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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.

https://brainmass.com/physics/gauss-law/cylindrical-capacitor-gauss-law-192855

## SOLUTION This solution is FREE courtesy of BrainMass!

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 : âˆ«E.ds = Q/Îµ where Q is the total charge enclosed by the Gaussian surface. Here, Q = charge on length L of the inner cylindrical plate = Î»L

Let E1 be the radially outwards electric field on the curved face of the Gaussian cylinder. The, applying Gauss theorem : âˆ«E.ds = E1x 2Î rL = Î»L/Îµ1

Or E1 = Î»/2Î Îµ1r

As D = ÎµE, D1 = Î»/2Î r

Similarly for the region b < r < c (region 2) : âˆ«E.ds = E2x 2Î rL = Î»L/Îµ2

Or E2 = Î»/2Î Îµ2r and D2 = Î»/2Î r

We know that energy density w = Â½ ÎµE2.

Energy density in region 1 = w1 = Â½ Îµ1E12 = Â½ Îµ1(Î»/2Î Îµ1r)2 = Î»2/8Î 2Îµ1r2

Total energy stored in a concentric hollow cylinder of length L, radius r and wall thickness dr = dW1 = Energy density x Volume = [Î»2/8Î 2Îµ1r2] x 2Î rdr L = [Î»2L/4Î Îµ1r] dr

b
W1 = [Î»2L/4Î Îµ1]âˆ«1/r dr = [Î»2L/4Î Îµ1] loge(b/a)
a
Similarly total energy stored in region 2 = W2 = [Î»2L/4Î Îµ2] loge(c/b)
Total energy stored in the capacitor = W = [Î»2L/4Î Îµ1] loge(b/a) + [Î»2L/4Î Îµ2] loge(c/b)

Or W = [Î»2L/4Î Îµ1Îµ2][Îµ2loge(b/a) + Îµ1loge(c/b)] ...........(1)

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

Here, Q = Î»L. Hence, W = Î»2L2/2C .........(2)

Equating (1) and (2) : Î»2L2/2C = [Î»2L/4Î Îµ1Îµ2][Îµ2loge(b/a) + Îµ1loge(c/b)]

Or L/C = [1/2Î Îµ1Îµ2][Îµ2loge(b/a) + Îµ1loge(c/b)]

Or C = 2Î LÎµ1Îµ2/[Îµ2loge(b/a) + Îµ1loge(c/b)]

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