Temperature Not Energy But Particles

May be it is best to start over...

Capturing a $T^{+}$ particle in a $B$ orbit due to a spinning electron.

We are getting warmer!

Consider a charge on approach to a disc of radius $r$,

The total perpendicular $E$ field component through the area of the disc is,

$E_A=\int_{0}^{r} {2\pi xtan(\theta). Ecos(\theta) }d\,r$

where the electric field from the charge is

$E=E_oe^{-\cfrac{x}{a_{e}}}$

the exponential form of an electric field analogous to the exponential form of the expression for gravity field.

$E_A=\int_{0}^{r} {2\pi xtan(\theta). E_oe^{-\cfrac{x}{a_{e}cos(\theta)} }cos(\theta) }d\,r$

the distance of the elemental ring from the charge is $\cfrac{x}{cos(\theta)}$

$E_{ A }=2\pi E_{ o }\int _{ 0 }^{ r }{ xtan(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} }cos(\theta) } d\, r$

and the change of $E_A$ with time $t$,

$\cfrac { \partial \, E_{ A } }{ \partial t } =2\pi E_{ o }\cfrac { \partial \, }{ \partial t }\int _{ 0 }^{ r }{ xtan(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} } cos(\theta)} d\, r$

this was in the post "Electron Orbit B Field II" dated 17 Oct 2014, corrected for the term $cos(\theta)$ to consider normal component of $E$ through the disc.

The reason why the normal component through the disc was not taken is because later the electron travels in a circle away from the axis through the center of the disc.  If the normal component is taken again by multiplying $cos(\phi)$, some $E$ field lines will be lost.  As those taken out previously by multiplying $cos(\theta)$, are now rotated and are normal to the disc.

I must be tired!

So, we are back to,  the time varying $E$ field through a disc is

$\cfrac { \partial \, E_{ A } }{ \partial t } =2\pi E_{ o }\cfrac { \partial \, }{ \partial t }\int _{ 0 }^{ r }{ xtan(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} } } d\, r$

And the diagram that causes strabismus is back.

Have a nice day.