Question

A cylindrical capacitor has two co-axial cylinder of length 20 cm and radii 1.8 cm and 1.82 cm. The outer cylinder is earthed. The net capacitance of the system is $N \times 10^{-9}$ F. The value of $N$ is ______

Answer 1

1.0

Answer 2

The capacitance of a cylindrical capacitor is calculated using the formula $C = \frac{2\pi\varepsilon_{0}L}{\ln\left( \frac{r_{o}}{r_{i}} \right)}$.

Given $L = 0.2 \text{ m}$, $r_i = 0.018 \text{ m}$, $r_o = 0.0182 \text{ m}$, and $\varepsilon_0 = 8.854 \times 10^{-12} \text{ F/m}$.

Substitute these values: $C = \frac{2\pi (8.854 \times 10^{-12}) (0.2)}{\ln\left( \frac{0.0182}{0.018} \right)}$

$C = \frac{11.12158 \times 10^{-12}}{\ln(1.011111...)}$

$C = \frac{11.12158 \times 10^{-12}}{0.0110505}$

$C \approx 1006.43 \times 10^{-12} \text{ F} \approx 1.00643 \times 10^{-9} \text{ F}$.

Since the capacitance is $N \times 10^{-9}$ F, $N \approx 1.00643$. Rounding to one decimal place, $N=1.0$.