Image Processing Reference
In-Depth Information
ð
z
0
r(
z
0
)d
z
0
þ
E
0
1
e
2
E
2
(
z
) ¼
(
10
:
39
)
The constant E
0
is determined using Gauss
is law at the boundary condition. The
'
electric
fields E
1
and E
3
at the boundary are given by
E
1
¼
e
2
e
1
E
2
d
ðÞ
E
3
¼
e
2
(
10
:
40
)
e
3
E
2
(
0
)
The applied potential, V
a
, is equal to the sum of the potentials across each layer that
is obtained from the
fields and distances:
d
2
ð
V
a
¼
E
1
d
1
E
2
(
z
)d
z
E
3
d
3
(
10
:
41
)
0
Equations 10.39 through 10.41 are solved to
nd the unknown
field, E
0
:
ð
ð
ð
d
2
0
r(
z
)d
z
d
2
z
0
r(
z
0
)d
z
0
d
z
V
a
e
2
D
D
1
e
2
D
1
E
0
¼
(
10
:
42
)
2
D
e
0
where
X
3
d
i
e
i
D
i
¼
i
¼
1,2,3
and
D
¼
D
i
(
10
:
43
)
i
¼
1
From Equation 10.39, since E
2
(z
s
)
¼
0, using this useful result and Equations 10.41
and 10.42, we obtain
ð
ð
ð
ð
z
s
0
r(
z
)d
z
¼e
2
E
0
¼
d
2
0
r(
z
)d
z
þ
d
2
z
0
r(
z
0
)d
z
0
d
z
V
a
D
þ
D
1
D
1
(
10
:
44
)
2
D
e
0
A solution for Equation 10.39 is obtained for three cases: (1) for constant charge
density in the layer 3 containing insulating particles (i.e.,
r¼r
0
), (2) for exponential
r¼r
0
e
g
z
where
charge distribution in the toner particles (i.e.,
represents the
charge-penetration parameter), and (3) for a more complex charge distribution with
two exponentials that can be found in a photoactive particle electrophotography
(i.e.,
g
r
2
e
g
(d
z)
). The two exponentials in the expression can be
thought of as the result of two processes: discharging the particle layer by exposure
to light through one dielectric layer in the presence of an electric
r¼r
1
e
g
z
field, and corona
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