Most present electrons around a proton nucleus as shown where the electrons are furthest apart from each other. But if we write down the force equation based on simple electrostatic then.\(\cfrac{{q}_{e}{q}_{e}}{4\pi \varepsilon_o {r}^2_{ee}}=\cfrac{2{q}_{p}{q}_{e}}{4\pi \varepsilon_o {r}^2_{pe}}\)
if the basic positive and negative charge magnitudes are the same \({q}_{e}={q}_{p}\) then,
\(\cfrac{1}{{r}^2_{ee}}=\cfrac{2}{{r}^2_{pe}}\)and
\(\sqrt{2}{r}_{ee}={r}_{pe}\)
This is possible only if the electrons are all on the same side.
But, we see that the center negative charge, experiences a net force towards the positive end.
The configuration is impossible unless, the center charge can fall no further towards the positive charge.
Consider a negative charge accelerated by the pull of a positive charge, moves towards the charge at greater and greater speed. What if the negative charge reaches light speed before colliding with the positive charge? The negative charge will go into circular motion around the positive charge where the attractive force between the charges provides for the centripetal force. And the negative charge is at light speed. In this case,
\(-\cfrac{{m}_{e}c^2}{{r}_{e}}=-\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e}}-\cfrac{q.q}{4\pi\varepsilon_o {r}^2_{ee}}\)
\(\because {r}_{pe}={r}_{ee}+{r}_{e}\)
We have then,
\(\cfrac{{r}_{ee}+{r}_{e}}{{r}_{ee}}=\sqrt{2}\)
\({r}_{ee}=\cfrac{1}{\sqrt{2}-1}.{r}_{e}\)
\(\cfrac{{m}_{e}c^2}{{r}_{e}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e}}+\cfrac{q.q}{4\pi\varepsilon_o {r}^2_{e}}.({\sqrt{2}-1})^2\)
\( { m }_{ e }c^{ 2 }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ r }_{ e } } (2+({ \sqrt { 2 } -1 })^{ 2 })\)
\({ m }_{ e }c^{ 2 }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ r }_{ e } } (5-2\sqrt { 2 } )\)
\({ r }_{ e }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ { m }_{ e }c^{ 2 } } } (5-2\sqrt { 2 } )\)
\({ r }_{ ee }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ { m }_{ e }c^{ 2 } } }\cfrac{ (5-2\sqrt { 2 } )}{\sqrt{2}-1}\)
\({ r }_{ pe }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ { m }_{ e }c^{ 2 } } }\cfrac{ (5-2\sqrt { 2 } )}{\sqrt{2}-1}.\sqrt{2}\)
We find that the resultant force on the furthest negative charge has a acceleration component towards the positive charge. This charge is still falling towards the positive charge. This configuration is stable only if this charge is also performing circular motion.
This invalidates the the first equation where the net force on this charge was considered zero. We will start again.
