Transcript Document

Firohman
135060300111012
Current is a flux quantity and is defined as:
Current density, J, measured in Amps/m2 , yields current in Amps when
it is integrated over a cross-sectional area. The assumption would be
that the direction of J is normal to the surface, and so we would write:
In reality, the direction of current flow may not be normal to the surface in question, so we treat
current density as a vector, and write the incremental flux through the small surface in the usual way:
where S = n da
n
Then, the current through a large surface
is found through the flux integral:
Consider a charge Q, occupying volume v, moving in the positive x direction at velocity vx
In terms of the volume charge density, we may write:
Suppose that in time t, the charge moves through a distance x = L = vx t
Then
The motion of the charge represents a current given by:
or
We now have
The current density is then:
So that in general:
Suppose that charge Qi is escaping from a volume through
closed surface S, to form current density J. Then the total
current is:
Qi(t)
where the minus sign is needed to produce positive outward
flux, while the interior charge is decreasing with time.
We now apply the divergence theorem:
so that
or
The integrands of the last expression
must be equal, leading to the
Equation of Continuity
Free electrons move under the influence of an electric field. The applied force on an electron
of charge Q = -e will be
When forced, the electron accelerates to an equilibrium velocity, known as the drift velocity:
where e is the electron mobility, expressed in units of m2/V-s. The drift velocity is used to find the
current density through:
The expression:
from which we identify the conductivity
for the case of electron flow :
In a semiconductor, we have hole current as well, and
S/m
is Ohm’s Law in point form
Consider the cylindrical conductor shown here, with voltage V applied across the ends. Current flows
down the length, and is assumed to be uniformly distributed over the cross-section, S.
First, we can write the voltage and current in the cylinder in terms of field quantities:
Using Ohm’s Law:
We find the resistance of the cylinder:
a
b
a
b
Consider a conductor, on
which excess charge has been placed
E
+ + +
++
+
+
+
+
+
+ s
solid conductor
E = 0 inside
+
+
+
+
+
+ + + + +
Electric field at the surface
points in the normal direction
1.
Charge can exist only on the surface as a surface charge density, s -- not in the interior.
2.
Electric field cannot exist in the interior, nor can it possess a tangential component at the surface
(as will be shown next slide).
3.
It follows from condition 2 that the surface of a conductor is an equipotential.
Over the rectangular integration path, we use
or
To find:
These become negligible as h approaches zero.
dielectric
Therefore
n
More formally:
conductor
Gauss’ Law is applied to the cylindrical surface shown below:
This reduces to:
as h approaches zero
dielectric
Therefore
n
More formally:
s
conductor
Tangential E is zero
At the surface:
Normal D is equal to the surface charge density
The Theorem of Uniqueness states that if we are given a configuration of charges and
boundary conditions, there will exist only one potential and electric field solution.
In the electric dipole, the surface along the plane of symmetry is an equipotential with V = 0.
The same is true if a grounded conducting plane is located there.
So the boundary conditions and charges are identical in the upper half spaces of both configurations
(not in the lower half).
In effect, the positive point charge images across the conducting plane, allowing the conductor to be
replaced by the image. The field and potential distribution in the upper half space is now found much
more easily!
Each charge in a given configuration will have its own image
In dielectric, charges are held in position (bound), and ideally there are no free charges that can move
and form a current. Atoms and molecules may be polar (having separated positive and negative charges),
or may be polarized by the application of an electric field.
Consider such a polarized atom or molecule, which possesses a dipole moment, p, defined as the charge
magnitude present, Q, times the positive and negative charge separation, d. Dipole moment is a vector
that points from the negative to the positive charge.
Q
d
p = Qd ax
A dielectric can be modeled as an ensemble of bound charges in free space, associated with
the atoms and molecules that make up the material. Some of these may have intrinsic dipole moments,
others not. In some materials (such as liquids), dipole moments are in random directions.
The number of dipoles is
expressed as a density, n
dipoles per unit volume.
The Polarization Field of the
medium is defined as:
v
[dipole moment/vol]
or
[C/m2]
Introducing an electric field may increase the charge separation in each dipole, and possibly re-orient dipoles so that
there is some aggregate alignment, as shown here. The effect is small, and is greatly exaggerated here!
E
The effect is to increase P.
= np
if all dipoles are identical
Consider an electric field applied at an angle  to a surface normal as shown. The resulting
separation of bound charges (or re-orientation) leads to positive bound charge crossing upward
through surface of area S, while negative bound charge crosses downward through the surface.
Dipole centers (red dots) that lie within the range (1/2) d cos above or below
the surface will transfer charge across the surface.
E
The total bound charge that crosses the surface is given by:
S
volume
E
P
The accumulation of positive bound charge within a closed
surface means that the polarization vector must be pointing
inward. Therefore:
S
-
+
+
+
+ qb
-
P
-
Now consider the charge within the closed surface
consisting of bound charges, qb , and free charges, q.
The total charge will be the sum of all bound and free
charges. We write Gauss’ Law in terms of the total
charge, QT as:
where
free charge
QT = Qb + Q
bound charge
E
QT
S
q+ + +
+
+
+
+
+q
b
P
We now have:
and
where
QT = Qb + Q
combining these, we write:
we thus identify:
which we use in the familiar form
of Gauss’ Law:
Taking the previous results and using the divergence theorem, we find the point form expressions:
Bound Charge:
Total Charge:
Free Charge:
A stronger electric field results in a larger polarization in the medium. In a linear medium, the relation
between P and E is linear, and is given by:
where e is the electric susceptibility of the medium.
We may now write:
where the dielectric constant, or relative permittivity is defined as:
Leading to the overall permittivity of the medium:
where
In an isotropic medium, the dielectric constant is invariant with direction of the applied electric field.
This is not the case in an anisotropic medium (usually a crystal) in which the dielectric constant will vary
as the electric field is rotated in certain directions. In this case, the electric flux density vector components
must be evaluated separately through the dielectric tensor. The relation can be expressed in the form:
We use the fact that E is conservative:
n
So therefore:
Region 1
1
Leading to:
Region 2
2
More formally:
We apply Gauss’ Law to the cylindrical volume shown here,
in which cylinder height is allowed to approach zero, and there
is charge density s on the surface:
n
Region 1
1
s
Region 2
2
The electric flux enters and exits only through the bottom and top surfaces, respectively.
More formally:
From which:
and if the charge density is zero:
We wish to find the relation between the angles 1 and 2, assuming no charge density on the surface.
The normal components of D will be continuous across
the boundary, so that:
Then, with tangential E continuous across the
boundary, it follows that:
Now, taking the ratio of the two underlined
equations, we finally obtain:
or…