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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.

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    SOLUTION This solution is FREE courtesy of BrainMass!

    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 Gaussian cylinder. The, applying Gauss theorem : ∫E.ds = E1x 2ΠrL = λL/ε1

    Or E1 = λ/2Πε1r

    As D = εE, D1 = λ/2Πr

    Similarly for the region b < r < c (region 2) : ∫E.ds = E2x 2ΠrL = λL/ε2

    Or E2 = λ/2Πε2r and D2 = λ/2Πr

    We know that energy density w = ½ εE2.

    Energy density in region 1 = w1 = ½ ε1E12 = ½ ε1(λ/2Πε1r)2 = λ2/8Π2ε1r2

    Total energy stored in a concentric hollow cylinder of length L, radius r and wall thickness dr = dW1 = Energy density x Volume = [λ2/8Π2ε1r2] x 2Πrdr L = [λ2L/4Πε1r] dr

    W1 = [λ2L/4Πε1]∫1/r dr = [λ2L/4Πε1] loge(b/a)
    Similarly total energy stored in region 2 = W2 = [λ2L/4Πε2] loge(c/b)
    Total energy stored in the capacitor = W = [λ2L/4Πε1] loge(b/a) + [λ2L/4Πε2] loge(c/b)

    Or W = [λ2L/4Πε1ε2][ε2loge(b/a) + ε1loge(c/b)] ...........(1)

    b) Energy stored in the capacitor is also given by : W = Q2/2C

    Here, Q = λL. Hence, W = λ2L2/2C .........(2)

    Equating (1) and (2) : λ2L2/2C = [λ2L/4Πε1ε2][ε2loge(b/a) + ε1loge(c/b)]

    Or L/C = [1/2Πε1ε2][ε2loge(b/a) + ε1loge(c/b)]

    Or C = 2ΠLε1ε2/[ε2loge(b/a) + ε1loge(c/b)]

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 7:28 pm ad1c9bdddf>