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Let's suppose we have a sphere but unlike theoretical ones it'll have some thickness say $\Delta r$ and inner radius $R$. What I was wondering about is how will it behave if we place some charge $q$ in the center? What would the field look like and what will be the charge density on either side. When it's grounded and when it's not.

My attempt:

When it's not grounded I though that outside if we create a spherical surface Gaussian with $r>R+\Delta r$ then there's only the charge $q$ to consider so it'll behave like a exactly like $q$ on it's own meaning $E=k \frac{q}{r^2} \hat{r}$ , and the same goes for $r<R$. inside the shell it'll be zero as it's a conductive surface so $\phi = const$. On the inner surface and the outer surface the sum of all charge will have to be $-q , q $ accordingly as they have to cancel each other so the density will be $\sigma = \frac{-q}{4\pi R^{2}} , \frac{q}{4\pi\left(R+\Delta r\right)^{2}} $ .

When it comes to the grounded version I'm a little confused as I don't really understand what's the difference apart from the fact that the initial $\phi =0 $ but electrical charge will still be pulled to the sphere from $\infty$ so it seems like there's no difference but I'm a beginner so I'm not sure whether my deduction are valid or not and it seems kind of fishy to me but I can't really point to what is essentially wrong.

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You say, "electrical charge will still be pulled to the sphere from $\infty$". - But this feature is new in the grounded case! You have yourself said, that in the ungrounded case the net charge has to be zero. – Ilja Apr 5 at 12:36

First of all, how can a sphere have a thickness $dr$. You must have a sphere of radius $R$ itself(no question of $dr$ unless its a shell)?

The charge $Q$ shall come out and get distributed over the outer surface of sphere only if its conducting. The thing you did for not ground condition is corrected except that the assumption of $+q$ and $-q$ is wrong: It wil be just $\sigma=\frac{Q}{4\pi\epsilon R^2}$. Remeber field inside a conductor is $0$ only. If grounded, all charge wil flow to earth as Earth is at $0$ potential and the sphere wil be almost completely discharged, $\sigma=0$.

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Actually, the charge $Q$ does not come out and distribute itself on the surface of the sphere. Rather, the electrons on the conducting sphere accumulate on the inner surface of the sphere so that there is a net charge of $-Q$ on the inner surface. And because of the excess of electrons on the inner surface, the outer surface has a deficit of electrons leading to a $+Q$ charge. When you ground the sphere, electrons will rush up from the earth to compensate for that deficit and the sphere will have a net charge of $-Q$ on it with the outer surface being charge free. – Bolt64 May 26 '13 at 13:11
That's what I said as well. Charge on the outer surface is $0$ and charge on the inner surface is $-Q$. Hence, net charge on the sphere is the sum of inner and outer charges, which is $-Q$. – Bolt64 Apr 22 '14 at 6:05

I think you have the right approach for the non-grounded. For the grounded case, we can use the uniqueness theorem, which says that given a charge distribution and the voltage on the boundaries, there is only one solution for the voltage. The grounded sphere has a $V = 0$ surface at $R + \delta R$ and at infinity, and no charge outside. I can solve this by positing $V = 0$ everywhere outside the sphere, no electric field. Therefore, by uniqueness, this is the only solution and there is no electric field outside the sphere. Since there's also no electric field in the conductor, we can see that there must be an induced charge on the inner surface, but that's it.

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correct, there will be charge (of opposite sign and the same magnitude as the central charge, by the way) on the inner surface only. – Ilja Apr 5 at 12:34

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