Electrostatics -- Dielectrophoresis

Part III: Dielectrophoretic force calculation

Things we must consider for the calculation of the dielectrophoretic force: as we've seen earlier, dielectrophoresis effect is generated by applying a non uniform electric field on a dielectric object (here we will use a dielectric sphere) in a dielectric environment to transform electrostatic energy in mechanical energy. We will model these hypothesis for our calculation of the dielectrophoretic force.
The key is to polarize the dielectric object, so that the sphere will behave as if it was charged in surface. Then, a non-uniform electric field will apply forces on the sphere following Coulomb's law. As the electric field is non-uniform, the force will be higher on one side than on another and we get a not null force balance: the micro-object is moved by dielectrophoresis!

Coulomb's force...

What we call Coulomb's force and we note $\vec{F_{C}}$ is the force applied on a charged particle $q$ by an electric field $\vec{E_{0}}$ . It is defined by the relation:

$\vec{F_{C}} = q . \vec{E_{0}}$

...applied to a dipole by an electric non-uniform field

Let's consider now a dipole ${+ Q — - Q}$ aligned along the $O_{z}$ .

Electrostatic dipole in non-uniform electric field
Schematic of an electrostatic dipole in an non-uniform electric field. The Coulomb's force are not symmetric, so the balance mechanical force applied on the dipole is not null.

So we have:

$\vec{F_{C}} = Q . \vec{E (z + \frac{d}{2})} - Q . \vec{E (z - \frac{d}{2})}$

where $δz$ is small before $\vec{E_{0} (z)}$ . Then we can write:

$\vec{E_{0} (z + δz)} \approx \vec{E_{0} (z)} + \vec{\nabla E_{0} (z)}$

that leads to:

$\vec{F_{C}} = Q . d . \vec{\nabla E_{0} (z)}$

We can recognize here $Qd \vec{u_{z}}$ that is a dipolar moment we will note $\vec{m}$ , a vector which direction is along $O_{z}$ . So we can note:

$\vec{F_{C}} = (\vec{m} . \vec{\nabla}) \vec{E_{0} (z)}$

To go further away in the development, we consider now that the electric field has an alternative amplitude and depends on not only $z$ , but also on all other spatial coordonnates.

$\begin{matrix} \vec{E (\vec{r}, t)} & = & \vec{E_{x,0} (\vec{r})} \cos (ωt + Φ_{x} (\vec{r})) \vec{u_{x}} \\ + & \vec{E_{y,0} (\vec{r})} \cos (ωt + Φ_{y} (\vec{r})) \vec{u_{y}} \\ + & \vec{E_{z,0} (\vec{r})} \cos (ωt + Φ_{z} (\vec{r})) \vec{u_{z}} \end{matrix}$

Back to the sphere case

We take back the case of a lossy dielectric sphere in a lossy dielectric environment. The dipolar moment become:

$\vec{m} = m_{x} (t) . \vec{u_{x}} + m_{y} (t) . \vec{u_{y}} + m_{z} (t) \vec{u_{z}}$

and the integration of the polarization vector seen for the sphere becomes:

$\vec{m} = 4 π ϵ_{0} ϵ_{ext} R^{3} [\begin{matrix} E_{x,0} (\vec{r}) . (ℜ [K (ω)] \cos (ωt + Φ_{x} (\vec{r})) + ℑ [K (ω)] \sin (ωt + Φ_{x} (\vec{r}))) \\ E_{y,0} (\vec{r}) . (ℜ [K (ω)] \cos (ωt + Φ_{y} (\vec{r})) + ℑ [K (ω)] \sin (ωt + Φ_{y} (\vec{r}))) \\ E_{z,0} (\vec{r}) . (ℜ [K (ω)] \cos (ωt + Φ_{z} (\vec{r})) + ℑ [K (ω)] \sin (ωt + Φ_{z} (\vec{r}))) \end{matrix}]$

We now have the electric field in spatial and temporal coordinates, $\vec{E (\vec{r}, t)}$ , the dipolar moment $\vec{m}$ . These two components can give us the expression of the force seen earlier:

$\vec{F_{c} (\vec{r}, t)} = \sum_{i = x, y, z} (m_{i} (t) \frac{\partial E_{x} (t)}{\partial i} \vec{u_{x}} + m_{i} (t) \frac{\partial E_{y} (t)}{\partial i} \vec{u_{y}} + m_{i} (t) \frac{\partial E_{z} (t)}{\partial i} \vec{u_{z}})$

These values are instantaneous, and so are not much interesting to calculate the displacement of a particle, the displacement being far less fast than the electric field variations. We need to calculate an average value on a temporal period to get the true force applied on the particle. To keep things simple, let's consider only one component of the vector:

$\begin{matrix} m_{i} \frac{\partial E_{j} (t)}{\partial i} & = & 4 π ϵ_{0} ϵ_{ext} R^{3} & (ℜ [K (ω)] \cos (ωt + Φ_{i}) & - & ℑ [K (ω)] \sin (ωt + Φ_{i})) \\ \times & (\frac{\partial E_{j, 0}}{\partial i} \cos (ωt + Φ_{i}) & - & \frac{\partial Φ_{i}}{\partial i} E_{j, 0} \sin (ωt + Φ_{i})) \end{matrix}$

We calculate the temporal average on one period:

$<m_{i} (t) \frac{\partial E_{j} (t)}{\partial i}> = 2 π ϵ_{0} ϵ_{ext} R^{3} (ℜ [K (ω)] E_{j, 0} \frac{\partial E_{i, 0}}{\partial i} + ℑ [K (ω)] E_{j, 0}^{2} \frac{\partial Φ_{i}}{\partial i})$

So we can calculate the average force:

$<\vec{F_{C}}> = 2 π ϵ_{0} ϵ_{ext} R^{3} (ℜ [K (ω)] \vec{\nabla E_{eff}^{2}} + ℑ [K (ω)] (E_{x, 0}^{2} \vec{\nabla Φ_{x}} + E_{y, 0}^{2} \vec{\nabla Φ_{y}} + E_{z, 0}^{2} \vec{\nabla Φ_{z}}))$

In which $E_{eff}$ is the root mean square value of $E_{0}$ , that is, for a sinusoidal function, $E_{0} / \sqrt{2}$ . In practice, dielectrophoretic excitation signals are often sinusoidal, and we use 180° phased signals to get a non-uniformity. We can take those points in account and simplify the expression (no phase gradient):

$<\vec{F_{C}}> = 2 π ϵ_{0} ϵ_{ext} R^{3} (ℜ [K (ω)] \vec{\nabla E_{eff}^{2}})$

The Clausius-Mossotti factor depends on the excitation frequency. It is the key of the dielectrophoretic force. In this using of the dielectrophoretic force, there are many available applications. There is another way of using the dielectrophoretic effect, this time considering the phase gradient, called Travelling Waves Dielectrophoresis (TWD) that allows design of a large stroke micro-object conveyer.

Electrostatics -- Dielectrophoresis

Part III: Dielectrophoretic force calculation

Coulomb's force...

...applied to a dipole by an electric non-uniform field

Back to the sphere case

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Electrostatics -- Dielectrophoresis

Part III: Dielectrophoretic force calculation

Coulomb's force...

...applied to a dipole by an electric non-uniform field

Back to the sphere case

Contact matthieu.lagouge AT free.fr

Contact
matthieu.lagouge AT free.fr