Dielectric in an electrostatic field

AI Thread Summary
When a small dielectric, like paper, is placed in an electrostatic field, it becomes polarized, resulting in a surface charge that creates an upward force if the field is positive. The force on the dielectric can be calculated using specific formulas, such as F=[(1-epsilon/2+epsilon)]2q^2R^3/d^5 in Gaussian units. In a uniform electric field, like that between two parallel plates, no net force acts on the dielectric, as it experiences equal forces on both sides. A non-uniform field is necessary to generate a force that can move the dielectric. Understanding these principles is crucial for analyzing the behavior of dielectrics in electrostatic fields.
Nakis
Messages
3
Reaction score
0
Hi all,

I am sorry this could sound like heard many times before, but I am trying to understand this problem and found nothing in my physics books.

If I put a small dielectric (bit of paper) into an electrostatic field (for example generated by a rubbed plastic rod), the paper will be attracted. In theory, the electric field causes a polarization in the paper, which amounts to a surface charge... If the rod is positive, a negative charge will appear in the paper surface, resulting in an "upward" force. How can his force be calculated?

Furthermore, how does this work if I use two parallel conductor plates, put the paper on a plate and apply some potential? Here, ithe charges would be different on each side of the paper, so that it would never leave one surface to find some equilibrium position...

I know it's weird, but anyway thanks for any help !

-Nakis
 
Physics news on Phys.org
The electric field produces an electric dipole moment in a small chunk of dielectric.
If the field is uniform (as between two parallel plates, there will be no force, as you suggested. You need a non-uniform field to get a force. For instance, the force on a dielectric sphere of radius R a distance d from a point charge q is
F=[(1-epsilon/2+epsilon)]2q^2R^3/d^5. This is in Gaussian units.
It is a bit more complicated in SI. The dielectric factor would be somewhat different
(and difficult to calculate) for other shapes
 
Thread 'Motional EMF in Faraday disc, co-rotating magnet axial mean flux'
So here is the motional EMF formula. Now I understand the standard Faraday paradox that an axis symmetric field source (like a speaker motor ring magnet) has a magnetic field that is frame invariant under rotation around axis of symmetry. The field is static whether you rotate the magnet or not. So far so good. What puzzles me is this , there is a term average magnetic flux or "azimuthal mean" , this term describes the average magnetic field through the area swept by the rotating Faraday...
Back
Top