Correct Answer - `0.2xx10^(9) F ; 18xx10^(4) V`
Here, `r_(a) = (0.2)/(2) = 0.1 m, r_(b) = 0.15 m`
`q = 6 mu C = 6xx10^(-6) C, K = 6`.
`C = 4pi in_(0) K = (r_(a) r_(b))/(r_(b) - r_(a))`
`C = (6)/(9xx10^(9)) (0.1xx0.15)/((0.15 - 0.1)) = 0.2xx10^(9) F`
`V = (q)/(4pi in_(0)) (r_(b) - r_(a))/(r_(b) r_(a))`
`V = 9xx10^(9)xx6xx10^(-6) xx (0.15 - 0.1)/(0.15xx0.1) = 18xx10^(4) V`