Even though dust particles are neutral, they tend to be attracted to a charged surface. I am guessing this is due to charge induction.
Is there a way I can compute the attraction? how will it vary based on:
-
separation distance.
I think the distance will affect as inverse square law (1/r^2) [Coulomb's law]. But to calculate the force, I need to know how much charge has been induced ( how do I find that?) and does that vary due to separation distance. -
type of dust.
Do all material act the same , or are some material gets more charge induced? ( metals, for example with more free electrons) -
the size of dust particles.
Intuition suggests that bigger particles will be induced less, but I am not sure.
Best Answer
The attraction is that between a charge and an induced dipole. If the charged object is a sphere, the field is $QE\over 4\pi r^2$, in units where $\epsilon_0=1$. This field is what induces the dipole and causes the attraction.
Once you get a dipole moment in the dust, the dust dipole is attracted to regions of stronger field with a force that goes like the derivative of the E field. The total dipole is neutral, so you get net opposite positive and negative forces which only fail to cancel out to the extent that the E field is stronger in the regions of induced positive charge. The magnitude of the force is (in the nearly perfect small-dust approximation) $ d \cdot \nabla E$, the dipole moment can be thought of as defined by this equation (although that's not the definition--- the dipole moment vector d is the sum of qx over all the little infinitesimal regions in the dust, where q is the charge of the region, which sums to zero over the dust particle, and x is the position).
So if the dipole moment is of fixed size, the force is $2Qd\over r^3$. But the dipole itself is proportional to E, with a coefficient that I'll call "p" for polarizability, so you get a force which is
$$ {2 Q \over 4\pi r^3} { pQ\over 4\pi r^2} = {2pQ^2\over 16\pi^2 r^5}$$
and this is the force between a sphere and a dipole (in units where $k={1\over 4\pi\epsilon_0} = 1$). You see it falls off as $1/r^5$, two powers of r for the induced dipole strength, proportional to the E field, and another three powers from the force, which is as the gradient of E.
To calculate p requires knowing something about the material, namely it's dielectric polarizability constant for DC electric fields. This is radically different for different materials, depending on whether the molecules are polar themselves, and how mobile they are, whether it's liquid or solid. In general, you can figure this out from the extent to which the electric field in the interior of the dust is reduced from the exterior, assuming a bulk block of dust material. This is the dielectric constant $\epsilon$ of the dust.
For a solid block of dust in a constant electric field E perpendicular to the plane surface of the dust-block, the electric field in the interior of the dust block is $E\over \epsilon$ where $\epsilon$ for DC fields is always bigger than 1. For a metal, the field in the interior is zero, and this corresponds to infinite $\epsilon$. I will calculate the polarization constant in an order of magnitude for metal dust, but for the largely nonpolar typical carbon-chain dust material, the actual answer will range from 1%-10% of the metal answer.
For the metal, there is a nice useful trick for getting the induced dipole moment magnitude, which is the method of images/conformal maps. A dipole at the origin is conformally equivalent by an inversion around a sphere of certain radius to a constant electric field at infinity, so if you place a dipole at the origin, and the equal electric field inverted, the result makes the sphere have the same potential.
The electric potential of a dipole is by differentiating the electric potential of a point charge:
$$ \phi = {dz \over 4\pi r^3}$$
The potential of constant z-direction electric field is
$$ \phi = Ez$$
adding the two, the surface at potential zero (only a zero potential stays an equipotential under inversion) is a sphere:
$$ Ez - {dz\over 4\pi r^3} = 0$$
This is solved by a sphere (away from the z=0 plane which is an accidental equipotential by symmetry)
$$ d = 4\pi E r^3 $$
and so the induced dipole moment is proportional to the cube of the sphere size. The coefficient p is $4\pi a^3$ for a metal spherical dust of radius a.
So the answer for an ideal resistive metal dipole (perfectly screening object, zero response time to adjust to a new field, these are good assumptions for this situation) is
$$ F = {Q^2 a^3 \over 4\pi \epsilon_0 r^5} $$
Where I restored $\epsilon_0$ in the final answer. For a non-metal resistive dipole, the induced moment is less by the facor $(1-{1\over \epsilon})$. For hydrocarbons, the dielectric constant is about 2, so you get 1/2 the force. Note than when r is of order a, the maximum force goes up as the object gets small.