12
1. INTRODUCTION
1.3 DEP MICROFLUIDIC CHIPS
1.3.1 THEORY OF DEP
When a particle is polarized in an electric eld, the charges in the particle are redistributed. e
positive and negative charges move in opposite directions, causing a heterogeneous induced charge
at the interface between the particle and the solution [101]. Due to the dierent polarizabilities of
the solution and particle, the amount of induced charges generated at their interface is dierent, as
shown in Figure 1.8.
8
8
(
3
3
S
8
8
(
3
3
S
(a)(b)
Figure 1.8: Schematic diagram of particle polarization: (a) the particle polarizability is lower than the
solution; and (b) the particle polarizability is larger than the solution.
If the polarizability of the solution is greater than that of the particle, the amount of induced
charges generated by the solution polarization is more than that of the particle, so the lower in-
terface of the particle will be positively charged and the upper interface will be negatively charged,
as shown in Figure 1.8(a). On the contrary, the lower side interface of the particle is negatively
charged and the upper side is positively charged, as shown in Figure 1.8(b).
A charged system consisting of two equal-point charge +q and -q at a distance d is called an
electric dipole, as shown in Figure 1.9. Under the action of an electric eld, the particle is polarized
to form an electric dipole. When the distance d between the set point charges +q and -q is much
smaller than u (the distance of the negative charge from the origin point), the dipole moment p
(C·m) can be expressed as:
p = qd
.
(1-1)
e direction is a negative charge pointing to a positive charge.
13
E
⃗
(u
⃗
+ d
⃗
)
d
⃗
E
⃗
(u
⃗
)
u⃗
O (0,0,0)
+
-
Figure 1.9: Schematic diagram of dipole moment.
Under the action of the non-uniform electric eld E, the resultant force F of the electric eld
of the electric dipole is
F
= qE
(u+d
) – qE
(u). (1-2)
e rst term on the right-hand side of Equation (1-2) is subjected to Taylor series expan-
sion and after omitting the high order term one can get
qE
(u+d
) = qE
(u) + qd
∇E
(u). (1-3)
Combining Equations (1-2) and (1-3) yields
F
= qd
∙ ∇E
.
(1-4)
Substituting Equation (1-1) into (1-4)
F
= p
∙ ∇E
. (1-5)
For a particle with radius r
p
, the dipole moment can be expressed as [102]
p
= 4πr
p
3
ε
m
∙ K
CM
∙ ∇E
, (1-6)
where K
CM
is the Clausius-Mossotti factor, dened as
K
CM
= , (1-7)
where ε
p
*
and ε
m
*
are the complex permittivity of the particle and the solution, respectively.
ε
p
*
= ε
p
– j
σ
p
, (1-8)
ε
m
*
= ε
m
– j
σ
m
, (1-9)
ε
p
*
– ε
m
*
ε
p
*
+ 2ε
m
*
ω
ω
1.3 DEP MICRODLUIDIC CHIPS
14
1. INTRODUCTION
where ε
p
and σ
p
are the permittivity and conductivity of particle, respectively; ε
m
and σ
m
are the
permittivity and conductivity of solution, respectively; and ω is the angular frequency of the electric
eld.
erefore, the DEP force can be expressed as:
F
DEP
=4πr
p
3
ε
m
∙ K
CM
∙ (E ∙ ∇) E
= 2πr
p
3
ε
m
Re[K
CM
]∇E
2
.
(1-10)
It can be seen from Equation (1-7) that the K
CM
coecient are related to the complex per-
mittivity of solution and particle and the frequency of electrical signal, as shown in Figure 1.10.
When Re[K
CM
]>0, the DEP force is positive DEP (pDEP), and under the pDEP force,
the particle will be attracted to the region with larger gradient of electric eld squared; when
Re[K
CM
]<0, the DEP force is negative DEP (nDEP), and under the nDEP force, the particle will
be pushed to the region with smaller gradient of electric eld squared; when Re[K
CM
] = 0, the DEP
force is 0, and the corresponding frequency of electrical signal is the crossover frequency.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
pDEP
(a)
nDEP
(b)
Figure 1.10: Schematic diagram of positive and negative DEP: (a) positive DEP; and (b) negative DEP.
When the particle is in a rotational electric eld, since the dipole generation takes certain
time, there is a phase shift θ between the direction of the dipole moment p and the direction of the
electric eld E, and the dipole moment is rotated parallel to the direction of the electric eld, so the
dipole moment p will continuously rotate with the electric eld.
e average torque in a rotational electric eld is the DEP torque [103]:
Γ
ROT
= – ½ Re[p×E
]. (1-11)
Substituting Equation (1-6) into (1-11) yields
Γ
ROT
= –4πε
m
r
p
3
Im[K
CM
] E
2
. (1-12)
..................Content has been hidden....................
You can't read the all page of ebook, please click here login for view all page.