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A cable consisting of a wire `3 mm` thick dielectric of relative permitively 10. Calculate the capacitance of `1 km` length of the cable.

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Correct Answer - `0.506 muF`
Here `C_(1) = 3 muF, C_(2) = 6 muF`
`r_(b) = 1.5 + 3 = 4.5 mmn = 4.5xx10^(3) m`
`L = 1 km = 10^(3) m , K = 10 , C = ?`
`C = (2pi in_(0) L K)/(2.303 log_(10) ((r_(b))/(r_(a)))) = (2pi xx 8.85xx10^(-12)xx10^(3)xx10)/(2.303 log ((4.5xx10^(-3))/(1.5xx10^(-3))))`
`= 50.6xx10^(-8) F = 0.506 mu F`

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