Field Equation Exponential Form

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.