where one of the electron is slightly displaced as shown. We see that resultant forces act on the electrons to keep them furthest apart at a distance of 2\({r}_{e}\).
So, for the case where the electrons are on opposite sides of the positive charge,
\(\cfrac { { m }_{ e }c^{ 2 } }{ { r }_{ e } } =\cfrac { q.2q }{ 4\pi \varepsilon _{ o }{ r }^{ 2 }_{ e } } -\cfrac { q.q }{ 4\pi \varepsilon _{ o }{ (2{ r }_{ e })^{ 2 } } } \)
\({ { r }_{ e } } =\cfrac { 7q^2 }{ 16\pi \varepsilon _{ o }{ { m }_{ e }c^{ 2 } } } \)
\({ { r }_{ e } }\) = 7*(1.60217657e-19)^2/(16*pi*8.8541878176e-12*9.10938291e-31*299792458^2)
\({ { r }_{ e } }\) = 4.9314e-15 m
If we check this against the atomic radius of He, 31e-12 m. This value is much lower. Mass of the electrons are too small to consider gravitational pull. If we consider the fact that the electron charge may be higher because of the dialectic effects of oil in the oil drop experiment, \(e \rightarrow 3e\).
\({ { r }_{ e } }\) = 4.4382e-14 m
is still lower, than the published He atomic radius.
