Question about capacitance on a double cylinder I had an exercise that I am struggling to solve, I have 2 concentric metal surfaces with $R_1= 1 \,\text{cm}$ and $R_2 = 2 \,\text{cm}$.
The cylinders have infinite extension along the direction normal to the paper and placed in free space. 
The inner cylinder carries a charge of $Q_1 = 5 \,\text{mC}$ per meter length, and the outer cylinder carries a charge of $2 = −5 \,\text{mC}$ per meter length.
I need to calculate the capacitance per meter length of the double-cylinder structure.
I tried by going with the formula of capacitance since I know the magnitude of $Q$ and I could calculate the voltage across the 2 surfaces of the cylinders but I started getting weird numbers and I kind of got lost.
I added a picture of the whole exercise and the scheme as well.

Do you have any suggestions on how to solve this question?
 A: Probably there are smarter ways to solve this problem, but I would do the following 


*

*Use Gauss law to compute the electric field between the cylinders. Let me define the linear charge density $\lambda = Q/l = 5\,m\mbox{C}/\mbox{m}$. 
Let me remember that Gauss law states that the flux of the electric field across a surface $S$ is equal to the total charge inside the surface, divided by $\epsilon_0$ (as I guess there is nothing between the cylinders), i.e.
\begin{equation}
\Phi_S(\textbf{E}) = \frac{Q_{in}}{\epsilon_0}
\end{equation}
Given the geometry of the system, we can imagine that the electric field between the cylinders is directed radially and is symmetrical for rotations around the central axis of the system. If we make the smart choice of considering $S$ as a cylinder with radius $r$ coaxial with the metal surfaces, we have that the flux $\Phi_S(\textbf{E})$ simply becomes
\begin{equation}
\Phi_S(\textbf{E}) = \int_{S(r)}\textbf{E}(r)\cdot d\textbf{S}= E(r) 2\pi r l 
\end{equation}
I would like to point out that the flux across the basis of the cylinder $S$ is zero.
Computing the charge inside $S$ is super easy as well, as we knon the linear charge density $\lambda$
\begin{equation}
Q_{in} = \lambda l 
\end{equation}
Therefore, plugging in the last two expressions into the Gauss law, we end up with
\begin{equation}
 E(r) 2\pi r l  = \frac{\lambda l }{\epsilon_0}
\end{equation}
Thence, the radial (and only component) of the electric field inside the double cylinder is 
\begin{equation}
 E(r) = \frac{\lambda}{2\pi\epsilon_0r}\qquad\mbox{for}\qquad R_1\le r\le R_2
\end{equation}

*I compute now the potential difference $\Delta V$ between the outer and inner cylinder
\begin{equation}
\Delta V = \int_{R_1}^ {R_2} E(r)\,dr = \frac{\lambda}{2\pi\epsilon_0}\int_{R_1}^ {R_2} \frac{1}{r}\,dr = \frac{\lambda}{2\pi\epsilon_0}\ln\left(\frac{R_2}{R_1}\right)
\end{equation}

*I use the definition of capacitance
\begin{equation}
C = \frac{Q}{\Delta V} = 2\pi\epsilon_0\frac{\lambda l}{\lambda\ln\left(\frac{R_2}{R_1}\right)}=2\pi\epsilon_0\frac{l}{\ln\left(\frac{R_2}{R_1}\right)}
\end{equation}

*Finally the capacitance per unit length is 
\begin{equation}
\mathcal{C} = \frac{C}{l} = \frac{2\pi\epsilon_0}{\ln\left(\frac{R_2}{R_1}\right)}
\end{equation}
