Firohman

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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 bethat the direction of J is normal to the surface, and so we would write:

n

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

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

or

The motion of the charge represents a current given by:

We now have

The current density is then:

So that in general:

Qi(t)

Suppose that charge Qi is escaping from a volume throughclosed surface S, to form current density J. Then the total current is:

where the minus sign is needed to produce positive outwardflux, 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 electronof 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:

from which we identify the conductivityfor the case of electron flow :

The expression:

is Ohm’s Law in point formS/m

In a semiconductor, we have hole current as well, and

Consider the cylindrical conductor shown here, with voltage V applied across the ends. Current flowsdown 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

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.

s

+

+

+++

+

+

++ +

+

+++

+

+

+ ++

+

+

E

solid conductor

Electric field at the surfacepoints in the normal direction

E = 0 inside

Consider a conductor, onwhich excess charge has been placed

conductor

dielectric

n

Over the rectangular integration path, we use

To find:

or

These become negligible as h approaches zero.

Therefore

More formally:

n

dielectric

conductors

Gauss’ Law is applied to the cylindrical surface shown below:

This reduces to: as h approaches zero

Therefore

More formally:

At the surface:

Tangential E is zero

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

Qd p = Qd ax

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 vectorthat points from the negative to the positive charge.

A dielectric can be modeled as an ensemble of bound charges in free space, associated withthe 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.

[dipole moment/vol] or [C/m2]

v

The number of dipoles isexpressed as a density, ndipoles per unit volume.

The Polarization Field of the medium is defined as:

E

Introducing an electric field may increase the charge separation in each dipole, and possibly re-orient dipoles so thatthere is some aggregate alignment, as shown here. The effect is small, and is greatly exaggerated here!

The effect is to increase P.

= npif all dipoles are identical

E

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 upwardthrough 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 belowthe surface will transfer charge across the surface.

E

The total bound charge that crosses the surface is given by:

S

volume

P

The accumulation of positive bound charge within a closedsurface means that the polarization vector must be pointing inward. Therefore:

S

P

+

+

+ +

-

-

-

-

qb

Now consider the charge within the closed surfaceconsisting of bound charges, qb , and free charges, q.The total charge will be the sum of all bound and freecharges. We write Gauss’ Law in terms of the total charge, QT as:

where

QT = Qb + Q

bound charge

free charge

S

P

+

+

+ +

qb

+

++

+

E

q

QT

QT = Qb + Q

We now have:

and

where

combining these, we write:

we thus identify: which we use in the familiar formof 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 relationbetween 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 componentsmust be evaluated separately through the dielectric tensor. The relation can be expressed in the form:

Region 1

Region 2

n

2

1

We use the fact that E is conservative:

So therefore:

Leading to:

More formally:

nRegion 11

Region 22

We apply Gauss’ Law to the cylindrical volume shown here,in which cylinder height is allowed to approach zero, and thereis charge density s on the surface:

The electric flux enters and exits only through the bottom and top surfaces, respectively.

s

From which:

and if the charge density is zero:

More formally:

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 acrossthe boundary, so that:

Then, with tangential E continuous across theboundary, it follows that:

or…

Now, taking the ratio of the two underlined equations, we finally obtain: