Electric Gravity - Appendix

27 July 2012
Magnetic forces are not shielded or reduced by a copper plate between magnets. I didn't speculate about photons. I tried two magnets and a penny. Magnets are related to electricity. Electromagnetic waves could not go through a penny but forces do. Are the forces between two charges shielded by a conductive foil? Shielding of electromagnetic waves may be easily accomplished inside of a conductive can. Is the force between charges also easily shielded? Is the charge the same as the electric field of the charge? Forces between the orbiting bodies in our solar system have no calculated delay in their interactions. Do forces between charges have a delay in their interactions?

I haven't heard of any experimental evidence of gravitational shielding being evaluated from the July 22, 2009 solar eclipse in Asia. The Sun and Earth interaction are blocked by the moon, during the eclipse, over a tiny area of the Earth.
A copper or aluminum foil might be used to demonstrate shielding of electrostatic repulsion or attraction forces, if any shielding exists, if you can figure a way to avoid interaction of the charges with the plate. I tried a simple experiment. An electroscope works by the deflection of charges by electrostatic repulsion. Two styrofoam balls are suspended by threads making a crude electroscope. They are charged with a "Fun-fly-stick" battery powered portable toy van de Graff generator. Having the same charge, the styrofoam balls repel. Then I tried to insert an aluminum foil between the balls. The aluminum foil attracts the balls because it has less charge. The balls touch and loose their charge. With sufficient charge, I speculate, the foil would become a neutral surface in an electrical gradient. It would be charged to the local gradient. The forces would be unaffected. There are simple experiments here somewhere but I have yet to find an experimental answer. The force between charged plates can be measured with Kelvin's absolute electrometer. It can be nearly as easily measured with a digital scale as was the force between
magnets in the appendix. We know the insertion of a dielectric between charged plates should decrease the force between the plates at low frequencies. What is the measurable effect? What measurable effect does the insertion of a conducting foil have on the force between the plates?

Dielectrics and capacitors
Dielectrics are insulators. When a dielectric is inserted in the charged plates of a capacitor, dipoles are induced in the dielectric.

Azaroff and Brophy in "Electronic processes in materials", "Another way of describing an insulator is to note that the electrons are so tightly bound to the atoms that at ordinary temperatures they cannot be dislodged either by thermal vibrations or with ordinary electric fields. The negative and positive charges in each part of the crystal can be considered to be centered at some point, and since no conductivity is possible, the localized charges remain that way essentially forever. When an electric field is applied to the crystal, the centers of positive charge in the crystal are slightly displaced in the direction of the applied field and the centers of negative charge are slightly displaced in the opposite direction. This produces local dipoles throughout the crystal, and the process of inducing such dipoles in the crystal is called polarization. The ratio of the induced dipole moment to the effective field is called the polarizability of the atom, and the dipole moment induced in a unit volume of a polarized insulator can be considered as the average of the dipole moments of all the atoms in that unit volume. It is possible that certain groups of atoms (complex ions or molecules) already possess permanent dipole moments. In crystals containing such atomic groups, an external field tends to orient the dipoles parallel to the field direction. In the absence of an external field, the dipoles are randomly oriented because of their thermal motion, so that the crystal has a zero net moment. The polarization of such polar crystals is strongly temperature-dependent since, even in the presence of an applied field, thermal motion tends to randomize the dipole orientation. On the other hand, the polarization of non polar crystals is independent of temperature since, in the absence of an external field, no dipoles exist in the crystal. The temperature dependence of polarization, therefore, can be used to distinguish polar insulators from non polar insulators." We will follow non polar insulators and the separation of charge mentioned above.
Origins of flux
Thomas L. Martin Jr. in the "Physical basis for Electrical Engineering"; "It now becomes convenient to assign a synthetic reality to the flux lines, although they are a creation of the mind only and do not exist physically, it is convenient to assume that flux lines do exist and to use them to describe the regions about charged bodies. Thus, we assume the following statements are true:
  1. Charged bodies are the sources of lines of electric flux.
    • Flux lines emanate from bodies carrying positive charge.
    • Flux lines terminate on bodies carrying negative charge.
  2. The flux lines are directed parallel to the force exerted on a positive test charge.
  3. The total number of flux lines associated with a charged body is proportional to the flux density and test charge.
The flux lines are used to represent systematically the flux about a charged body. The flux and charge are really just two different ways to describing the same phenomenon. The charge q describes the properties at a point. The flux and flux density describe the properties some distance away from the point occupied by q. Thus flux and q are just different manifestations of the same physical quantity, so electric flux = q."

In the figure above
"A charge of +q coulombs is produced on one plate in vacuum, and a charge of -q coulombs on the other plate when a potential difference of V volts is established. If the potential difference is doubled it is found that the number of charges on each metal plate doubles. It appears that the ratio of the charge q to the potential difference V is a constant of the system. This constant is defined as the capacitance C. A physical structure having capacitance is called a capacitor."
C = q/V, charge2/energy, farads, coulombs/volt, a2*s4/(kg*m2),

Adding a dielectric
"When some insulating material is included between the plates of a capacitor, the situation depicted above results. It is convenient to picture the insulator as being full of dipoles randomly oriented by thermal Brownian motion under normal conditions. When an electric field is applied, we assume that it causes the dipoles to rotate until they come into alignment with the electric field by an amount sufficient to bring them into alignment with the electric field. In so doing they produce new lines of flux, and induce additional charges on the capacitor plates. This causes the total accumulated charge q' on each plate to be larger than the vacuum case without the dielectric even though the potential difference remains the same. We spoke of dipoles rotating when the electric field was applied. Actually, all of the dipoles in the material may be induced and present only when the electric field is applied.
qi = induced charges
q' = q + qi, the new value of the charge.
C' = q'/V = (q + qi)/V, the new value of the capacitance.
This capacitance is larger than that obtained with vacuum between the plates.
You can easily see that the extent of increase of the capacitance is controlled by the character of the insulating material between the plates. The capacitance increases as the density of dipoles increase because this raises the number of induced charges.
D = charge/area = q/(4*pi*r2), the flux density imposed by q, coulombs/meter2, a*s/m2, on the surface of a sphere which surrounds the charge q.
E = D/e0, the electric field strength. e0 is the dielectric constant of vacuum or the permittivity of vacuum.
F = q2*D/e0 = q2*E,
F = q2*q/(4*pi*e0*r2), substituted for D, this is Coulomb's law.
Dp = n*p/v, the flux density added by dipoles, dielectric polarization, dipole moment per unit volume. n is the number of dipoles in the material between the plates. n*p is the total dipole moment in coulombs*meters. v is the volume of the insulating material.
Dv = e0*E, flux density in vacuum, charge/area = charge2/(force*area) *force/charge.
Dm = em*E, flux density in dielectric, em is the dielectric constant of the dielectric material between plates.
Dm = Dv + Dp, flux density in vacuum + flux density from dipoles.
em*E = e0*E + Dp,
em = e0 + Dp/E, divided by E the electric field intensity.
kr = em/e0 = 1 + Dp/Dv, divided by e0.
The ratio of the capacitance obtained with some insulating material between plates to the capacitance with vacuum insulation is the relative dielectric constant kr."
kr = C'/C = (q + qi)/q = 1 + qi/q = em/e0 = 1 + Dp/Dv
For vacuum the relative dielectric constant kr is equal to one. For water the relative dielectric constant kr = 78 at zero frequency. The force between capacitor plates is divided by kr = 78 when the space between the plates is filled with water. At optical frequencies of 10E14 to 10E15 hertz the kr = 2 for water. Conductors are infinitely polarizable so their kr approaches infinity at zero frequency. At the high frequencies of the atomic dipoles, 6E15 hertz, there is a force between the charges. Dielectric constants have no obvious relationship to mass.

We take from this that charge is induced by the presence of a dielectric in a potential field and that a dielectric induces these charges as the dielectric becomes polarized with dipoles. Some of these induced charges, those that are caused by the polarization of gravity, are the gravitational charges we have been looking for. Dipoles have an attractive force proportional to the inverse forth power of distance so dipoles can not be the source of gravity. Charges induced by dipoles have an attractive force proportional to the inverse square of distance and can therefore be the source of gravity.

Tidal bulges
Tidal bulge equations and exposition, tides and other interesting items are at Johnson's website.
We calculate the tidal bulge of the Earth using accelerations. We calculate the accelerations both using mass and using gravitational charge. The accelerations have the same values. Tidal bulges in atoms, caused by charges, may use the same procedure.
mearth = 5.972E24_kg, the mass of the Earth.
rearth = 6.378E6_m, the radius of the Earth.
mmoon = 7.348E22_kg, the mass of the moon.
dmoon = 384.4E6_m, the distance from the center of the Earth to the center of the moon.
gcearth = mearth*(4*pi*e0*G).5 = 5.146E14_a*s, the gravitational charge of the Earth.
gcmoon = mmoon*(4*pi*e0*G).5 = 6.3317E12_a*s, the gravitational charge of the moon.

Accelerations using mass
ge = G*mearth/rearth2 = 9.79822007_m/s2, the gravitational acceleration of the Earth at the surface of the Earth, using mass.
ged = G*mearth/(rearth+Δr)2 = 9.79821898_m/s2, the gravitational acceleration of the Earth at the surface of the Earth +Δr.
gmd = G*mmoon/(dmoon-rearth-Δr)2 = 3.431742494E-5_m/s2, the gravitational acceleration of the moon at the surface of the Earth +Δr.
gms = G*mmoon/(dmoon-rearth)2 = 3.341742487E-5_m/s2, the gravitational acceleration of the moon at the surface of the Earth.

Accelerations using gravitational charge
ge = gcearth/(rearth2 )*(G/(4*pi*e0)).5, the gravitational acceleration of the Earth at the surface of the Earth, using gravitational charge.
ge = gcearth/(rearth2)*.77448_m3/(a*s3) = 9.7979_m/s2, since (4*pi*e0*G).5/(4*pi*e0) = (G/(4*pi*e0)).5 = .77448_m3/(a*s3), using gravitational charge.
ged = gcearth/(rearth+Δr)2*(G/(4*pi*e0)).5, the gravitational acceleration of the Earth at the surface of the Earth +Δr, using gravitational charge.
gmd = gcmoon/(dmoon-rearth-Δr)2*(G/(4*pi*e0)).5, the gravitational acceleration of the moon at the surface of the Earth +Δr, using gravitational charge.
gms = gcmoon/(dmoon-rearth)2*(G/(4*pi*e0)).5, the gravitational acceleration of the moon at the surface of the Earth, using gravitational charge.

Δg = ge-gms-(ged-gmd) = 1.09689E-6_m/s2, the net amount of acceleration applied to something on the surface of the Earth by the gravity of the moon when the moon is overhead. The moon overhead reduces your weight slightly. It pulls more on your head than your feet. Does this improve your posture?

ge - Δg = G*mearth/(rearth + Δrearth)2, the gravitational acceleration of the Earth and the pull of the moon above the surface of the Earth.
ge - Δg = gcearth/(rearth + Δrearth)2*.77448_m3/(a*s3), using gravitational charge.
(ge - Δg)/ge = G*mearth/(rearth + Δrearth)2*rearth2 /G*mearth, divided by ge.
(ge - Δg)/ge = gcearth/(rearth + Δrearth)2*.77448_m3/(a*s3)*rearth2 /(gcearth*.77448_m3/(a*s3)), using gravitational charge.
1 - Δg/ge = rearth2/(rearth + Δrearth)2, canceled mass or gravitational mass.
1 + Δg/ge = (rearth+ Δrearth)2/rearth2, inverted.
1 + Δg/ge = (rearth/rearth+ Δrearth/rearth)2,
1 + Δg/(2*ge) = 1+ Δrearth/rearth, square root.
_ We just used these two handy approximations which I discovered by accident.
_ Check these with your calculator.
_ (1.00008)-1 = 1+(.00008*-1) = 1-.00008, inverted
_ (1.00008).5 = 1+(.00008*0.5) = 1.00004, square root
_ The integer and fractional parts are separated which simplified this problem.

Δg*rearth/(2*ge) = Δrearth, canceled the ones.
Tidal bulge = Δrearth = 0.3570_m, This is the amount of the tidal bulge. The Earth is egg shaped or ellipsoid shaped by the pull of the moon. It is so small because the acceleration imposed by the moon, Δg is so small, 1.096E-6_m/s2. The Earth is also shaped by its spin on its axis producing a much larger equatorial bulge.

Equatorial bulge
The Earths equatorial radius is 6378165_m.
The Earths polar radius is 6356785_m.
The Earths equatorial bulge is 21380_m. The tidal bulge was a tiny 0.3570_m. The Earth rotation and centrifugal force makes it an oblate spheroid with an eccentricity of e = 0.08181.

Tidal bulge ratio
Δg/(2*ge) = Δrearth/rearth = 5.59750E-8 = Δratom/ratom,
5.59750E-8 *ratom = Δratom = 2.962E-18_m, tidal bulge, on each side of the atom. The atoms have a radius of about 5.292E-11_m so the tidal bulge of the Earth or atom is,
1/17.285E6, one part in 17 million. The Earth is ellipsoidal under the gravitational effects of the Sun and moon. The Earths atoms are ellipsoidal under the gravitational effects of the Sun and moon and their neighbor atoms and any other accelerations. This bulge in the atoms is too small to be seen with a scanning tunneling microscope. The atoms still appear spherical because they are egg shaped by only one part in 17 million. Both the atom and Earth are bulged a similar small proportional amount by the same external forces.

Eccentricity of atoms and ellipses
a = ratom+Δratom = 5.292E-11_m +2.962E-18_m = 5.2920002962E-11_m, x radius of the elliptical atom,
b = ratom-Δratom = 5.292E-11_m - 2.962E-18_m = 5.2919997038E-11_m, y radius of the elliptical atom,
e = (a2-b2).5/a = 4.7316E-4, eccentricity or,
e = ((r+Δr)2-(r-Δr)2).5/(r+Δr), eccentricity or,
e = (4*Δr/r).5 = 4.7318E-4,
e = rapogee-rperigee/(rapogee+rperigee), 2*a*e/(2*a) = e
We can move on to discover the charge density at the ellipsoid ends of the atoms. This might relate to the gravitational charge of the atoms.

Charge separation
The red electron and blue proton ellipses share a common focus and eccentricity.
The radius of the Bohr atom is re + rp. The center of the red ellipse of the electron from the focus is re. The center of the blue ellipse of the proton from the focus is rp. The amount of charge separation is proportional to the eccentricity as is the amount of opposite charge on each end of the atom. e is the eccentricity of both ellipses but the electron ellipse is much bigger. e*(re + rp), is the separation of the centers of the electron and proton ellipses. If the eccentricity is zero then the atom is spherical and free of bipolar end charges.

Click figure to animate! Electron and proton binary atom with elliptical orbits
r = a*(1-e2)/(1+cos(angle)*e), the polar form of the ellipse equation.
rapogee = a*(1-e2)/(1-e) = a*(1+e), when the angle is pi, r is maximum, at apogee, at its farthest from the center of mass focus.
rperigee = a*(1-e2)/(1+e) = a*(1-e), when the angle is 0, r is minimum, at perigee, at its nearest to the center of mass.
Kepler tells us planets sweep out equal areas in equal time. Their velocities must vary in elliptical orbits.
area = r2*angle/2, the area of each wedge of the ellipse is defined by r and angle.
angle = 2*area/r2, the angle advanced for a fixed area of wedge. A constant equal to 2*area can be used with the r from the ellipse equation to calculate an angular increment to trace out equal areas in equal times. This traces out the ellipse as it shows the changes in velocity along its orbit. See the figure above. This change in velocity as the proton and electron pair move around their elliptical orbits is what interest us with respect to polarized atoms. As the velocity varies so does the charge density. The positive and negative atomic charges are visible at the opposite ends of the atom. The atom is polarized.
See Orbit 101 for several methods of predicting future orbital locations.

Charge per radian
Planets and electrons trace out equal areas in equal times in their orbits. Electrons on elliptical orbits tracing out equal areas in equal times have a variable velocity which leads to a variable charge density and polarization of the atom. The area is,
Δareaapogee = Δangleapogee*a2*rapogee2/2 = Δangleapogee*a2*(1+e)2/2, on the electron end of the atom.
Δareaperigee = Δangleperigee*a2*rperigee2/2 = Δangleperigee*a2*(1-e)2/2, on the proton end of the atom.
Δareaapogee = Δareaperigee
Δangleapogee*a2*(1+e)2/2 = Δangleperigee*a2*(1-e)2/2,
Δangleapogee*(1+e)2 = Δangleperigee*(1-e)2,
Δangleapogee*(1+e)2/(1-e)2 = Δangleperigee,
e = 0.0004732 for the tidal bulge of the Earth or the atom.
Δangleapogee*(1.001894) = Δangleperigee,
Δangleapogee*(1+e*4) = Δangleperigee, The electron traces out equal areas in equal time, but the angle the electron traces out at the proton end of the elliptical atom is bigger. The velocity is greater and the charge per angle or charge per radian is less. The charge per radian is greater where the electrons move slower, as they do, at the extended portions of the elliptical orbit where the electron radius is greater.

Elliptical velocity graph

On the left is shown, the tangent velocity of an electron or proton on an elliptical orbit, at equal time intervals, around the center of mass of the electron-proton system. In the previous figure we saw both particles. Here we only look at one particle at a time. The particles slow down and speed up. They are accelerated. Like anything on an elliptical orbit there is an oscillating transfer between potential and kinetic energy of the particles along their orbit. There is equilibrium. Vp is the perigee velocity. Va is the apogee velocity. The apogee velocity is smaller than the perigee velocity. The orbiting particles spend more time near the apogee end of their orbits. The probability of an electron or proton being near the apogee end of its orbit is greater. The electron and proton each have their greatest charge per radian at the opposite ends of the atom. The atom is polarized. On the right is the hodograph. It is an interesting way of looking at velocity in elliptical orbits. The velocity is least at Va. The points along the orbit of equally spaced time intervals are closest together at Va. The charge density is greatest at Va. This charge density is much greater than at Vp. The atom is polarized by differences in charge density. See the original graph and papers by Butikov.
rperigee = a*(1-e), the minimum radius, closest to the center of mass focus.
rapogee = a*(1+e), the maximum radius, farthest from the center of mass focus.
vapogee*rapogee = vperigee*rperigee, this holds for all orbits.
vapogee*a*(1+e) = vperigee*a*(1-e), substituted for the radii.
vapogee*(1+e)/(1-e) = vapogee*(1+e)*(1+e) = vperigee, isolated vperigee.
vapogee*(1+2e+e2) = vperigee, another approximation where e is small.
Δv = vperigee - vapogee = vapogee*(1+2e+e2) - vapogee = vapogee*((1+2e+e2) - 1) = vapogee*(2e+e2)

Circle to ellipse
An electron in an circular orbit goes to an elliptical orbit by a change of velocity, Δv. This is an increase in momentum and energy as well as velocity.
rcircle = rperigee = a*(1-e), the minimum radius, closest to the center of mass focus.
rapogee = a*(1+e), the maximum radius, farthest from the center of mass focus.
vcircle*(1-e)/(1+e) = vapogee, isolated vapogee.
vcircle*(1-2*e+e2) = vapogee, or
vcircle/vapogee = (1+2*e+e2),
vcircle/vapogee = (1+Δv), substituted Δv = 2*e+e2,

Another clue
Atomic mass is measured in AMU, atomic mass units, 1.660E-27 kilograms, about the mass of a proton and electron dipole. Mass is proportional to the number of dipoles. Gravity is proportional to mass.

One AMU atomic mass unit = 1.66053886E-27 kilograms
Wiki says it is one twelfth of the mass of an isolated atom of carbon-12 (12C) at rest and in its ground state.
1 AMU or U = 1.660538782E-27 kg = 931.494028 MeV/cē
electron plus proton, (1.0078 amu) 1.6735E-27_kg = (1.0073 amu) 9.1094E-31_kg + (0.0005 amu) 1.6726E-27_kg
hydrogen atom, (1.00794 amu)*1.66054E-27_kg/amu = 1.67372E-27_kg,
neutron, (1.0087 amu), 1.6749E-27_kg,
1 amu or u = 1_gram/(Avogaodro's number) = 1_kg/ (1000*(Avogadro's number))
We noticed before that (c3*e0/hp).5 = 6.000359E23_a*s/(kg*m) is very close to Avogadro's number. Using it as Avogadro's number,
1_kg/ (1000*(Avogadro's number)) = 1.66656E-27_m/(A*s) = 1_AMU * 1_kg*m/(A*s)
This is at least an odd and interesting coincidence which links Avogadro's number, AMU and meters per charge.
The gravitational charge per AMU is,
1.66656E-27_kg * (4*pi*e0*G).5 = 1.4361E-37_A*s, or
1.66656E-27_kg*m/(A*s) * (4*pi*e0*G).5 = 1.4361E-37_m,

Earth in Bohr atoms
ratom = rc/(alpha2) = 5.292E-11_m, the minimum radius of the Bohr atom.
ratom*2 = 1.058E-10_m, the diameter of the atom or,
9.448E9_Bohr atoms/meter, if the atoms are cheek to jowl.
9.448E9_atoms/meter * 6.37E6_m (radius of the Earth) = rearth*alpha2/(2*rc) = 6.026E16_atoms, the radius of the Earth in diameters of series atoms or capacitors.
4/3*pi*6.018E163 = 9.133E50 atoms, the volume of the Earth in Bohr atoms.

What is the composition of the Earth?
Wiki and Google say, Earth's solid mass is about
32% iron * 55.845 amu = 17.8704 amu.
30% oxygen * 15.9994 amu = 4.79982 amu.
15% silicon * 28.0855 amu = 4.212825 amu.
14% magnesium * 24.305 amu = 3.4027 amu.
3% sulfur * 32.066 amu = 0.96198 amu.
2% nickel * 58.6934 amu = 1.173868 amu.
1.5% calcium * 40.078 amu = 0.60117 amu.
1.4% aluminum * 26.9815 amu = 0.377741 amu.
98.9% at 33.400 amu or
100% at an average of 33.7719 amu per atom for the Earth.
9.133E50 atoms at 33.7719 amu = 3.08438E52_amu or dipoles, the volume of the Earth in amu or dipoles.

Gravitational charge per AMU or dipole
gcearth = mearth*(4*pi*e0*G).5 = 5.146E14_a*s, the gravitational charge of the Earth.
gcamu = mamu*(4*pi*e0*G).5 = 1.430929E-37_a*s, the gravitational charge per AMU or dipole.
ce/gcamu = 1.6021E-19_a*s /1.430929E-37_a*s = 1.11962E18, The gravitational charge of the dipole is much, much smaller than the charge of the electron. Were does this tiny charge reside?
gcearth/gcamu = 5.146E14_a*s/1.430929E-37_a*s = 3.59626E51 gravitational charges or dipoles.
mearth*(4*pi*e0*G).5/(mamu*(4*pi*e0*G).5) =
mearth/(mamu) = 3.59626E51,
but this is only a ratio of masses.

3.08438E52 /3.5956E51 = 8.5782067, ratio of amu from volume to amu from gravitational charge. This is of the right magnitude.

Atomic dipoles
An electric dipole consists of two charges of equal magnitude separated by a distance.
q*d = electric dipole moment, q is one of the two charges. d is the distance between the charges. d is constant in a circular orbit. d varies in an elliptical orbit.
Torque = q*d*sin(angle)*E, a torque is created by the dipole as it tries to align itself with an external charge or electric field. This is the cross product of q*d and E. This torque is why objects tend to orbit in a plane and why solar systems and some galaxies form disks.
Ey = q*d/(4*pi*e0*y3), the electric field perpendicular to the dipole. y is the perpendicular distance far from the dipole.
Ex = 2*q*d/(4*pi*e0*x3), the electric field along the axis of the dipole. x is the distance far from the dipole. The electric field along this axis is twice that which is perpendicular to the axis. See Tatum.
Ex = 2*a/(b*x3), substituted a = q*d. b = 4*pi*e0.
dEx/dx = -(2*a*(b*(3*x2))/(b*x3)2,
dEx/dx = -(6*a/(b*x4),
q*d *dEx/dx = -(6*q2*d2/(4*pi*e0*x4), the force on a dipole in the x direction is proportional to q*d.
q*d *dEy/dy = -(3*q2*d2/(4*pi*e0*y4), the force on a dipole in the y direction is proportional to q*d.
The force between dipoles decreases with the inverse forth power of the distance between the dipoles. Gravitational force decreases with the inverse square of the distance between the masses. The interaction of solitary dipoles can not explain gravity. Long series of dipoles have large q*d products and proportionally large forces.

Magnetic forces on series of dipoles
Dipoles in rows stick together. The ends of rows of dipoles are oppositely charged. They are a larger version of a single dipole. The forces at the end of rows of dipoles are greater than with single dipoles. The attractive and repulsive forces between the ends of rows of dipoles are much greater. We will look at the force between rows of series magnets to try to understand the rules of series dipoles.

Calculations using inverse square force
force1 = m*m*k/x2 = m2*k/x2, for the force between two magnets. One m is a magnet on a digital scale. The other m is a magnet in a stack of magnets on a platform above the magnet on the digital scale. k is the constant of proportionality. x is the distance between the magnet on the scale and the first magnet on the platform. All the magnets appear the same but they vary somewhat in strength.
force2 = m2*k/(x)2 + m2*k/(x+d)2, for the first two magnets on the stack. d is the distance between the magnets on the stack. Each additional magnet is stacked and added at a greater distance in multiples of d from the scale as it is stacked onto the previous magnet. Each addition identical magnet adds a decreasing incremental force at the scale because it is at an increasing distance from the scale.
force4 = m2*k/(x+0*d)2 + m2*k/(x+1*d)2 + m2*k/(x+2*d)2 + m2*k/(x+3*d)2, for the first four magnets on the stack.

Using one inch for the distance between the magnet on the scale and the stack of magnets
force4 = m2*k/(4/4)2 + m2*k/(4/4+1/4)2 + m2*k/(4/4+2/4)2 + m2*k/(4/4+3/4)2, x = 4/4 and d = 1/4 for quarter inch magnets, for the first four magnets on the stack.
force4 = m2*k/(4/4)2 + m2*k/(5/4)2 + m2*k/(6/4)2 + m2*k/(7/4)2, collect terms.
force4 = m2*k*(4/4)2 + m2*k*(4/5)2 + m2*k*(4/6)2 + m2*k*(4/7)2, inverted the denominator and multiplied.
force4 = m2*k*16*( (1/4)2 + (1/5)2 + (1/6)2) + (1/7)2),
force10 = m2*k*16*(1/16 +1/25 +1/36 +1/49 +1/64 +1/81 +1/100 +1/121 +1/144 +1/169),
force10 = m2*k*16*0.20978 = m2*k*3.3565, collected terms for a calculated series, for a ten magnets stack.

Weighing magnetic force
A gram scale may be used to determine the attractive and repulsive force between series magnets. Within the scale, a steel screw which supports a reference magnet is threaded into the small aluminum beam which measures forces by its deflection. When the small aluminum beam deflects it stretches a strain gage. The strain gage operates a digital display. The results are non-linear increases in force with increasing numbers of series magnets. First, put one magnet on the scale. Second, zero the scale with this magnet. Then, measure the force of attraction or repulsion, of a series of magnets stacked on a platform which is at a fixed distance above the magnet on the scale. Each additional magnet adds a decreasing force as it is stacked one quarter of an inch farther from the scale as they are stacked one quarter of an inch higher on the platform.
The internal batteries were wired and moved to a distance. The interaction of their steel ruined careful measurements of forces between magnets.

We measured the force between two magnets as 10.68_g using x = one inch as the distance between the magnets. The force between one magnet and a ten magnet stack equals 21.77_g.
force10 = 21.77_g = m2*k*3.3565, The total force is now correct but we seek the increment added by each magnet.
m2*k = 6.4859_g
m2*k*16 = 103.77_g
103.77_g *(1/16 +1/25 +1/36 +1/49 +1/64 +1/81 +1/100 +1/121 +1/144 +1/169),
103.77_g/16 +103.77_g/25 +103.77_g/36, for the first three terms in the series calculated.
6.48_g +4.15_g +2.88_g, for the three terms in the inverse square series calculated versus,
10.68_g +4.94_g +2.09_g, for the first three terms of the series measured with one inch as the distance between the magnets and the scale. Measurements enforce some reality in any process. The inverse square calculation does have pretty math but it does not seem to be a good fit to the measured reality. We must look for a better fit to the measurements.

Using a binary fraction
When you look at a binary number such as,
0.1111111111b, an integer which is zero, a "decimal point" and a fractional part in binary notation with ten ones, or
0. +1/2 +1/4 +1/8 +1/16 +1/32 +1/64 +1/128 +1/256 +1/512 +1/1024 = 1023/1024 = 210- 1/210, in decimal notation. The fractional part adds up to a little less than one depending on how close you want to go to one with a decreasing increment of the fraction by adding more ones on the right of the binary fraction.

We seek a simple rule to estimate the increment of force added by each magnet so that we might know the total force of a series of n magnets from the force between the first few magnets in the long series. The force on a series of magnets can be considered like a weight multiplied by a binary fraction which adds up to a little less than one.
21.77_g*(1/2 +1/4 +1/8 +1/16 +1/32 +1/64 +1/128 +1/256 +1/512 +1/1024) =
21.77_g *1023/1024,
21.77_g/2 +21.77_g/4 +21.77_g/8, for the first three terms of the series calculated using the binary procedure.
10.88_g +5.44_g +2.72_g, for the first three terms of the series calculated.
10.68_g +4.94_g +2.09_g, for the first three terms of the series measured with one inch as the distance between the magnets and the scale.

Twice the first measured term of 10.68_g estimates 21.36_g for the sequence of ten magnets. We see that this approach is near the correct overall force between a single magnet and a series of ten magnets with 21.77_g measured, but the first three calculated individual terms are a little too big so the last calculated terms will be too small. These last terms will be too small to measure. This approach is ongoing with gradual improvements in experimental technique.

The first magnet had a force of 10.68_g. The first ten magnet stack had a force of 21.77_g. The force with a twenty magnet stack equals 22.74_g. The force with a forty magnet stack equals 22.92_g. The second ten magnets only added 0.97_g. The second twenty magnets only added 0.18_g.

This is only a start in research by weighing magnetic repulsion using quarter inch cube magnets. I also have eighth inch cube magnets. Cylinder magnets may be better. This is a series with decreasing increments which too quickly become too small for my 0.01_g display scale and non-uniform neodymium magnets which vary by more than 0.25_g in force when separated by one inch.

Twenty magnets makes a dipole twenty times longer than a dipole made with one magnet. The first dipole had 10.68_g on one end and -10.68_g on the other end separated by one magnet and 6.35_mm. The dipole moment of one magnet q*d is 67.82_g*mm. Twenty magnets makes a dipole with a force of 22.74_g on one end and -22.74_g on the other end separated by twenty magnets and 127_mm. The dipole moment of twenty magnets q*d is 2888_g*mm. This force is more complex than the analysis of isolated dipoles would suggest. A long series of dipoles has an accumulated force on their oppositely charged ends and a lot of distance between their ends for a large q*d.


When the two parallel plates of a capacitor are connected across a battery they store energy. The plates become charged and attract each other with an electric field between them. Some say the energy is stored in the electric field between the plates. If you disdain the abstraction and distraction of the field then the forces between the plates may be analyzed just as well as any fields.
Q*E = Q2/(4*pi*e0*r2), the electrical force equals the Coulomb force.
The charged plates are only held apart by some type of mechanical structure so the electrical force, Q*E measures the force of attraction of the plates. The actual force between the plates is measured in this electric forces between charged plates lab and in this Coulomb balance lab.

Useful details are in this current balance lab..
A circuit is not necessary for electrostatic gravity
Polarized atoms storing their tiny gravitational charge do not need to be wired together. If there is a conductive path their electrostatic gravity is unimpeded. The bipolar atoms are like capacitors and are arranged in series, along each row, so
Qseries = Q1 = Q2 = Q3, they all have the same small gravitational charge.
Cseries = 1/(1/C1 + 1/C2 + 1/C3), 1/(n*(1/C)) = C/n, since n, the number of capacitors is very large, the capacitance of the row is very low.
Vseries = V1 + V2 + V3, This is a voltage divider over a very large number of capacitors where there is a conductive path.
The capacitors are arranged in parallel rows, so
Qparallel = Q1 + Q2 + Q3, is the sum of the charge in all the rows, which is the sum of all the charge. The total charge is undiluted by series or parallel effects. All the individual atomic gravitational charges participate in the gravitational effect if there is a conductive path or not.
Cparallel = C1 + C2 + C3, this is the sum of the capacitance of the rows. A large sum of small capacitance is what? The capacitance is unknown.
Vparallel = V1 = V2 = V3, The voltage drop along the parallel rows is the same where there is a conductive path.

Spherical capacitors
Q*E = Q2/(4*pi*e0*r2), the electrical force equals the Coulomb force.
E = Q/(4*pi*e0*r2) = volts/meter = force/charge = kg*m/(a*s3), divided by Q.
V/r = Q/(4*pi*e0*r2), substituted V/r for E.
V = Q/(4*pi*e0*r) = volts = energy/charge = kg*m2/(a*s3), multiplied by r.
Q/V = 4*pi*e0*r = C = Farads = charge2/energy = a2*s4/(kg*m2), this is the capacitance of a isolated sphere with a radius r. As r increases the capacitance goes up.

Parallel plate capacitor using force and charge
Q = C*V, charge = charge2/energy * energy/charge, energy = Q*V
Q = C*E*d, substituted E*d for V, V = E*d, E = V/d, charge = charge2/energy * force/charge * distance, energy = Q*E*d
Q = C*d*Q/(area*e0), substituted Q/(area*e0) for E, E = D/e0 = charge density/permittivity: D = charge density = Q/area: E = Q/(area*e0), so force/charge = charge/(area)* force*area/(charge2), so e0 = charge2/(force*area) = a2*s4/(kg*m3)
C = e0*area/d, canceled Q, charge2/energy = charge2/(force*area) *area/(distance), capacitance in a plate capacitor. If the area goes up, or the distance between the plates decreases, the capacitance goes up.
Q = e0*area *V/d, substituted e0*area/d for C, in Q = C*V, charge = charge2/(force*area) *area *energy/(charge)*1/(distance), Combined the above equations. Even something as complex as the capacitor can be understood in terms of charge, force, energy, distance and action-at-a-distance without any reference to electric fields.

Atoms as capacitors
Two oppositely charged ellipsoid ends of polarized bipolar atoms, work the same as the parallel plate capacitors. If the voltage (energy/charge) is constant, while the area of the plates or the charged area of the ellipsoid ends increases, then the charge goes up. As an atom becomes more polarized the opposite charges at each end of the ellipsoid increases as the atoms charges become more separated. The charged area increases. The distance between the ends of neighbor atoms decreases as the atoms become more ellipsoid. An increase in the area of the charge or a decrease in the distance apart of the charges on neighbor atoms both increase the capacitance. The energy required is converted to additional charge stored between the plates or ellipsoids. Where we see forces or accelerations we see charge. Charge is starting to look like a placeholder for force on the plates. Energy is conserved.

Charged concentric spherical shell atoms
Binary orbits in a plane are stable in isolation. Stable atoms, which we see in photos, require the spherical symmetry which they acquire by the precession of their orbits out of the plane. The electrons and protons are not confined to the two dimensions of a plane. They orbit in three dimensions. Atoms are spherical spinning precessing dipoles. This is how medical scanners and microwave ovens work. In a proton-electron binary atom, the proton is close to the center of mass where it orbits and precesses. This is seen as a charge spread over a spherical surface traced out by the orbiting and precessing proton. The electron across the center of mass from the proton and farther out but orbiting with the same angular velocity as the proton is also like a charge spread over a spherical surface traced out by the orbiting and precessing electron. Is this an electron cloud? How can a cloud-like electron become a negative ion particle? What you see depends on the metaphors you use. The surface of the atom, the electron orbit, is negative for circular orbits. Negative spherical surfaces of atoms repel each other and do not clump together without being polarized. The positive surface of a positive ion would be attracted and pulled into the negative surface of an atom. They would merge until the ion and atom were at equilibrium, until the ion reached the region of charge neutrality within the atom which is halfway between the electron spherical surface and the much smaller proton spherical surface. The positive ion would be repelled by the proton within the atom beyond the region of charge neutrality. This leaves us with the familiar image of bonding as overlapping spherical atoms.

An atom as a concentric spherical capacitor
This is another approach to charge separation. Aren't the electron and proton in the Bohr atom somewhat like the oppositely charged parallel plates of a capacitor? Or the oppositely charged concentric spherical plates of a capacitor? The proton orbit is like a sphere of charge surrounded by the much larger sphere of charge of the electron orbit. While they are neutral they are spherical and concentric. As the charges separate the spheres become ellipsoids. The ellipsoid ends become oppositely charged. Their area increases. Their capacitance increases.
The electron orbit is considered as a negatively charged sphere with a radius of re = 5.2889E-11_m.
The proton orbit is considered as a positively charged sphere with a radius of rp = re*me/mp = 2.8804E-14_m.
The distance between the spheres is re-rp = 5.28602E-11_m apart.
Q/V = C = Farads = 4*pi*e0*r = charge2/energy = a2*s4/(kg*m2), This is the capacitance of a isolated spherical capacitor. We can use this formula for a concentric spherical capacitor with r defined in a special way. Here,
r = 1/(1/(rp) - 1/(re)) = re*rp/(re-rp) = re*me/(mp-me) = 2.88159E-14_m, this r is only slightly larger than the proton sphere rp.
C = 4*pi*e0*r = 4*pi*e0*2.88159E-14_m = 3.2061E-24_Farads, additional charge could be imposed by electrostatic gravity with or without a change in geometry.

Charged ends of atoms
The positive end of an atom has a positive _a*s charge and the negative end of an atom has a negative _a*s charge. This is the very small polarized charge of the atom. This is the slight separation of the much larger charges of the electron and proton in the atom.

Force between the ends of the polarized atom
Q*E = Q2/(4*pi*e0*d2) = _kg*m/s2,
E = Q*E/Q = _kg*m/s2/(_a*s) = _volts/meter,

Gravitational energy stored between the ends of the polarized atom
Q2/(4*pi*e0*D) = _kg*m2/s2,

Voltage drop across the gravitational capacitor
Q = e0*a *V/d, formula for a plate capacitor.
V = Q*d/(e0*A), isolate V. We are solving for the voltage drop across each gravitational capacitor.
Q = _a*s, gravitational charge of an atom.
d = ratom*e = 2 *5.292E-11_m * e = _m, spacing of the capacitor plates is the distance between the foci. This is the separation between the centers of the electron and proton ellipses. The centers are no longer concentric.
ratom2*pi/2 = 5.292E-11_m2*pi/2 = 4.3987E-21_m2, area of a ellipsoid end or plate is half the surface of the Bohr atom.
V = Q*d/(e0*A) = _kg*m2/(a*s3), volts/atom

Capacitance of an atom:
Q/V = C =_a*s/(_volts) = _Farad per atom. These tiny capacitors depend on the electro negativity and electro positivity of the atoms to work. A very small eccentricity, in an almost perfectly circular orbit, allows atoms to become polarized.
Useful equations
me = 9.1094E-31_kg, mass of the electron
ce = 1.6021E-19_a*s, charge of the electron
mp = 1.6726E-27_kg, mass of the proton
c = 299792458_m/s, speed of light
alpha = .007297 or 1/137.036, fine structure constant
hp = 2*pi*me*c*rc/alpha = 6.626E-34 _kg*m2/s, Plank's constant
e0 = 1E7/(4*pi*c2)_a2*s2/(kg*m) = 8.854E-12_a2*s4/(kg*m3), e0 = charge2/(force*area) , permittivity of vacuum.
Avogadro's number? = (c/(u0*hp)).5 = (c3*e0/hp).5 = 6.000359E23_a*s/(kg*m)
G = 6.6739 E-11_m3/(kg*s2) is the gravitational constant.
rc = ce2/(4*pi*e0*me*c2) = 2.82E-15_m, the classical radius of the electron or
me*c2 = ce2/(4*pi *e0 *rc), The rest energy of the electron equals the energy associated with the charge of the electron of ce and a radius of rc. If the rest energy increases then rc decreases.

These papers go some way in the right direction
Accelerating superconducting disk produces reversing pulses of gravitation
Spinning super conductors - Gravitational attraction when accelerated in one direction and repulsion when accelerated in the other has been experimentally detected by Tajmar.

Electromagnetic drive
emDrive - This looks like the drive which will be used in space craft of the future. It is an impulse drive where microwaves in a truncated cone magnetron push harder on the small end of the cone than the larger. I hope this is real and can compete with plasma drive. New Scientist magazine had a wonderful article showing a flying car on the cover a few years ago. http://www.emdrive.com/

This is a nifty HTML calculator DANIWEB has made available to copy and paste. There is much to be learned by studying its code. It is an example of products available for free that are as good as any you can buy.