InterviewSolution
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InterviewSolution
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The gap between the plates of a parallel-plate capacitor is filled up with an inhomongeous poorly conducting medium whose conductivity varies lineraly in the direaction perpendicular to the plates form the `sigma_(1) = 1.0 pS//m to sigma_(2) = 2.0 pS//m`. Each plate has an area `S = 230 cm^(2)`, and the separation between the plates is `d = 2.0 mm` . Find teh current flowing thorugh the capacitor due to a voltage `V = 300 V`. |
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Answer» The resistance of a layer of the medium, of thickness `dx` and at a distance `x` from the first plate of the capacitor is given by, `dR = (1)/(sigma (x)) (dx)/(S)` ....(1) Now, since `sigma` varies linerly with the distance from the plane. It may be represented as, `sigma = sigma_(1) + ((sigma_(2) - sigma_(1))/(d)) x`, at a distance `x` from any one of plate. From Eq. (1) `dR = (1)/(sigma_(1) + ((sigma_(2) - sigma_(1))/(d))x) (dx)/(S)` or, `R = (1)/(S) int_(0)^(d) (dx)/(sigma_(1) + ((sigma_(2) - sigma_(1))/(d))x) = (d)/(S(sigma_(2) - sigma_(1))) In (sigma_(2))/(sigma_(1))` Hence, `i = (V)/(R) = (S V (sigma_(2) - sigma_(1)))/(d In (sigma_(2))/(sigma_(1))) = 5 nA` |
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