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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 (D)
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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Electric Flux Density (D)
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:
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Electric Flux Density (D)
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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Electric Flux Density (D)
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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Applications of Gauss's Law 1
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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Applications of Gauss's Law 1
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
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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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Applications of Gauss's Law 2
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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Applications of Gauss's Law 2
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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Applications of Gauss's Law 2
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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Applications of Gauss's Law 2
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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Applications of Gauss's Law 2
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
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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
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