aem ece 351 e-flux density(2)

SC

IEN

CE

NEE

DS

YO

U

Dept. of ECE

Dr. M. Reyhani

[email protected]

Tel: (701) 231 Ext. 8816

Thursday, 02 February 2012

Department of Electrical and Computer Engineering North Dakota State University

North Dakota State University

Applied Electromagnetics

(Ref. ECE 351 2011 - 2012)

Electric Flux Density

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smsreyhani 2 Dept. of ECE

1) Electric Flux Density

2) Gauss’s Law Applications

SOME SYMMETRICAL CHARGE DISTRIBUTIONS

DIFFERENTIAL VOLUME ELEMENT

3) Divergence Divergence & Maxwell’s First Equation

The Vector Operator ∇ & the Divergence Theorem

AGENDA

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Michael Faraday, (22 September 1791 – 25 August 1867)

an English chemist and physicist (or natural philosopher, in the terminology of the time) who contributed to the fields of electromagnetism and electrochemistry. Although Faraday received little formal education and knew little of higher mathematics such as calculus, he was one of the most influential scientists in history;

historians of science refer to him as having been the best experimentalist in the history of science.

It was on account of his research regarding the magnetic field around a conductor carrying a DC electric current that Faraday established the basis for the concept of the electromagnetic field in physics

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Faraday Experiment

Michael Faraday started with a pair of metal spheres of different sizes; the larger one consisted of 2 hemispheres that could be assembled around the smaller sphere

Michael Faraday

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+Q

Faraday Apparatus, Before Grounding

The inner charge, Q, induces an equal and opposite charge, -Q, on the inside surface of the outer sphere, by attracting free electrons in the outer material toward the positive charge. This means that before the outer sphere is grounded, charge +Q resides on the outside surface of the outer conductor.

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Faraday Apparatus, After Grounding

q = 0

ground attached

Attaching the ground connects the outer surface to an unlimited supply of free electrons, which then neutralize the positive charge layer. The net charge on the outer sphere is then the charge on the inner layer, or -Q.

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Interpretation of the Faraday Experiment

q = 0

Faraday concluded that there occurred a charge “displacement” from the inner sphere to the outer sphere.

Displacement involves a flow or flux, Ψ, existing within the dielectric, and whose magnitude is equivalent to the amount of “displaced” charge.

Specifically:

D = Displacement flux density or displacement density

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q = 0

The density of flux at the inner sphere surface is equivalent to the density of charge there (in C/m2)

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Vector Field Description of Flux Density

q = 0

A vector field is established which points in the direction of the “flow” or displacement. In this case, the direction is the outward radial direction in spherical coordinates. At each surface, we would have:

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Radially-Dependent Electric Flux Density

q = 0

r

At a general radius r between spheres, we would have:

Expressed in units of C/m2, and defined over the range (a ≤ r ≤ b)

D(r)

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Faraday’s Experiment Cont.

Faraday concluded there was a “displacement” from the charge on the inner sphere through the inner sphere through the insulator to the outer sphere.

The electric displacement (or electric flux) is equal in magnitude to the charge that produces it, independent of the insulating material and the size of the spheres.

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Visualization of Electric Fields

Flux lines are suggestive of the flow of some

fluid emanating from positive charges (source) and terminating at negative charges (sink).

Although electric field lines do NOT represent fluid flow, it is useful to think of them as describing the flux of something that, like fluid flow, is conserved.

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Point Charge Fields

If we now let the inner sphere radius reduce to a point, while maintaining the same charge, and let the outer sphere radius approach infinity, we have a point charge. The electric flux density is unchanged, but is defined over all space:

C/m2 (0 < r <∞ )

We compare this to the EFI in free space:

V/m (0 < r <∞ )

..and we see that:

EFI = electric field intensity

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Finding E and D from Charge Distributions

It now follows that:

For a general volume charge distribution in free space,

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Example

Determine D at (4, 0, 3) if there is a point charge —5π mC at (4, 0, 0) and

a line charge 3π mC/m along the y-axis.

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Example Cont.

Determine D at (4, 0, 3) if there is a point charge —5π mC at (4, 0, 0) and a line charge 3π mC/m along the y-axis.

Let D = DQ + DL where DQ and DL are flux densities due to the point charge and line charge, respectively, as shown in Figure below:

where r - r' = (4, 0, 3) - (4, 0, 0) = (0, 0, 3). Hence,

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Example Cont.


Flux density D due to a point charge and an infinite line charge.

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Example Cont.


Also

In this case

Hence,

Thus

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Gauss’ Law

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Johann Carl Friedrich Gauss

a German mathematician and physical scientist who contributed significantly to many fields, including number theory, statistics, analysis, differential geometry, geodesy, geophysics, electrostatics, astronomy and optics. Sometimes referred to as the Princeps mathematicorum (Latin, "the Prince of Mathematicians" or "the foremost of mathematicians") and "greatest mathematician since antiquity", Gauss had a remarkable influence in many fields of mathematics and science and is ranked as one of history's most influential mathematicians. He referred to mathematics as "the queen of sciences".

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Gauss’ Law

The electric flux passing through any closed surface is equal to the total charge enclosed by that surface

The generalizations of Faraday’s experiment

D = Displacement flux density or displacement density

The electric flux density DS at P arising from charge Q. The total flux passing through ∆S = DS .∆S.

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Development of Gauss’ Law

We define the differential surface area (a vector) as

where n is the unit outward normal vector to the surface, and

where dS is the area of the differential spot on the surface

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Mathematical Statement of Gauss’ Law

Line charge:

Surface charge:

Volume charge:

in which the charge can exist in the form of point charges:

For a volume charge, we would have:

or a continuous charge distribution:

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Using Gauss’ Law to Solve for D Evaluated at a Surface

Knowing Q, we need to solve for D, using Gauss’ Law:

The solution is easy if we can choose a surface, S, over which to integrate (Gaussian surface) that satisfies the following 2 conditions:

The integral now simplfies:

So that:

where

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Gauss’s Law

Gauss’s law states that “the net electric flux

emanating from a close surface S is equal to the total charge contained within the volume V bounded by that surface.”

enclS

QsdD =⋅∫

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Example

The cylindrical surface ρ = 8 cm contains the surface charge density, ρS = 5e−20|z| nC/m2. (a)What is the total amount of charge present?

(b) How much electric flux leaves the surface

ρ = 8 cm, 1 cm< z < 5 cm, 30◦ < φ < 90◦?

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Example Cont.

The cylindrical surface ρ = 8 cm contains the surface charge density, ρS = 5e−20|z| nC/m2. (a) What is the total amount of charge present? (b) How much electric flux leaves the surface ρ = 8 cm, 1 cm< z < 5 cm, 30◦ < φ < 90◦?

integrate over the surface to find:

Just integrate the charge density on that surface to find the flux that leaves it.

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Example

An electric field in free space is Find the total charge contained within a sphere of 3-m radius, centered at the origin.

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Example

An electric field in free space is Find the total charge contained within a sphere of 3-m radius, centered at the origin.

Using Gauss’ law, we set up the integral in free space over the sphere surface, whose outward unit normal is ar:

where in this case z = 3 cos θ and (in all cases) az · ar = cos θ.

These are substituted to yield

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Example

Let

D = 4xyax + 2(x2 + z2)ay + 4yzaz nC/m2

and evaluate surface integrals to find the total charge enclosed in the rectangular parallelepiped 0 < x < 2,

0 < y < 3,

0 < z < 5 m.

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Example Cont.

Let D = 4xyax + 2(x2 + z2)ay + 4yzaz nC/m2 and evaluate surface integrals to find the total charge enclosed in the rectangular parallelepiped 0 < x < 2, 0 < y < 3, 0 < z < 5 m.

Of the 6 surfaces to consider, only 2 will contribute to the net outward flux. Why? First consider the planes at y = 0 and 3. The y component of D will penetrate those surfaces, but will be inward at y = 0 and outward at y = 3, while having the same magnitude in both cases. These fluxes will thus cancel. At the x = 0 plane, Dx = 0 and at the z = 0 plane, Dz = 0, so there will be no flux contributions from these surfaces. This leaves the 2 remaining surfaces at x = 2 and z = 5.

The net outward flux becomes:

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Example Cont.

Let D = 4xyax + 2(x2 + z2)ay + 4yzaz nC/m2 and evaluate surface integrals to find the total charge enclosed in the rectangular parallelepiped 0 < x < 2, 0 < y < 3, 0 < z < 5 m.

The net outward flux becomes:

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APPLICATION OF

GUASS’ LAW

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SOME SYMMETRICAL CHARGE DISTRIBUTIONS

APPLICATION OF

GUASS’ LAW SOME SYMMETRICAL CHARGE DISTRIBUTIONS

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Example: Point Charge Field

Begin with the radial flux density:

and consider a spherical surface of radius a that surrounds the charge, on which:

On the surface, the differential area is:

and this, combined with the outward unit normal vector is:

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Point Charge Application (continued)

Now, the integrand becomes:

and the integral is set up as:

=

=

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Another Example: Line Charge Field Consider a line charge of uniform charge density ρL on the z axis that extends over the range −∞ < z < ∞ .

We need to choose an appropriate Gaussian surface, being mindful of these considerations:

We know from symmetry that the field will be radially-directed (normal to the z axis) in cylindrical coordinates:

and that the field will vary with radius only:

So we choose a cylindrical surface of radius ρ, and of length L.

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Line Charge Field (continued)

Giving: So that finally:

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Another Example: Coaxial Transmission Line

We have 2 concentric cylinders, with the z axis down their centers. Surface charge of density ρS exists on the outer surface of the inner cylinder.

A ρ-directed field is expected, and this should vary only with ρ (like a line charge). We therefore choose a cylindrical Gaussian surface of length L and of radius ρ, where a < ρ < b.

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Another Example: Coaxial Transmission Line

The left hand side of Gauss’ Law is written:

…and the right hand side becomes:

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Coaxial Transmission Line (continued)

We may now solve for the flux density:

and the electric field intensity becomes:

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Coaxial Transmission Line: Exterior Field

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Coaxial Transmission Line: Exterior Field

If a Gaussian cylindrical surface is drawn outside (ρ > b), a total charge of zero is enclosed, leading to the conclusion that

or:

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Coaxial Transmission Line

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Example

Select a 50-cm length of coaxial cable having an inner radius of 1 mm and an outer radius of 4 mm.

The space between conductors is assumed to be filled with air.

The total charge on the inner conductor is 30 nC.

We wish to know the charge density on each conductor, and the E and D fields.

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Example Cont.

Select a 50-cm length of coaxial cable having an inner radius of 1 mm and an outer radius of 4 mm. The space between conductors is assumed to be filled with air. The total charge on the inner conductor is 30 nC. We wish to know the charge density on each conductor, and the E and D fields.

finding the surface charge density on the inner cylinder,

The negative charge density on the inner surface of the outer cylinder is

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Example Cont.


The internal fields may therefore be calculated easily:

and

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Example Cont.


The internal fields may therefore be calculated easily:

Both of these expressions apply to the region where 1 < ρ < 4 mm. For ρ < 1 mm or ρ > 4 mm, E and D are zero.

and

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51

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DIFFERENTIAL VOLUME ELEMENT

APPLICATION OF

GUASS’ LAW DIFFERENTIAL VOLUME ELEMENT

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Electric Flux Within a Differential Volume Element

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Taking the front surface, for example, we have:

nearly Equal, Or about equal. Can also be

written with a ~ on top.

equality under an integral

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Unicode Characters in the 'Symbol, Math' Category


Nearly Equal, Or about equal. Can also be written with

a ~ on top.

APPROACHES THE LIMIT ≐

Approximate Symbol

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We now have:

and in a similar manner:

nearly Equal, Or about equal. Also can be written with a ~ on top.


APPROACHES THE LIMIT

≐

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and in a similar manner:

Therefore:

minus sign because Dx0 is inward flux through the back surface.


Nearly Equal, Or about equal. Can also be


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Charge Within a Differential Volume Element

Now, by a similar process, we find that:

and Nearly Equal, Or about equal.

Can also be written with a ~ on top.


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Charge Within a Differential Volume Element

Now, by a similar process, we find that:

and

All results are assembled to yield:

∆v

= Q (by Gauss’ Law)

where Q is the charge enclosed within volume ∆v

Nearly Equal, Or about equal. Can also be



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Example

Find an approximate value for the total charge enclosed in an incremental volume of 10−9 m3 located at the origin,

if

D = e−x sin y ax − e−x cos y ay + 2zaz C/m2.

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Example Cont. Find an approximate value for the total charge enclosed in an incremental volume of 10−9 m3 located at the origin, if D = e−x sin y ax − e−x cos y ay + 2zaz C/m2.

Evaluate the 3 partial derivatives from

APPROACHES THE LIMIT

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Example Cont. Find an approximate value for the total charge enclosed in an incremental volume of 10−9 m3 located at the origin, if D = e−x sin y ax − e−x cos y ay + 2zaz C/m2.

At the origin, the first 2 expressions are zero, and the last is 2. Thus, we find that the charge enclosed in a small volume element there must be approximately 2∆ν.

If ∆ν is 10−9 m3, then we have enclosed about 2 nC.

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Example

A charge distribution with spherical symmetry has density

Determine E everywhere.

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Gaussian surface for a uniformly charged sphere when: (a) r >= a and (b) r ≤ a.

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Example Cont.

A charge distribution with spherical symmetry has density

Determine E everywhere.

The charge distribution is similar to that in Figure above. Since symmetry exists, we can apply Gauss's law to find E.

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Example Cont.

or

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Example

Given that D = Zρcos2ϕaz C/m2,

calculate the charge density at (1, π/4, 3)

and the total charge enclosed by

the cylinder of radius 1 m with -2 ≤ z ≤ 2 m .

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Example Cont.

Given that D = Zρcos2ϕaz C/m2, calculate the charge density at (1, π/4, 3) and the total charge enclosed by the cylinder of radius 1 m with -2 ≤ z ≤ 2 m .

The total charge enclosed by the cylinder can be found in 2 different ways.

At (1, TT/4, 3),

Method 1: This method is based directly on the definition of the total volume charge.

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Example Cont.


Method 1: This method is based directly on the definition of the total volume charge.

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Example Cont.


Method 2: Alternatively, we can use Gauss's law.

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Example Cont.



where Ψs, Ψt and Ψb are the flux through the sides, the top surface, and the bottom surface of the cylinder, respectively (see Figure). Since D does not have component along aρ, Ψs = 0, forΨt , dS = ρ dϕ dρ az so

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Example Cont.



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Example Cont.



and for Ψb, dS = -ρ dϕ dρ az, so

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Example Cont.



and for Ψb, dS = -ρ dϕ dρ az, so

Thus

as obtained previously.

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Example

2 point charges - 4 μC and 5 μC are located at ( 2 , - 1 , 3) and (0, 4, - 2 ) , respectively. Find the potential at (1, 0, 1) assuming zero potential at infinity.

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Example Cont.

Let

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Example Cont.

Hence

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79

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Divergence

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Vectors – Gradient Assume that we have a scalar field F(x), we want to know how it changes with respect to the coordinate axes, this leads to a vector called the gradient of F

Φ∂Φ∂Φ∂

=Φ∇

z

y

x

With the nabla operator

∂∂∂

=∇

z

y

x

and xx ∂∂

=∂

The gradient is a vector that points in the direction of maximum rate of change of the scalar function F(x).

What happens if we have a vector field?

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Vectors – Divergence + Curl

The divergence is the scalar product of the nabla operator with a vector field V(x). The divergence of a vector field is a scalar!

zzyyxx VVV ∂+∂+∂=•∇ V

Physically the divergence can be interpreted as the net flow out of a volume (or change in volume). E.g. the divergence of the seismic wavefield corresponds to compressional waves.

The curl is the vector product of the nabla operator with a vector field V(x). The curl of a vector field is a vector!

∂−∂∂−∂∂−∂

=∂∂∂=×∇

xyyx

zxxz

yzzy

VVVVVV

VVV zyx

zyx

kjiV

The curl of a vector field represents the rotational part of that field (e.g. shear waves in a seismic wavefield)

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Divergence and Maxwell’s First Equation

Mathematically, this is:

Applying our previous result, we have: div A =

and when the vector field is the electric flux density:

= div D

Maxwell’s first equation

zzyyxx VVV ∂+∂+∂=•∇ V

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Divergence and Maxwell’s First Equation

= div D

Maxwell’s first equation

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Divergence Expressions in the 3 Coordinate Systems

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Example

Find div D at the origin if D = e−x sin y ax − e−x cos y ay + 2zaz .

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Example

Find div D at the origin if D = e−x sin y ax − e−x cos y ay + 2zaz .

using div A =

The value is the constant 2, regardless of location. If the units of D are C/m2, then the units of div D are C/m3. This is a volume charge density.

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The Del Operator

= div D =

The del operator is a vector differential operator, and is defined as:

Note that:

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Divergence Theorem

We now have Maxwell’s first equation (or the point / differential-equation form of Gauss’ Law) which states:

and Gauss’s Law in large-scale form reads:

leading to the Divergence Theorem:

differential-equation form of Gauss’s law

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Divergence Theorem

The integral of the normal component of any vector field over a closed surface

is equal to the

integral of the divergence of this vector field throughout the volume enclosed by the closed surface.

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Divergence Theorem

Consider the divergence of D in the region about a point charge Q located at the origin

using

Because Dθ = Dφ = 0, we have

Thus, ρν = 0 everywhere except at the origin, where it is infinite.

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Statement of the Divergence Theorem

The divergence of the vector flux density A is the outflow of flux from a small closed surface per unit volume as the volume shrinks to zero.

The volume is shown here in cross section.

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Example

Evaluate both sides of the divergence theorem for the field D = 2xyax + x2ay C/m2

and the rectangular parellelepiped formed by the planes x = 0 and 1, y = 0 and 2, and z = 0 and 3.

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Example Cont.

note that D is parallel to the surfaces at z = 0 and z = 3, so D· dS = 0 there

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Example Cont.

However, (Dx )x=0 = 0, and (Dy )y=0 = (Dy )y=2, which leaves only

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Example Cont.

the volume integral becomes

Since

and the check is accomplished.

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Example

Calculate ∆ · D at the point specified if (a)D = (1/z2)[10xyz ax + 5x2z ay + (2z3 − 5x2 y) az] at P(−2, 3, 5); (b) D = 5z2 aρ + 10ρz az at P(3,−45° , 5); (c) D = 2r sin θ sin φ ar + r cos θ sin φ aθ + r cos φ aφ

at P(3, 45◦ , −45°).

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Example

Calculate ∆ · D at the point specified if (a)D = (1/z2)[10xyz ax + 5x2z ay + (2z3 − 5x2 y) az] at P(−2, 3, 5); (b) D = 5z2 aρ + 10ρz az at P(3,−45° , 5); (c) D = 2r sin θ sin φ ar + r cos θ sin φ aθ + r cos φ aφ at P(3, 45◦ , −45°).

2

3( 2,3,5)

10 100 2 8.96D y x yz z −

∇ ⋅ = + + + =

In rectangular coordinates

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Example


2

(3, 45 ,5)

1 1 5( ) 10 71.67D zD D zDz

φρρ ρ

ρ ρ ρ φ ρ−

∂∂ ∂∇ ⋅ = + + = + = ∂ ∂ ∂

In cylindrical coordinates

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Example


22

(3, 45 , 45 )

1 1 1( ) (sin )sin sin

cos2 sin sin6sin sin 2sin sin

D rD

r D Dr r rr

φθθ

θ θ θ φθ φ φθ φ

θ θ −

∂∂ ∂∇ ⋅ = + +

∂ ∂ ∂

= + − = −

In spherical coordinates

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Example

(a)A point charge Q lies at the origin.

Show that div D is zero everywhere except at the origin.

(b) Replace the point charge with a uniform volume charge density ρv0 for 0 < r < a.

Relate ρv0 to Q and a so that the total charge is the same.

Find div D everywhere.

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Example Cont. (a) A point charge Q lies at the origin. Show that div D is zero everywhere except at the origin. (b) Replace the point charge with a uniform volume charge density ρv0 for 0 < r < a. Relate ρv0 to Q and a so that the total charge is the same. Find div D everywhere.

2/(4 )D arQ rπ=

For a point charge at the origin we know that

Using the formula for divergence in spherical coordinates, we find in this case that

22 2

1 04

D d Qrdrr rπ

∇ ⋅ = =

rovided

The above is true provided r > 0. When r = 0, we have a singularity in D, so its divergence is not defined.

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Example Cont. (a) A point charge Q lies at the origin. Show that div D is zero everywhere except at the origin. (b) Replace the point charge with a uniform volume charge density ρv0 for 0 < r < a. Relate ρv0 to Q and a so that the total charge is the same. Find div D everywhere.

Q

32 3

0 3443r v

Qrr D ra

π π ρ= =

32

3 2 3 31 3C/m and

4 4 4π π π

= ∇ ⋅ = =

rQr d Qr QD

dra r a aD

as expected. Outside the charged sphere, as before, and the divergence is zero.

2/(4 )D arQ rπ=

Thus

so. Gauss’ law tells us that inside the charged sphere

3 30 3 /(4 ) C/mρ π=v Q a

To achieve the same net charge, we require that, 30(4 / 3) va Qπ ρ =

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Example

In the region of free space that includes the volume 2 < x, y, z < 3, (a)Evaluate the volume integral side of the

divergence theorem for the volume defined here.

(b) Evaluate the surface integral side for

the corresponding closed surface.

22

2 ( 2 ) C/m= + −x y zyz xz xyz

D a a a

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Example Cont.

In the region of free space that includes the volume 2 < x, y, z < 3, (a) Evaluate the volume integral side of the divergence theorem for the volume

defined here. (b) Evaluate the surface integral side for the corresponding closed surface.

22


D a a a

In cartesian, we find ∇ ·D = 8xy/z3. The volume integral side is now

3 3 3

32 2 2

8 1 1(9 4)(9 4) 3.47 C4 9vol

xydv dxdydzz

∇ ⋅ = = − − − = ∫ ∫ ∫ ∫D

38 /D xy z∇ ⋅ =

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Example Cont.



22


D a a a

We call the surfaces at x = 3 and x = 2 the front and back surfaces respectively, those at y = 3 and y = 2 the right and left surfaces, and those at z = 3 and z = 2 the top and bottom surfaces.

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Example Cont.



22


D a a a

To evaluate the surface integral side,

we integrate D·n over all six surfaces and sum the results.

Note that since the x component of D does not vary with x, the outward fluxes from the front and back surfaces will cancel each other.

The same is true for the left and right surfaces,

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Example Cont.



22


D a a a

The same is true for the left and right surfaces, since Dy does not vary with y.

This leaves only the top and bottom surfaces, where the fluxes are:

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Example

a) Use Maxwell’s first equation, ∇ ・D = ρv, to describe the variation of the EFI with x in a region in which no charge density exists and in which a nonhomogeneous dielectric has a permittivity that increases exponentially with x.

The field has an x component only;

b) repeat part (a), but with a radially directed E- field (spherical

coordinates), in which again ρv = 0, but in which the permittivity decreases exponentially with r .


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Example


The field has an x component only; b) repeat part (a), but with a radially directed E- field (spherical coordinates), in

which again ρv = 0, but in which the permittivity decreases exponentially with r .


The permittivity can be written as where ϵ1 and α1 are constants. Then

1 1( ) exp( )x xε ε α=

11 00 ( )x x

x xdE E E x E edx

αα −+ = ⇒ =

This reduces to

1 1 11 1 1[ ( ) ( )] ( ) 0D E xx x x

x xd dEx x e E x e E edx dx

α α αε ε ε α ∇ ⋅ = ∇ ⋅ = = + =

where E0 is a constant.

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Example





In this case, the permittivity can be written as

where ϵ2 and α2 are constants. Then

2 2( ) exp( )r rε ε α= −

This reduces to 2

2 0rr

dE Edr r

α + − =

2 222 2 22 22 2

1[ ( ) ( )] 2 0D E rr rr r r

d dEr r r e E rE r E r edr drr r

α αεε ε α− − ∇ ⋅ = ∇ ⋅ = = − + =

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Example





whose solution is

[ ] 200 2 0 2 2

2( ) exp exp 2ln rr

EE r E dr E r r er r

αα α = − − = − + = ∫

where E0 is a constant.

This reduces to

22 0r

rdE Edr r

α + − =

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Example

Given the flux density

use 2 different methods to find the total charge within the region

1 < r < 2 m,

1 < θ < 2 rad,

1 < φ < 2 rad.

216 cos(2 ) C/m ,θθ=r

D a

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Example

Given the flux density D = 16/r cos(2θ) aθ C/m2, use 2 different methods to find the total charge within the region 1 < r < 2 m, 1 < θ < 2 rad, 1 < φ < 2 rad.

216 cos(2 ) C/m ,θθ=r

D a

We use the divergence theorem and first evaluate the surface integral side. We are evaluating the net outward flux through a curvilinear “cube”, whose boundaries are defined by the specified ranges. The flux contributions will be only through the surfaces of constant θ, however, since D has only a θ component.

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Example


216 cos(2 ) C/m ,θθ=r

D a

The flux contributions will be only through the surfaces of constant θ, however, since D has only a θ component.

On a constant-theta surface, the differential area is da = r sin θdrdφ,

where θ is fixed at the surface location.

Our flux integral becomes

sinda r drdθ φ=

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Example


216 cos(2 ) C/m ,θθ=r

D a

We next evaluate the volume integral side of the divergence theorem, where in this case,

21 1 16 16 cos2 cos(sin ) cos2 sin 2sin 2

sin sin sinD d dD

r d r d r rθ

θ θθ θ θ θθ θ θ θ θ

∇ ⋅ = = = −

Evaluate: 2 2 2 2

21 1 1

16 cos2 cos 2sin 2 sinsinvol

dv r drd dr

θ θ θ θ θ φθ

∇ ⋅ = − ∫ ∫ ∫ ∫D

Then integral simplifies to 2 2 2 2

1 1 1 116[cos2 cos 2sin 2 sin ] 8 [3cos3 cos ] 3.91Cdrd d dθ θ θ θ θ φ θ θ θ− = − = −∫ ∫ ∫ ∫

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117

Abstract Data Types

We are intelligent too. We prefer to live it’s a beautiful world

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Summary

Summary Summary

Summary Summary

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Electric (Displacement) Flux Density

The density of electric displacement is the electric (displacement) flux density, D. In free space the relationship between

flux density and electric field is

ED 0ε=

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Electric (Displacement) Flux Density Cont.

The electric (displacement) flux density for a point charge centered at the origin is

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Gauss’s Law

Gauss’s law states that “the net electric flux emanating from a close surface S is equal

to the total charge contained within the volume V bounded by that surface.”

enclS

QsdD =⋅∫

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Gauss’s Law Cont.

V

S ds By convention, ds

is taken to be outward from the volume V.

∫=V

evencl dvqQ

Since volume charge density is the most

general, we can always write Qencl in this way.

enclS

QsdD =⋅∫

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Applications of Gauss’s Law

Gauss’s law is an integral equation for the unknown electric flux density resulting from a given charge distribution.

enclS

QsdD =⋅∫known

unknown

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Applications of Gauss’s Law Cont.

In general, solutions to integral equations must be obtained using numerical techniques. However, for certain symmetric

charge distributions closed form solutions to Gauss’s law can be obtained.

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Applications of Gauss’s Law Cont.

Closed form solution to Gauss’s law relies on our ability to construct a suitable family of Gaussian surfaces.

A Gaussian surface is a surface to

which the electric flux density is normal and over which equal to a constant value.

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Electric Flux Density of a Point Charge Using Gauss’s Law

Consider a point charge at the origin:

Q

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Electric Flux Density of a Point Charge Using Gauss’s Law Cont.

(1) Assume from symmetry the form of the field

(2) Construct a family of Gaussian

surfaces

( )rDaD rrˆ=

spheres of radius r where

∞≤≤ r0

spherical symmetry

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(3) Evaluate the total charge within the volume enclosed by each Gaussian surface

∫=V

evencl dvqQ

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Q

R

Gaussian surface

QQencl =

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(4) For each Gaussian surface, evaluate the integral

DSsdD

S

=⋅∫

( ) 24 rrDsdD rS

π=⋅∫

magnitude of D on Gaussian

surface.

surface area of Gaussian

surface.

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(5) Solve for D on each Gaussian surface

SQD encl=

24ˆ

rQaD r π

= 200 4

ˆr

QaDE r πεε==⇒

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Electric Flux Density of a Spherical Shell of Charge Using Gauss’s Law

Consider a spherical shell of uniform charge density:

≤≤

=otherwise,0

,0 braqqev

a

b

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Electric Flux Density of a Spherical Shell of Charge Using Gauss’s Law Cont.



surfaces

( )RDaD rrˆ=

spheres of radius r where

∞≤≤ r0

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Here, we shall need to treat separately 3 sub-families of Gaussian surfaces:

ar ≤≤01)

bra ≤<2)

br >3)

a

b

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Gaussian surfaces for which

ar ≤≤0


bra ≤<


br >

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∫=V

evencl dvqQ

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0=enclQ

For

For

ar ≤≤0

bra ≤<

( )330

30

300

34

34

34

arq

aqrqdvqQr

aencl

−=

−== ∫

π

ππ

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For

( )330

30

30

34

34

34

abq

aqbqdvqQb

aevencl

−=

−== ∫

π

ππ

br >

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DSsdD

S

=⋅∫

( ) 24 rrDsdD rS

π=⋅∫


surface.


surface.

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SQD encl=

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Notice that for r > b

24ˆ

rQaD tot

r π=

Total charge contained in spherical shell

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0 1 2 3 4 5 6 7 8 9 10 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

R

D r (C

/m)

m 2m 1C/m 1 3

0

===

baq

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Electric Flux Density of an Infinite Line Charge Using Gauss’s Law

Consider a infinite line charge carrying charge per unit length of qel:

z

elq

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Electric Flux Density of an Infinite Line Charge Using Gauss’s Law Cont.



surfaces

( )ρρρ DaD ˆ=

cylinders of radius ρ where

∞≤≤ ρ0

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Electric Flux Density of an Infinite Line Charge Using Gauss’s Law Cont


∫=L

elencl dlqQ

lqQ elencl =cylinder is

infinitely long!

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DSsdDS

=⋅∫

( ) lDsdDS

ρπρρ 2=⋅∫


surface.


surface.

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SQD encl=

ρπρ 2ˆ elqaD =

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Gauss’s Law in Integral Form

∫∫ ==⋅V

evenclS

dvqQsdD

V S

sd

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15

QUESTIONS

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15

Questions?

? ?

aem ece 351 e-flux density(2)

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