From the post "May The Hunt Begin" dated 01 Jun 2016,
We know that \(\Delta q\) decreases to zero at \(r=a_{\psi}\), as such,
\(-\Delta q=m.4\pi r^{ 2 }.\Delta r\)
\({q}=m.4\pi a^2_{\psi}=- \cfrac { dq }{ d\,r}|_{r=a_{\psi} }\)
previously \(q\) has retained the minus sign. If we solve for
\( \cfrac { dq }{ d\,r }=-q\)
we have,
\(\int { \cfrac { 1 }{ q } } dq=\int { -1} dr\)
\(ln(q)+C= -r+B\)
\(ln(Aq)= -( r-D) \)
\(Aq=e^{- (r-D ) }\)
\(q=E.e^{-(r-D )}\)
where at \(r=a_{\psi}\), \(\cfrac { dq }{ d\,r}|_{r=a_{\psi} }=-q_{a_{\psi}}\)
we let \(D=a_{\psi}\), \(E={q_{a_{\psi}}}\)
\(q=q_{a_{\psi}}.e^{-(r-a_{\psi}) }\)
\(\because\)
\(\cfrac { dq }{ d\, r } =q_{ a_{ \psi } }.e^{ -(r-a_{ \psi }) }=-q\)
And,
\(\cfrac { q }{ \varepsilon _{ o } } =\cfrac { q_{ a_{ \psi } } }{ \varepsilon _{ o } } e^{(r-a_{ \psi }) }\)
This is the change in flux \(\cfrac { q }{ \varepsilon _{ o } }\) along a radial line \(r\), which is also the Newtonian force along the line joining two particles center to center.
It is the change in \(q\), not \(q\) itself but \(\Delta q\) decreasing at \(r=a_{\psi}\) that invites a minus sign in the exponent of the expression for force in exponential form.
The direction of the force in expression (*) depends on the sign of \(q_{a_{\psi}}\) or \(q\).
This expression will satisfy the wave equation, whereas the expression,
\(F_{\small{G}}\cfrac{1}{4\pi a^2_{\psi}}=\cfrac{q}{4\pi\varepsilon_o a^2_{\psi}}=2{ mc^{ 2 } }.ln(cosh(\pi))\)
does not; as shown previously.
Expression (*) should be interpreted as the decreasing effects of \(q\) over \(r\) beyond \(r=a_{\psi}\) and not the spread of \(q\) beyond \(a_{\psi}\). The particle extends up to \(r=a_{\psi}\) and ends abruptly, the gradient of which is given by,
\( \cfrac { dq }{ d\,r }=-q\)
And the force is exponential. May the force extends beyond you exponentially.
