A cylinderical capacitor is made up of a conducting cylinder or wire of radius a surrounded by another concentric cylinderical shell of radius b (b>a).

Let L be the length of both the cylinders and charge on inner cylender is +Q and charge on outer cylinder is -Q.

For calculate electric field between the conductors using Gauss's law consider a Gaussian surface of radius r and length L^{1} as shown in figure 4.

According to Gauss's law flux through this surface is $ \frac {q}{\epsilon _0}$ where q is net charge inside this surface.

We know that electric flux is given by
$\phi =E.A$
$=EA cos \theta$
$=EA$
since electric field is constant in magnitude on the Gaussian surface and is perpendicular to this surface. Thus,
$\phi=E(2 \pi rL)$
since $\phi=\frac {q}{\epsilon _0}$
=> $ E(2 \pi rL)=\frac {\lambda L}{\epsilon _0}$
where $\lambda = Q/L$ = charge per unit length
So,
$E=\frac{\lambda }{2\pi \epsilon _{0}r}$

If potential at inner cylinder is V_{a} and V_{b} is potential of outer cylinder then potential difference between both the cylinders is
V=V_{a} and V_{b}=∫Edr
where limits of integration goes from a to b.

Potential of inner conductor is greater then that of outer conductor because inner cylinder carries positive charge. Thus potential difference is
$$V=\frac {Qln(b/a)}{2\pi \epsilon _{0}L}$$

Thus capacitance of cylinderical capacitor is
$C=\frac {Q}{V}
or,
$C=\frac{2\pi \epsilon _{0}L}{ln (b/a)}$ ------(5)

From equation 5 it can easily be concluded that capacitance of a cylinderical capacitor depends on length of cylinders.

More is the length of cylinders , more charge could be stored on the capacitor for a given potential difference.

Question
A cylindrical capacitor is constructed using two coaxial cylinders of the same length 10 cm of radii 5 mm and 10 mm.
(a) calculate the capacitance
(b) another capacitor of the same length is constructed with cylinders of radii 8 mm and 16 mm. Calculate the capacitance . Solution
Capacitance is given by
$C=\frac{2\pi \epsilon _{0}L}{ln (b/a))}$
(a) Here L=10 cm = .1 m
b=10 mm=, a=5 mm
$C= \frac {2 \times 3.14 \times 8.88 \times 10^{-12} \times .10}{ln (10/5}$
$=8 pF$
(b) Now b=16 mm and a=8 mm
Since ln 16/8 = ln 10/5
C= 8pF