to Laplace’s equation in some volume is uniquely determined if the potential is specified on the boundary surface.

Corollary

The potential in a volume is uniquely determined if (a) the charge density throughout the region, and (b) the value of the potential on all boundaries, are specified.

Dr. Champak B. Das (BITS, Pilani)

Second Uniqueness Theorem :

In a volume surrounded by conductors and containing a specified charge density, the electric field is uniquely determined if the total charge on each conductor is given.

(The entire region can be unbound/bounded by another conductor).

Dr. Champak B. Das (BITS, Pilani)

SPECIAL TECHNIQUES

• The Method of Images
• Multipole expansion

Dr. Champak B. Das (BITS, Pilani)

METHOD OF IMAGES

P

( x, y, z)

z

q

d

y

Grounded conducting plane

x

To find out the potential in the region above the plane

Dr. Champak B. Das (BITS, Pilani)

Solution of Poisson’s equation:

(in the region z > 0 ),

WITH

• A point charge q at (0,0,d)
• Boundary conditions:
• V = 0 when z = 0
• V  0 when x2+y2+z2 >> d2

Dr. Champak B. Das (BITS, Pilani)

 one function that meets the requirement

Guaranteed by :

Corollary of First Uniqueness Theorem

The potential in a volume is uniquely determined if (a) the charge density throughout the region, and (b) the value of the potential on all boundaries, are specified.

Dr. Champak B. Das (BITS, Pilani)

Z

P

( x, y, z)

+q

d

Y

d

-q

X

A new problem:

Dr. Champak B. Das (BITS, Pilani)

Z = 0

x2+y2+z2 >> d2

Final answer

(By virtue of Uniqueness Theorem)

Dr. Champak B. Das (BITS, Pilani)

Induced Surface Charge

Total induced charge :

Dr. Champak B. Das (BITS, Pilani)

Force

Force of attraction on q towards the plane

Force of attraction on +q towards -q

Dr. Champak B. Das (BITS, Pilani)

ENERGY

Two point charges and no conductor :

Single point charge and conducting plane :

Dr. Champak B. Das (BITS, Pilani)

R

a

q

V=0

Another example :

A point charge and a grounded conducting sphere :

Dr. Champak B. Das (BITS, Pilani)

Image charge :

Location of image charge :

(to the right of the centre of the sphere)

Dr. Champak B. Das (BITS, Pilani)

rs

r

rs´

b

q

q'

a

Two point charges q and q and no conductor

Dr. Champak B. Das (BITS, Pilani)

rs

r

rs´

Prob. 3.7(a):

θ

z

q

b

q'

a

 V=0

r = R

Dr. Champak B. Das (BITS, Pilani)

Prob. 3.7(b) :

Induced surface charge on the sphere :

Total Induced surface charge :

Dr. Champak B. Das (BITS, Pilani)

Prob. 3.7(c) :

Force on q :

Energy of the configuration :

Dr. Champak B. Das (BITS, Pilani)

rs

d'

r

θ'

r'

MULTIPOLE EXPANSION

To characterize the potential of an arbitrary charge distribution, localized in a rather small region of space

Dr. Champak B. Das (BITS, Pilani)

Law of cosines 

Dr. Champak B. Das (BITS, Pilani)

Legendre polynomials

More on this next sem. in Maths - III

Dr. Champak B. Das (BITS, Pilani)

Systematic expansion for the potential of an arbitrary localized charge distribution, in powers of 1/r

Multipole expansion of V in powers of 1/r

Dr. Champak B. Das (BITS, Pilani)

Monopole term

Dipole term

Quadrupole term

Dr. Champak B. Das (BITS, Pilani)

The Monopole Term:

…… is the most dominant term for r >>

Potential of any distribution  Vmon ,

(if looked from very far point)

For a point charge at origin,

V = Vmon, everywhere

Dr. Champak B. Das (BITS, Pilani)

The Dipole Term:

…… is the most dominant term if total charge is zero

Dr. Champak B. Das (BITS, Pilani)

dipole moment of the distribution

Dr. Champak B. Das (BITS, Pilani)

z

-q

d

r'_

+q

r'+

y

x

For a collection of point charges,

For aphysicaldipole:

Dr. Champak B. Das (BITS, Pilani)

P

rs+

+q

r

d

rs-

-q

Potential of a physical dipole

Dr. Champak B. Das (BITS, Pilani)

Potential due to a point charge ~ 1/r

Potential due to a dipole ~ 1/r2

Dr. Champak B. Das (BITS, Pilani)

physical

dipole

Dr. Champak B. Das (BITS, Pilani)

Potential for a pure dipole (d  0)

Physical dipole

Pure dipole

for d  0, q   , with p=qd kept fixed

Dr. Champak B. Das (BITS, Pilani)

z

r

rs

y

d

O

q

x

Role played by ORIGIN of coordinate system in multipole expansion

A point charge away from origin :

 Posses a non zero dipole contribution

Dr. Champak B. Das (BITS, Pilani)

Dipole moment changes when origin is shifted :

d'

y

r'

a

x

Dr. Champak B. Das (BITS, Pilani)

If Q = 0, then

If net charge of the configuration is zero, then the dipole moment is independent of the choice of origin.

Dr. Champak B. Das (BITS, Pilani)

3q

a

a

a

-2q

-2q

a

q

Prob. 3.27:

In the charge configuration shown, find a simple approximate formula for potential, valid at points far from the origin. Express your answer in spherical coordinates.

Answer:

Dr. Champak B. Das (BITS, Pilani)

Field due of a dipole

Potential at a point due to a pure dipole:

z

r

p

y

x

Dr. Champak B. Das (BITS, Pilani)

Field due of a dipole (contd.)

Recall:

Dr. Champak B. Das (BITS, Pilani)

z

r

y

x

Prob 3.33:

Electric field in a coordinate free-form :

p

Dr. Champak B. Das (BITS, Pilani)

z

y

Field lines of a pure dipole

Dr. Champak B. Das (BITS, Pilani)

Field lines of a physical dipole

Dr. Champak B. Das (BITS, Pilani)