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

Given, a cylindrical capacitor has two co-axial cylinders, of which, the outer cylinder is earthed.

Length of cylinder-l= 15 cm = $15 \times 10^{- 2} m$
Radius of inner cylinder-a = 1.4 cm= $1.4 \times 10^{- 2} m$
Radius of outer cylinder-b = 1.5 cm= $1.5 \times 10^{- 2} m$
Charge on the inner cylinder-q = $3.5 μC = 3.5 \times 10^{- 6} C$

Capacitance of the sysyem=

$C = \frac{2 {πε}_{0} l}{2.303 \log_{10} (b / a)}$

$= \frac{2 \times π \times 8.854 \times 10^{- 12} \times 15 \times 10^{- 2}}{2.303 \log_{10} \frac{1.5 \times 10^{- 2}}{1.4 \times 10^{- 2}}}$
$C = 1.21 \times 10^{- 10} F$

Since, the outer cylinder is earthed the potential of inner cylinder will be equal to the potential difference between inner and outer cylinder.

Hence, potential of inner cylinder

$V = \frac{q}{C} = \frac{3.5 \times 10^{- 6}}{1.21 \times 10^{- 10}} = 2.89 \times 10^{4} V .$

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