3 a zh em i flux density gauss

8/10/2019 3 a ZH EM I Flux Density Gauss

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Dr. Zuhair M. Hejaz Set 3 a Electric Flux Density 1

Electric Flux DensityGauss's Law

Set 3 a


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Electric Flux Density



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Electric Flux Density (D)

Flux lines show the direction and density of the

flux.



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Electric flux density is a vector field describing

the number of flux lines crossing an area normal

to the lines.

It is denoted as D (originally from the word

Displacement).

The direction of D at a point is the direction of

the flux lines at that point.

The D at a point (r) meters from a point charge

can be given as:



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The flux lines are symmetrically directed

outward from the point and pass through an

imaginary spherical surface of area .

Comparing with the radial electric field intensity

E (discussed earlier) of a point charge in free

space and given as:

Therefore, we can relate E and D as:

Note thatDr. Zuhair M. Hejaz Set 3 a Electric Flux Density 5


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D is independent of the medium:

The Electric flux can be defined in terms

of D as:

D is measured in .

All formulas derived for E from Coulomb's law

can be used in calculating D, except that we

have to multiply those formulas by .Dr. Zuhair M. Hejaz Set 3 a Electric Flux Density 6


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For example, for an infinite sheet of charge, the

eqn. for field intensity:

gives D

and for a volume charge distribution, the eqn.

for field intensity:

gives

Later in this course, D will be applied to dielectric

materials, rather than free space only.Dr. Zuhair M. Hejaz Set 3 a Electric Flux Density 7


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Example 1

Recall: D is a function of charge and position

only; it is independent of the medium.

Example 1: Determine D at P(4, 0, 3) if there is

a point charge at P1(4, 0, 0) and a line

charge along the as shown:



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Example 1- Cont.

Solution:

Now: For the line charge



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Example 1- Cont.

In this case:

And

So

Thus, the total density will be:

End of Example 1



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Gauss's Law



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Gauss's Law Gauss's law is one of the fundamental laws of

electromagnetism.

Gauss's law provides an easy means of finding E

or D for symmetricalcharge distributions such as

a point charge, an infinite line charge, an infinitecylindrical surface charge, and spherical charge.

Gauss's law is an alternative statement of

Coulomb's law as we will see later, when

applying the divergence theorem to Coulomb's

law. It results in Gauss's law.



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Gauss's Law

Gauss's law states that: The total electric flux

through any closed surface is equal to the total

charge enclosed by that surface.

In general . From the fig, what is

the total flux

leaving the

closed surfaces(or Volumes)

and ??



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Gauss's Law

From , the total flux leaving is

From , the total flux leaving is 0. Why?

Because no charge is enclosed in that volume.

Note that according Gauss's law, the net fluxignores those charges outside and .

A continuous charge distribution has rectangular

symmetry if it depends only on ,cylindrical symmetry on , or spherical

symmetry on (independent of and ).



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Gauss's Law

Considering the

volume shown in

the fig. enclosing

a set of point

charges .

The flux crossing is then the product of the

normal component of and , so we canexpress :

The totalflux passing through the closedsurface is:Dr. Zuhair M. Hejaz Set 3 a Electric Flux Density 15


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Gauss's Law

Note that: the surface element always

involves the differentials of two coordinates,

such as , or

Now, we can present the mathematical

formulation of Gauss's law:

The enclosed surface is called Gaussian Surface.Dr. Zuhair M. Hejaz Set 3 a Electric Flux Density 16


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Gauss's Law

The enclosed charge can be several point charges

, a line, surface, or volume charge.

For a line charge

For asurface

charge

For a volume charge

The last form is usually used as it can represent

the other forms., so Gauss's law can be written interms of charge distribution as:

What does this formula mean?? Dr. Zuhair M. Hejaz Set 3 a Electric Flux Density 17


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Applications of Gauss's Law 1

The above formula means that: The total electric

flux through any closed surface is equal to the

charge enclosed.

Applications of Gauss's Law.

First, Check Faradays Experiment: Place a point

charge at the origin of a SCS. The Gaussian

Surface is a sphere with radius (see next fig.)

Recall: The electric field intensity of a point

charge: and flux density:



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So,

Now, at the surface of

the sphere:

The differential element

of area on a spherical

surface, in sphericalcoordinates (see page 11

in slide 1_d) is



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The integrand is then:

The closed surface integral can be written as:

this proves



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Example 2

Example 2: The cylindrical surface

cm contains the surface

charge density

(a) What is the total amount of

charge present?

(b) How much flux leaves the surface cm,

where and ?

Solution


x


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Example 2

Solution: (a) We integrate over the surface to

find the charge: recall

(b) We just integrate the charge density on thatsurface defined by and

to find the flux that leaves it.



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To apply Gauss's law, to calculate the electric

field, It must involve the following:

1) Knowing whether symmetry exists.

2) Once symmetric charge distribution exists, we

construct a mathematical closed surface

(Gaussian surface).

3) The surface is chosen such that D is normal or

tangential to theGaussian surface.

4) When D is normal to the surface,

because D is constant everywhere on the surface.Dr. Zuhair M. Hejaz Set 3 a Electric Flux Density 24


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5) When D is tangential to the surface,

For example, we, revisit the case of a point charge

in the origin of a sphere.

A. Point Charge:

To determine D at a point P,

it is easy to see that choosing

a spherical surface containingPwill satisfy symmetry conditions. Thus, a spherical

surface as shown in the fig. is the Gaussian surface.



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B. Infinite Uniform Line Charge:

Suppose an infinite line of

uniform charge lies

along the .

To determine D at a point

P, we choose a cylindrical surfacecontaining Pto satisfy symmetry condition as

shown.



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D is constant on and normal to the cylindrical

Gaussian surface, so: apply Gauss's law to anarbitrary length of the line:

Or as , finally . Note that



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D has no on the top and bottom

surfaces, which means that D is tangential tothose surfaces.

The electric field can then be found:

The same result is achieved with much less work.I.e. the integration is only over the area of the

chosen symmetric surface at which D is normal.



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You have learned about


The flux density vector D and how it relates to

the charge Q and the E vector.

Gauss law of electrostatics.

The application ofGauss law to the solution of

symmetrical problems.


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Suggested Problems & HW

Suggested Problems from text book:

3.2, .3.4, 3.7, 3.8, 3.10, 3.13, 3.15

HW (3): 3.7, 3.13, 3.15

Submission date: The week after concludingthis slide.



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End of Set 3 aElectric Flux Density

Gauss's Law

Thank You for Your

Attention


3 a zh em i flux density gauss

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