Many Internet articles solve for the E field on the axis of a spinning disk of charge. But few (if any) discuss the field in the plane of such a spinning disk. In this article the radial gravitational field is solved for a rotating disk of mass, and then the transformation rules are applied “in reverse” to find the radial electric field of a rotating disk of charge.
Let us begin by modeling a spiral galaxy as a disk of single-valued mass density σ=M/πR2, lying in the xy-plane and rotating with an angular velocity, ω, that points in the positive z-direction. In the article “Spinning Spiral Galaxies and the Gravitomagnetic Force” it was found that such a galaxy theoretically has an imaginary uniform gravitomagnetic field O that points in the z-direction:
O=2πGσωR/c2. (1)
=2GMω/Rc2.
Given a star of mass m, momentarily on the x-axis and at a fixed distance r from the galaxy’s center, the gravitomagnetic force on the star is then (according to the gravitomagnetic theory version of the Lorentz force law)
Fgravmag =mvO (2)
= mωrO.
And the total force on the star is theoretically the sum of the gravitomagnetic force plus the gravitational force (which also points toward the galactic center):
Ftotal =Fgravmag +Fgrav. (3)
Now Newton 2 indicates that Ftotax is:
Ftotal = |m|a (4)
=|m|ω2r.
Assuming that
Fgrav = mg (5)
where g is the gravitational field of the galaxy at the star’s position, we find that
g = Fgrav /m (6)
= (Ftotal – Fgravmag) /m
=-(|m|ω2r+2GmMω2r)/Rc2)) /m.
=iω2r-2GMω2r/Rc2.
Using the gravitomagnetic theory rules that (1) g transforms to E, (2) G transforms to 1/4πεo, and (3) O transforms to B, we find that the electric field in the disk’s plane points away from the center and has the magnitude
E = ω2r+qω2r /2πεoRc2. (7)
Note that in the case of a non-rotating disk the radial component of E is zero in the disk’s plane, but it is a linear function of r when the disk spins. And, applying the same transformations to Eq. 1, we find that
Bz=μoσωR/2. (8)
Eq. 8 agrees with the value for B specified on the Internet.