Question

plz help fast...need urgently

Answer 1

Final Answer : (1) 1nF

We are assuming r <<l.
r(1) = outer radius
r(2) = inner radius
l= length of cylinder




Steps and Understanding :
1) Cylindrical Capacitance is given by :
$\frac{2\pi \times 8.85 \times {10}^{ - 12} \times l}{ ln( \frac{r(1)}{r(2)} ) } \\ = \frac{2\pi \times 8.85 \times {10}^{ - 12} \times 0.2}{ ln( \frac{1.82}{1.8} ) } \\ = \frac{2\pi \times 8.85 \times 0.2 \times {10}^{ - 9} }{11} \\ = 1.01 \: \times {10}^{ - 9}$
=> 1.01nF

This is approx to 1 nF