Answer:
(i) Electric field outside the shell:
For point r>R; draw a spherical gaussian surface of radius r.
Using gauss law, ∮E.ds=q0qend
Since E is perpendicular to gaussian surface, angle betwee E is 0.
Also E being constant, can be taken out of integral.
So, E(4πr2)=q0q
So, E=4πε01r2q
In the first shell there can be a maximum of 2 electrons, then 8o in the second one and 8 in the third one.
i believe your looking for this word ?? "fundamental".
