3- ELECTRIC FLUX DENSITY AND
GAUSS’S LAW
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3- ELECTRIC FLUX DENSITY AND GAUSS’S LAW
3.1 Electric Flux Density
3- ELECTRIC FLUX DENSITY AND GAUSS’S LAW
3.1 Electric Flux Density
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The electric field is dependent on the medium in which
the charge is placed.
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A new vector called the Electric flux density D measured
in ( C/m2 ) which is independent of the medium and
defined for free space by:
and for any material as:
where
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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.
Solution:
Let D = DQ + DL where DQ and DL are flux densities due
to the point charge and line charge, respectively, as
shown in the figure
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3.2- Gauss's Law
3.2- Gauss's Law
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Gauss’s law states that:
The electric flux density (D) passing through any closed
surface is equal to the total charge enclosed ( Qenc ) by
that surface.
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The closed surface (known as Gaussian surface) is
chosen such that:
D is normal or tangential to the Gaussian surface and
ІDІ is constant on that surface.
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When D is normal to the surface, D • dS = D dS because
D is constant on the surface.
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When D is tangential to the surface, D • dS = 0.
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The dS is defined as the differential surface (or
area) defined as
dS = dSa
n
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where dS is the area of the surface element
and a
n
is a unit vector normal to the surface
dS and directed away from the volume.
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The differential normal area dS in:
A- Cartesian Coordinates is shown in the figures.
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B- Cylindrical Coordinates is shown in the figures.
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C- Spherical Coordinates is shown in the figures.
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Example: Point Charge
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A point charge Q is located at the origin. Determine
D at a point P
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Since D is everywhere normal to the Gaussian
surface, that is, D = Dr ar . Applying Gauss's
law ( Qenclosed = Q ) gives:
where
is the area of the Gaussian surface. Thus
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Example: Infinite Line Charge
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An infinite line of uniform charge pL C/m lies
along the z-axis. Determine D at a point P.
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we choose a cylindrical surface containing P to
satisfy symmetry condition as shown in Figure. D is
constant on and normal to the cylindrical Gaussian
surface; that is, D = Dp ap. If we apply Gauss's law
to an arbitrary length l of the line:
where is the surface area of the
Gaussian surface. Note that
evaluated on the
top and bottom surfaces of the cylinder is zero since D has
no z-component; that means that D is tangential to those
surfaces. Thus
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Example: Uniformly Charged Sphere
A sphere of radius a with a uniform volume charge
density pv C/m3. Determine D everywhere.
Solution:
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We construct Gaussian surfaces for r < a and r > a
separately.
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Since the charge has spherical symmetry, it is
obvious that a spherical surface is an appropriate
Gaussian surface.
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For r < a: The total charge enclosed by the spherical
surface of radius r, as shown in figure (a), is
and
Hence, or
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For r > a, the Gaussian surface is shown in
figure (b).
and then or
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Thus,
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and |D| is as sketched in the figure
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Example: A charge distribution with spherical
symmetry has density
Determine E everywhere.
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Solution:
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For r < R:
or
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For r > R:
or
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Example: A coaxial cable
Two coaxial cylindrical conductors of inner radius a
and outer radius b are infinite in extent . Assume a
total charge +Q is placed on the outer surface of the
inner cylinder and –Q on the inner surface of the
outer cylinder, find D everywhere.
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We choose Gaussian surface as a circular
cylinder of length L and radius ρ
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For Gaussian surface of ρ < a:
Qenc = 0 , then: D = 0 for ρ < a.
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For Gaussian surface a < ρ < b:
We will assume a charge distribution of ρS on the
outer surface of the inner conductor
from which we have:
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Because the charge on the inner cylinder must
terminate on a negative charge on the inner surface
of the outer cylinder, the total charge on that
surface must be
and the surface charge on the outer cylinder is
found as
or
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For Gaussian surface of ρ > b:
The total charge enclosed would then be zero,
for there are equal and opposite charges on
each conducting cylinder. Hence
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Example:
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. Find the charge density on each conductor,
and the E and D fields.
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