Now we have to review some field theory. We will be using
fields from time to time in this course, and when we need some
aspect of field theory, we will introduce what we need at that
point. This seems to make more sense than spending several
weeks talking about a lot of abstract theory without seeing how
or why it can be useful.

The first thing we need to remember is Gauss' Law.
Gauss' Law, like most of the fundamental laws of
electromagnetism comes not from first principle, but
rather from empirical observation and attempts to match
experiments with some kind of self-consistent mathematical
framework. Gauss' Law states that:

∮s,Dd
S
=
Q
encl
=∮v,ρvd
V
s
S
D
Q
encl
v
V
ρ
v

(1)
where

D D is the

electric
displacement vector, which is related to the

electric field vector,

E
E, by the relationship

D=εE
D
ε
E
.

εε is called the

dielectric constant. In silicon it has a value of

1.1×10-12Fcm
1.1-12
F
cm
. (Note that

D D must have
units of

Coulombscm2
Coulombs
cm
2
to have everything work out OK.)

Qencl
Qencl
is the total amount of charge
enclosed in the volume

V V, which is
obtained by doing a volume integral of the charge density

ρv
ρ
v
.

Equation 1 just says that if you add up the
surface integral of the displacement vector
DD over a closed surface
SS , what you get is the sum of
the total charge enclosed by that surface. Useful as it is, the
integral form of Gauss' Law, (which is what Equation 1 is) will not help us much in understanding
the details of the depletion region. We will have to convert
this equation to its differential form. We do this by first
shrinking down the volume VV until
we can treat the charge density
ρv
ρ
v
as a constant ρρ,
and replace the volume integral with a simple product. Since we
are making VV small, let's call it
ΔV
Δ
V
to remind us that we are talking about just a small
quantity.

∮Δv,ρvd
V
→ρΔv
Δ
v
V
ρ
v
ρ
Δ
v

(2)
And thus, Gauss' Law becomes:

∮s,Dd
S
=ε∮s,Ed
S
=ρΔV
s
S
D
ε
s
S
E
ρ
Δ
V

(3)
or

1ΔV(∮s,Ed
S
)=ρε
1
Δ
V
s
S
E
ρ
ε

(4)
Now, by

*definition* the limit of the LHS of

Equation 4 as

ΔV→0
Δ
V
0
is known as the divergence of the vector

EE,

divE
E
. Thus we have

limit
ΔV
→
0
1ΔV(∮s,Ed
S
)=divE=ρε
Δ
V
0
1
Δ
V
s
S
E
E
ρ
ε

(5)
Note what this says about the divergence. The divergence of the
vector

EE is the
limit of the surface integral of

EE over a volume

VV, normalized by the volume
itself, as the volume shrinks to zero. I like to think of as a
kind of "point surface integral" of the vector

EE.

If EE only varies in one dimension,
which is what we are working with right now, the expression for
the divergence is particularly simple. It is easy to work out
what it is from a simple picture. Looking at Figure 2 we see that if EE is only pointed along one
direction (let's say xx) and is
only a function of xx, then the
surface integral of EE over the volume
ΔV=ΔxΔyΔz
Δ
V
Δ
x
Δ
y
Δ
z
is particularly easy to calculate.

∮s,Ed
S
=Ex+ΔxΔyΔz−ExΔyΔz
s
S
E
E
x
Δ
x
Δ
y
Δ
z
E
x
Δ
y
Δ
z

(6)
Where we remember that the surface integral is defined as being
positive for an outward pointing vector and negative for one
which points into the volume enclosed by the surface. Now we
use the definition of the divergence

divE=limit
ΔV
→
0
1ΔV(∮s,Ed
S
)=limit
ΔV
→
0
(Ex+Δx−Ex)ΔyΔzΔxΔyΔz=limit
ΔV
→
0
Ex+Δx−ExΔx=∂Ex∂
x
E
Δ
V
0
1
Δ
V
s
S
E
Δ
V
0
E
x
Δ
x
E
x
Δ
y
Δ
z
Δ
x
Δ
y
Δ
z
Δ
V
0
E
x
Δ
x
E
x
Δ
x
x
E
x

(7)
So, we have for the differential form of Gauss' law:

Thus, in our case, the rate of change of
EE with
xx,
dd
x
E
x
E
, or the *slope of
Ex
E
x
* is just equal to the charge density,
ρx
ρ
x
, divided by εε.

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