**Question:
**

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 µC. 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).

**Solution:**

Length of a co-axial cylinder,* l* = 15 cm = 0.15 m

Radius of outer cylinder, *r*1 = 1.5 cm = 0.015 m

Radius of inner cylinder, *r*2 = 1.4 cm = 0.014 m

Charge on the inner cylinder, *q* = 3.5 µC = 3.5 × 10−6 C

Capacitance of a co-axial cylinder of radii $r_{1}$ and $r_{2}$ is given by the relation,

$C=\frac{2 \pi \epsilon_{0} l}{\log _{e} \frac{r_{1}}{r_{2}}}$

Where,

$\epsilon_{0}=$ Permittivity of free space $=8.85 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \mathrm{C}^{2}$

$\therefore C=\frac{2 \pi \times 8.85 \times 10^{-12} \times 0.15}{2.3026 \log _{10}\left(\frac{0.15}{0.14}\right)}$

$=\frac{2 \pi \times 8.85 \times 10^{-12} \times 0.15}{2.3026 \times 0.0299}=1.2 \times 10^{-10} \mathrm{~F}$

Potential difference of the inner cylinder is given by,

$V=\frac{q}{C}$

$=\frac{3.5 \times 10^{-6}}{1.2 \times 10^{-10}}=2.92 \times 10^{4} \mathrm{~V}$

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