Calculate $\phi$, satisfying $\nabla^2 \phi=0$ between the two cylinders $r=a$, on which $\phi=0$, and $r=b>a$, on which $\phi=V$.
I calculate it and found the solution is $$\phi=\frac{V}{\log b-\log a}\log r-\frac{V}{\log b-\log a}\log a.$$However, I am not sure if it is right.
Actually, I am having trouble in determine its dimension. Is it $2$ or $3$ if I use the following formula: $v(r)=b\log r +c$, when $n=2$; $v(r)=\frac{b}{r^{n-2}}$, when $n\ge3$?
But they all seem not right, since this is for a radially symmetric boundary condition. In this case, it is not radially symmetric.
Could someone kindly help me look at it? Thanks!
If the function is considered to be constant along one direction, then the function needs only to be harmonic in the other transverse directions. For instance if we are working in $\mathbb{R}^3$ and the cylinder extends in the $z$ direction, then we should look for some $\phi(x,y,z)=\phi(x,y)$ (that is, not really depending on $z$), and then $\nabla^2 \phi=\dfrac{\partial^2\phi}{\partial x^2}+\dfrac{\partial^2\phi}{\partial y^2}+\dfrac{\partial^2\phi}{\partial z^2}=\dfrac{\partial^2\phi}{\partial x^2}+\dfrac{\partial^2\phi}{\partial y^2}$. For this reason, we may think the problem as stated in $\mathbb{R}^2$ and the solution you propose is right, although we get a 'cylindrical' function and not a 'spherical' one (if we should deal with spherical function, then we would take $n=3$).
I hope these remarks are useful.