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    Cylindrical Capacitor: Gauss Law

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    A capacitor consists of two concentric long cylindrical conductors of radii a and c, where c > a, and each is of negligible thickness. The space between the conductors is filled with two layers of dielectric materials. Using cylindrical co-ordinates with z-axis along the axis of the cylindrical conductors, the space a<r<b is filled with an LIH (linear, isotropic and homogeneous) dielectric of relative permittivity e1 and the space b<r<c with a second LIH dielectric of relative permittivity e2 . The cylinders are sufficiently long that end effects can be neglected. The charge per unit length of the inner conducting cylinder (r=a) is lembda, and that on the outer conducting cylinder (r=c) is -lembda.

    a) Use the integral version of Gauss's law to find the electric field E and the electric displacement D in the region a < r < c. Hence calculate the electrostatic field energy stored in a length l of the capacitor.

    b) Hence find the capacitance of a length l of the capacitor.

    (Please see the attachment also)

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    Solution Preview

    Please refer to the attachment.

    a) For the region a < r < b (region 1) : Let us consider a cylindrical Gaussian surface of length L shown in dotted red in the fig. and apply Gauss theorem : ∫E.ds = Q/ε where Q is the total charge enclosed by the Gaussian surface. Here, Q = charge on length L of the inner cylindrical plate = λL

    Let E1 be the radially outwards electric field on the curved face of the ...

    Solution Summary

    The expert examines cylindrical capacitors for Gauss Laws.