Neutral particle, \(a_{\psi}=a_{\psi\,ne}\)
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,ne} }=\left.3 \cfrac { \partial V\, }{ \partial \, x } \right|_{x=a_{\psi\,\pi}}\)
Negative particle, \(a_{ \psi \,c }\lt a_{\psi}\lt a_{\psi\,ne}\)
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,c } }=\left[3 \cfrac { \partial V\, }{ \partial \, x } -1.49824\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi\,\pi}}\)
Positive particle, \(a_{\psi\,ne}\lt a_{\psi} \lt a_{ \psi \,\pi }\)
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,\pi } }=\left[3 \cfrac { \partial V\, }{ \partial \, x } +\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi\,\pi}}\)
Between the same charge, the \(\small{\cfrac { \partial T\, }{ \partial \, x }}\) are always in the opposite directions for a test particle placed in the middle the two particles. A positive \(\small{\cfrac { \partial T\, }{ \partial \, x }}\) increases \(KE\) as the test point moves away from the particle. The particles are driven apart as they move away from the test particle.
This component is in the same direction in the interaction between particle of opposite charges. A test placed in the middle is pulled closer to one of the particle and the particles move closer.
Between neutral and other charges, because of the singular \(\cfrac { \partial T\, }{ \partial \, x }\) component (only one direction as the case for opposite charges), it is always attractive. This was explained at a macro level with induced charges or surface charge polarization.
\(\psi\) is in circular motion. The change in \(T\) along its path is manifested as a force along the radial line, along the direction as the centripetal force. An increase in \(T\) is aided by a force in the same direction as the centripetal force, inwards. A decrease in \(T\) is a force opposite to the centripetal force. This force is outwards along the radial line. This reversal in sign is already accounted for in the derivation for \(\psi\) and \(\cfrac{dq}{dx}\).
A positive change in \(V\) will drive both particles closer as a test charge placed in the middle will experience a drop in potential if the particles move closer. A negative change in \(V\) will drive the interacting particles apart, as a test charge placed in the middle of these particles will experience a drop in potential if the particles moves apart. The particles, however do not move without a change in \(T\).
\(\cfrac { \partial \, T }{ \partial \, x }\) decides whether the particle move closer or apart, ie are attractive or repulsive
\(\cfrac { \partial \, V }{ \partial \, x }\) determines the change in \(V\) as the particle move relative to each other. It is likely that,
\(\cfrac { \partial V\, }{ \partial \, x }=0\)
because the force \(\cfrac{d\psi}{dx}=0\) on the plateau just beyond \(a_{\psi\,\pi}\). If,
\(\cfrac { \partial V\, }{ \partial \, x }=0\)
then there is no change in \(V\).
For \(\cfrac { \partial \, V }{ \partial \, x }\) positive, particles moving closer experience a drop in \(V\). For \(\cfrac { \partial \, V }{ \partial \, x }\) negative, particles moving apart experience a drop in \(V\). And vice-versa.
Furthermore, it does not matter which force comes from which particle acting on the test particle in the middle of the particles. It is possible to swap ownership of the forces as long the same forces still acts on the test particle.
In the case of two interacting negative charge particles, the two negative forces can swap ownership and be equivalent to the interaction of two positive particles with two positive forces. This is because,
\(\cfrac { \partial T\, }{ \partial \, x }\)
changes the energy (a scalar) of the test particle directly,
In a vector force, a pair of forces in tension is different from a pair of forces in compression.
What happen to \(a_{\psi\,c}\) particle coalescing? It is awkward here because we are using wave equation results to explain charged big body interactions. In big body interaction we can use test particle or test point between them and deduce the movement of the charge bodies from the movement of the test particle. Wave however, superimpose and interact directly.
As waves, the negative particles act on each other directly and force ownership swap cannot happen as the forces are vectors. Negative particles are attractive to each other because of the negative
\(\cfrac { \partial \, T }{ \partial \, x }\)
component. A positive
\(\cfrac { \partial \, T }{ \partial \, x }\)
component drives the particles apart.
"Like charges repel and unlike charges attract" is not true at the quantum level where waves dictate interactions. The use of a test particle or test point between the bodies accounts for the presence of an intervening medium between the charged bodies.
Have a nice day.
