Answer:
The capacitance is 1.75 nF
Explanation:
From the question we are given that
The inner radius is [tex]r_{in} = 0.001[/tex]
The outer radius is [tex]r_{out} = 0.0011 \ m[/tex]
Length of the capacitor is [tex]L = 1m[/tex]
The dielectric constant is [tex]Di = 2 \ for \ 0 < \phi < \pi[/tex]
The dielectric constant is [tex]Di_2 = 4 \ for \ \pi < \phi < 2\pi[/tex]
Generally the capacitance of a capacitor can be mathematically represented as
[tex]C = \frac{\pi \epsilon_0 Di_1 L}{ln\frac{r_{out}}{r_{in}} } + \frac{\pi \epsilon_0 Di_2L}{ln\frac{r_{out}}{r_{in}} }[/tex]
[tex]= \frac{\pi \epsilon_0 L (Di_1 + Di_2)}{ln\frac{r_{out}}{r_{in}} }[/tex]
[tex]= \frac{(3.142)(8.85*10^{-12})(1)(2+4)}{ln\frac{0.0011}{0.001} }[/tex]
[tex]=1.75*10^{-9} F[/tex]
[tex]1.75nF[/tex]