Find the temperature field, given that the temperature satisfies $\nabla^2 T = 0$, and T is a function only of r (working in cylindrical polars)
I'm assuming I'm supposed to solve the Laplacian. So I've solved this and got T=A.ln r + B Is this the temperature field or do I have to do something else?
In two dimensions, you fit $A\ln r+B$ to the boundary conditions (whatever they may be).
In three dimensions, you would do the same with $Ar^{-1}+B$. In general, it's $Ar^{2-n}+B$ in $n\ge 3$ dimensions.
By the way, an easy way to see that any radial solution ($T=T(r)$) of $\nabla^2 T =0$ has to be of the above form is to consider the flux of gradient across the sphere of radius $r$. For a harmonic function, the flux must be independent of $r$. Hence, $r^{n-1}T'(r)$ is constant, which after integration gives the above.