### Electric Force

The absolute value of potential energy between two quantum electric particles (electrons or protons), separated by the classical electron radius, is equal to the rest mass energy of the electron. Since potential energy is inversely proportional to the distance of separation, the equation for potential energy between each unique pair of electric particles is:

$u_{p}=\frac{r_{e}m_{e}c^{2} }{r}(\pm 1)(\pm 1)$

The factors, plus or minus one, represent single particles at each end of r, with a plus assigned to protons and a minus assigned to electrons. For point concentrations of Y particles at one end of r and Z particles at the other end, the total number of unique quantum pairs is the product, YZ, giving the total potential energy as:

$u_{p}=\pm \frac{r_{e}m_{e}c^{2} }{r}YZ$

#### Force between point concentrations of quantum particles

A spatial gradient of energy is equivalent to a force directed toward the region of lower energy. Taking the derivative of the above equation with respect to r gives the electric force between two point concentrations of quantum particles:

$f=\pm \frac{r_{e}m_{e}c^{2} }{r^{2}}YZ$

#### Force between a single electron and an infinite straight line of electrons

Imagine a line of uniformly distributed electrons on the 'x' axis, and the subject single electron on the 'y' axis, distance, d. from the origin. Partitioning the 'x' axis into small, equal incremental distances, dx, the number of electrons in each partition is:

$Z_{x}=\lambda _{Z}dx$
(lambda_Z is the linear density in electrons per unit length)

For each dx at +x, there is a corresponding dx at -x. Since the horizontal components of force from each paired group of electrons are equal and opposite, they cancel each other, leaving the net force always perpendicular to the 'x' axis:

$\Delta F=2\cos \phi \frac{r_{e}m_{e}c^{2}}{L^{2}}\lambda _{Z}dx$
(L is the distance from the electron to the dx located at +x, and phi is the acute angle between the 'y' axis and L.)

$L=\frac{d}{\cos \phi }$

Substituting and re-grouping,

$\Delta F=2\lambda _{Z}r_{e}m_{e}c^{2}\frac{\cos \phi }{d}\frac{\cos \phi dx}{L}$

The differential angle subtended by dx is the component of dx perpendicular to L, divided by L:

$d\phi =\frac{\cos \phi dx}{L}$

$\therefore \Delta F=\frac{2\lambda _{Z}}{d}r_{e}m_{e}c^{2}\cos \phi d\phi$

Integrating the incremental forces while phi goes from zero to PI/2, is equivalent to integrating the same incremental forces as x goes from zero to infinity. Since:

$\int_{0}^{\frac{\pi }{2}}\cos \phi d\phi =1$

The total vertical force on the electron is:

$F=\frac{2\lambda _{Z}}{d}r_{e}m_{e}c^{2}$

#### Force between parallel lines of particles

By considering the above subject electron to be part of a parallel line of uniformly distributed electrons, with lambda_Y being its linear density, a general equation emerges for the attractive or repulsive force per unit length between two parallel linear quantum densities:

$\frac{dF}{dL}=\pm 2\frac{r_{e}}{d}m_{e}c^{2}\lambda _{Y}\lambda _{Z}$
(L here represents length in the same direction as the parallel lines of particles.)

The above equations describe static electric forces which are repulsive if the signs of both quantum lines of particles are the same and attractive if their signs are opposite.

#### Force between parallel currents

Now consider two long, straight, parallel, electrically neutral conductors.  If the conductors carry currents in which the rates of electron flow are given by $\dot{Y}$ and $\dot{Z}$, respectively, the equivalent electron densities for light-speed currents are given by:

$\lambda_{Y}=\frac{\dot{Y}}{c};\lambda_{Z}=\frac{\dot{Z}}{c}$
(Particles per second divided by meters per second is equal to particles per meter.)

Substituting these equivalent electron densities into the static equation:

$\frac{dF}{dL}=\pm 2 \frac{r_{e}}{d}m_{e}\dot{Y}\dot{Z}$

This equation expresses the quantum origin of magnetic force. If both currents consist of electrons moving in the same direction, the magnetic force is attractive; otherwise it is repulsive.

Equivalent linear electron density is the hypothetical electron density for a light-speed current. Since electric signals travel at light-speed, the equivalent electron densities in the two conductors are invisible to each other.

Consider two parallel electron currents. The repulsive force between the two equivalent electron densities were an essential component of equilibrium in the static case, absent the currents. The loss of this repulsive force in the presence of the currents can be interpreted as the cause of the observed attraction between the two conductors.

This is the basic model for magnetic force which Nand tried to explain to John Wheeler in the “Matterhorn Paradox”. He wasn’t ready.

#### Determining the number of electrons in a coulomb

In the previous section, the magnetic force per unit length between two parallel currents was derived by integrating small contributions over an infinite length of the parallel currents:

$\frac{dF}{dL}=\pm 2 \frac{r_{e}}{d}m_{e}\dot{Y}\dot{Z}$

$\dot{Y}$and $\dot{Z}$ are the electron flow rates in the two currents, re is the classical electron radius, me is the electron mass, and d is the distance separating the two currents.

In 1820, Andre-Marie Ampere defined his namesake unit of current such that one unit of current in both of two parallel straight conductors, separated by one meter, would produce 2 X 10-7N of force between each meter length of the conductors. While his definition was totally arbitrary, it was obviously chosen to give a simple base-ten value for the magnetic force in newton per meter. His choice of two times a power of ten suggests that Ampere had solved the above integral in a general form, without knowing the product of the classical electron radius and the electron mass. When one meter is substituted for d, and one coulomb per second is substituted for both $\dot{Y}$and $\dot{Z}$, the factors of two cancel, assuring a simple relationship between the coulomb and the unknown (in 1820) constant, reme, whatever its value might be.

$\frac{dF}{dL}=2\frac{r_{e}}{1 meter}m_{e}\frac{coulomb^{2}}{second^{2}}=2\times 10^{-7}newton/meter$

$1 coulomb=\sqrt{\frac{10^{-7}newton\cdot second^{2}}{r_{e}m_{e}}}$

$\therefore 1 coulomb=\sqrt{\frac{10^{-7}}{(2.817 940 3e-15 )(9.109 383 7e-31)}}=6.2415091\times 10^{18}$

This was seventy-seven years before J.J. Thomson measured the mass/charge ratio of the electron, and eighty-seven years before Robert Millikan is credited with the measurement of the electron charge. If Ampere had known the electron's mass, he undoubtedly would have calculated the number of electrons in a coulomb from the above equation.

As it happened, however, fourteen years after Ampere defined the unit of current, Michael Faraday discovered that the coulomb could be measured experimentally from the electroplating mass-rate caused by one ampere of current. For nearly two hundred years physicists have relied on this empirical definition of the coulomb. There has been little interest in exploring the underlying quantum interactions responsible for its value.

#### Force between a single electron and a plane of uniformly distributed protons

Dropping a line from the electron, perpendicular to the plane, finds the closest point on the plane. Using that point as the center, consider a circular ring with radius, r, and width, dr. The number of protons in the ring is:

$Z_{r}=\sigma _{Z}2\pi r dr$

All of the protons in the ring are on a cone, equal distance, L, from the electron:

$L=\frac{r}{\sin \phi }$

From symmetry, it is clear that any force component parallel to the plane will be cancelled by equal proton concentrations on opposite sides of the ring. The total force contribution from the ring is, therefore, perpendicular to the plane:

$F_{r}=r_{e}m_{e}c^{2}\frac{Z_{r}}{L^{2}}\cos \phi =r_{e}m_{e}c^{2}\frac{\sigma _{Z}2\pi rdr}{L^{2}}\cos \phi$

The differential angle subtended by dr is the component of dr perpendicular to L, divided by L:

$d\phi =\frac{dr }{L}\cos \phi$

$\therefore F_{r} =2\pi r_{e}m_{e}c^{2}\sigma _{Z}\frac{r}{L}d\phi =2\pi r_{e}m_{e}c^{2}\sigma _{Z}\sin \phi d\phi$
Since,

$\int_{0}^{\frac{\pi }{2}}\sin \phi d\phi =1$
, the total force on one electron, any distance from an infinite plane of protons is:

$F_{e}=2\pi r_{e}m_{e}c^{2}\sigma _{Z}$

#### Pressure between two parallel planes of uniformly distributed electric particles

If the above single electron happens to be part of a uniformly distributed parallel plane of electrons, the force per area (pressure) between the two planes is:

$F/A=2\pi r_{e}m_{e}c^{2}\sigma _{Z}\sigma _{Y}$

Comments welcome.  CharlesdeAnne@sbytes.com

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