Empty Sphere field and potential

An empty sphere is charged. (the inner and outer radii of the sphere are $R_0$ and $R$). Please find the empty sphere field and potential as functions of $R$. What is the difference between the cases when the sphere is free in space and when the sphere is grounded?

Charged Empty Sphere Field

When the sphere is not grounded, the inside of the sphere is charged by induction with $–Q$ (from the central charge $+Q$) and outside of the sphere gets $+Q$ (from the neutrality condition: the entire sphere has to stay neutral).

When the sphere is grounded electrons come from the ground and neutralize the $+Q$ charge present on the outside of the sphere. Therefore the outside of the sphere has no charge on it, after it had been neutralized. It remains a total $–Q$ charge distributed over the inside of the sphere, from the initial charging, when the sphere was not grounded.

The reason of why $–Q$ on the inside of the sphere does not spread out in the metal volume is that in this case there will be a net field inside the metal that will move the charge back on its surface. Usually all virtual electric fields inside move electrical charges until equilibrium. Now we can solve to find the empty sphere field and the potential.

For $R>R_{inside}$ ($R>R_0$)  one has $Q_{inside}=+Q-Q=0$ so that

$E(R)S(R)=Q_{inside}/ϵ=0$  so $E(R)=0$ for $R>R_0$

For $R<R_{inside}$  one has $Q_{inside}=Q$  so that

$E(R)*4πR^2=Q/ϵ$    or $E(R)=Q/(4πϵR^2)$

If we make the convention $V_{ground}=V_{sphere}=0$  then

For $R>R_{inside}$     $V_{ouside} (R)=-∫_(R_{inside})^∞ E(R)dR= 0 =V_{sphere}$

For $R<R_{inside}$     $V(R)= -∫ E(R)dR=Q/4πϵR+C$


$V(R_{inside} )=0 V$ so that $C=-Q/(4πϵR_i )$  so that $V(R)=Q/4πϵ (1/R-1/R_i )$