Determining potential in coaxial cylinders If I have two conducting, coaxial cylinders as shown:

The potential at the outer cylinder is equal to zero. And I apply a potential difference across both cylinders in the form  $V_a - V_b = V_{ab}$, how can I find an expression of the potential everywhere, in terms of $V_{ab}$?
So this is what I have so far...
We place a $+Q$ on the inner cylinder and a -Q on the outer one. From Gauss' law, I know that $Q$ enclosed = $+Q$ so $\frac{Q}{\epsilon}$ holds.
I can set that equal to $2\pi rl E$ because $dA$ over all areas is just the surface area of the curved part of a cylinder. So $\frac{Q}{\epsilon} = 2rl\pi E$ and $E = \frac{Q}{\epsilon 2\pi rl}$ is the electric field between cylinders
Now, $E = \frac{dV}{dr}$ so the integral (from a to b) $dV$ is equal to $\Delta V_{ab} = -\frac{\lambda}{2\pi \epsilon}$
After doing all the integrals and bounds I got $\Delta V){ab} = -\frac{\lambda}{2\pi \epsilon}ln(\frac{b}{a})$
But this is only the potential  between the two cylinders. I need the potential everywhere. So I still need to find the potential at the inside of the smaller cylinder and the potential on the outside of the bigger cylinder. Now I'm stuck!
 A: Consider starting with Laplace's equation in cylindrical, as this will give you the potential directly:
$$
\frac{1}{r}\frac{d}{d r}
\left(r\frac{d V}{d r}\right)=0
$$
since the space between the cylinder is charge-free.  Moreover, you can use
$d/dr$ rather than $\partial/\partial r$ since by symmetry $V=V(r)$ only.
It follows from this that 
$$
r\frac{dV}{dr}=C_1\qquad\Rightarrow\qquad
V(r)=C_1\log(r)+C_2
$$
with $C_1$ and $C_2$ two integrating constants.  You can find $C_1$ and $C_2$ using $V(r)$ at $a$ and $b$, and then use your expression for the potential difference to convert to an expression containing $\lambda$ etc.
I think you probably approached it right and found the correct potential difference between the cylinders but the next step is to integrate your $\vec E$-field from $a$ to $r$, where $r$ is a point inside the cylinder, so,as to get $V(r)-V(a)$.  Since you are given a potential difference solving Laplace's equation is a little more direct.
A: As you have figured out, $V_{ab} \equiv V_a - V_b = - \int^a_b \frac{\lambda}{2 \pi \epsilon_0r} = \frac{\lambda}{2 \pi \epsilon_0} ln(\frac{a}{b})$.  But we don't know what $\lambda$ is yet!  In fact, we must use this equation to derive its value in terms of $V_{ab}$, which is assumed known.  Then, to get the potential at any radius $r$, simply do your integral of the field again, but this time let the limit be $r$ instead of $a$.  Therefore, $V(r)=\int_b^r\frac{\lambda}{2 \pi \epsilon_0r}$ (no minus sign due to change of integration direction) $=\frac{\lambda}{2 \pi \epsilon_0} ln(\frac{r}{b})$, where $\lambda = \frac{2 \pi \epsilon_0 V_{ab}}{ln(\frac{a}{b})}$.  Hope this helps!
