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Here we will animate nanoscopic reactions of dipole distributions due to monopole charge distributions. These reactions are usually expressed macroscopically using a term called the polarizability. This leads to a concept called the permittivity, `epsilon` which expresses Maxwell's displacement field `bbD=epsilonbbE` where `bbE` is the electric intensity. Even in vacuum the value of `epsilon` is non-zero and is called `epsilon_0`. In order to discuss reactions of atoms to electric fields, we first have to discuss the origin of `epsilon_0`.

By Coulomb's Law the repelling force between two charges, `q_1` and `q_2` (coulombs) is measured to be

`bbF=bbhatr(q_1q_2)/(4piepsilon_0r^2)`

where `r` is the distance between the charge centers and `bbhatr` is a unit vector in the `r` direction. So `epsilon_0` arises as a factor in the Coulomb force equation.For the derivation we will make one charge `Q` much larger than the other `q` i.e.

`bbF=bbhatr(Qq)/(4piepsilon_0r^2)`

Now we invoke Gauss's Law for the electric field, `bbE`, of `Q`. where the integral is over the surface of a sphere surrounding `Q` and `dbbs` is an outward vector surface increment of the sphere and the centered `*` dot represents the dot product of `bbE` with `dbbs`. This equation for `bbE` is also valid in a dipolar medium where `epsilon` can replace `epsilon_0`. Obviously, if `bbE` is parallel to `dbbs` everywhere the the integral is just `4pir^2|bbE|`, where `r` is the distance from the sphere center, so the equation becomes: and`bbE=Qbbhatr/(4piepsilon_0r^2)`

The force on `q` due to this electric field is `qbbE` and becomes:`F=bbhatr(Qq)/(4piepsilon_0r^2)`

`epsilon_0=(qQ)/(4pi|bbF|r^2)`

To measure `bbF` due to charges, one might resort to placing known changes `q` on each of two spheres just as was done for measuring the gravitational constant `G`. A big problem with this measurement is being sure of the value of `q`. Charging the spheres is done by providing a known current for a known time. But other parts of the apparatus may have charge too. The main importance of this section is to determine the units of `epsilon_0` at least in the MKSA system of units. First, from the table below note that the units of `|bbF|` are `kgms^-2`.`epsilon_0=q^2/(4pi|bbF|r^2) ~ Q^2m^(-3)kg^(-1)s^2`

It is well known that the speed of light in vacuum is

so, for completeness we should also disuss how `mu_0` can be measured. It is a constant in the equation for the force between 2 wires carrying currents `I_1` and `I_2`. This is called Ampere's Force Law. So it can be measured from this force in a manner very similar to the measurement of `epsilon_0`.For completeness we will include a list of MKS units and their acronyms which will aid in understanding
the relation between `bbD` and `bbE` and polarizability.

time:seconds:s

length:meters:m

mass:kilograms:kg

force:newton:N:`kg*m*s^(-2)`

pressure:pascal:Pa:`kg*m^(-1)*s^(-2)`

energy:joule:J:`kg*m^2*s^(-2)`

power:P:watt:W:`kg*m^2*s^(-3)`

Time:seconds:s

Length:meters:m

Current:Amperes:A

Charge:`"Coulombs":Q=A*s`

Voltage:V:`kg+m^2*sec^(-3)*A^(-1)`

Farad:F:`kg^(-1)m^(-2)sec^4A^2`

Canvas 1 shows the response of an atom's electron cloud to a sinusoidally varying electric field. The time variation of the electric field is achieved by changing the charges on a pseudo capacitor. On this canvas black is negative charge and red is positive charge. The electron cloud's mass is a factor of 1/2000 times that of the nucleus so essentially only the position of the electron cloud is changed by the electric field of the capacitor. Note the realism that the charge density of the electron cloud decreases with radius.

`F_q?=(qQ)/(4piepsilon_repsilon_0r^2)`

The animation below shows that, except for very small radii, the effect of the dipoles is much smaller than this equation would compute. This reCanvas 3 shows a spherical array of atom dipole charges with a smaller single layer spherical array of positive (red) monopole charges at the center. Of course, the dipole layers next to the monopole layer respond the most because there the electric field due to the monopole charges is largest. The negative electron cloud is pulled inward toward the red monopole. This has the effect of reducing the electric field exerted by the monopole layer especially at small layer radii where the field is lowest.

Of course we can use Gauss's Law to compute the electric field Vs position

`8piepsilon_0int_0^(r_i)bbErdr=sum_(i=0)^(i=n)q_i`

where `sum_(i=0)^(i=n)q_i` is the sum of both fixed (monopole) and bound (dipole) charges out to radius `r_i`.As is evident from the plot of the charge displacement spikes, the negative charge clouds are displaced inward and their displacement is negligible after a few layers. So the Electric field `bbE` and the force, `qbbE` on a very small test charge, 'q', due to that field is not significantly affected at large radii from center either. No difference in the effect would be seen if the outer boundaries of the dielectric were not spherical. This is very different from that of fields and forces in a parallel plate capacitor which is effectively a one dimensional system. In the capacitor, the field and test charge force due to charge on a plate is reduced across the entire thickness of the dielectric slab.