$A$ cylindrical capacitor has two co-axial cylinders of length $15 \; cm$ and radii $1.5 \; cm$ and $1.4 \; cm$. The outer cylinder is earthed and the inner cylinder is given a charge of $3.5 \; \mu C$. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e.,bending of field lines at the ends).

(N/A) Length of the co-axial cylinder,$l = 15 \; cm = 0.15 \; m$.
Radius of the outer cylinder,$r_1 = 1.5 \; cm = 0.015 \; m$.
Radius of the inner cylinder,$r_2 = 1.4 \; cm = 0.014 \; m$.
Charge on the inner cylinder,$q = 3.5 \; \mu C = 3.5 \times 10^{-6} \; C$.
The capacitance $C$ of a cylindrical capacitor is given by $C = \frac{2 \pi \epsilon_0 l}{\ln(r_1/r_2)}$.
Using $\epsilon_0 = 8.854 \times 10^{-12} \; F/m$:
$C = \frac{2 \times 3.14159 \times 8.854 \times 10^{-12} \times 0.15}{\ln(1.5/1.4)} \approx \frac{8.347 \times 10^{-12}}{0.06899} \approx 1.21 \times 10^{-10} \; F$.
The potential $V$ of the inner cylinder is $V = q/C$.
$V = \frac{3.5 \times 10^{-6}}{1.21 \times 10^{-10}} \approx 2.89 \times 10^4 \; V$.