Continuum Electromechanics Exam Help
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Problem 1 Prob. 2.16.5 A planar region, shown in Table 2.16.1, is filled by an inhomogeneous dielectric, with a permittivity that depends on x:
The free charge density is zero. (a) Show that the potential distribution is
where (b) Show that the transfer relations are
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Problem 2 Prob. 4.3.3 The moving member of an EQS device takes the form of a sheet, supporting the surface charge of and moving in the z direction, as shown in Fig. P4.3.3. Electrodes on the adjacent walls constrain the potentials there.
Courtesy of MIT Press. Used with permission. Problem 4.3.3 in Melcher, James R. Continuum Electromechanics. Cambridge, MA: MIT Press, 1981, p. 4.57. ISBN: 9780262131650. Problem 3 A general right-handed orthogonal curvilinear coordinate system is described by variables (u, v, w), where liveexamhelper.com
Since the incremental coordinate quantities du, dv, and dul do not necessarily have units of length, the differential length elements must be multiplied by coefficients that generally are a function of u, v, and w:
(a) What are the h coefficients for the Cartesian, cylindrical, and spherical coordinate systems?
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(b) What is the gradient of any function f(u, v, w)? (c) What is the area of each surface and the volume of a differential size volume element in the (u, v, w) space? (d) What are the curl and divergence of the vector
(e) What is the scalar Laplacian (f) Check your results of (b)-(e) for the three basic coordinate systems. 1. DIFFERENTIAL OPERATORS IN CYLINDRICAL AND SPHERICAL COORDINATES If r, φ, and z are circular cylindrical coordinates
and are unit vectors in the directions of increasing values of the corresponding coordinates,
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If r, θ, and φ are spherical coordinates and
are unit vectors in
the directions of increasing values of the corresponding coordinates,
If r, θ, and φ are spherical coordinates
are unit vectors in
the directions of increasing values of the corresponding coordinates,
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2. SOLUTIONS OF LAPLACE’S EQUATIONS A. Rectangular coordinates, two dimensions (independent of z):
B. Cylindrical coordinates, two dimensions (independent of z):
C. Spherical coordinates, two dimensions (independent of φ):
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Problem 1 Prob. 2.16.5 Gauss’ law and E =
requires that if there is no free charge
For the given exponential dependence of the permittivity, the x dependence of the coefficients in this expression factors out and it again reduces to a constant coefficient expression
In terms of the complex amplitude forms from Table 2.16.1, Eq. 2 requires that
Thus, solutions have the form exp(px) where p =
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The linear combination of these that satisfies the conditions that on the upper and lower surfaces respectively is as given in the problem. The displacement vector is then evaluated as
Evaluation of this expression at the respective surfaces then gives the transfer relations summarized in the problem. Problem 2 Prob. 4.3.3
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Figure 1: Force per unit area on middle charged surface (c/d) is found using the Maxwell stress tensor (Image by MIT OpenCourseWare.) With positions as designated in the sketch, the total force per unit area is
With the understanding that the surface charge on the sheet is a given quantity, boundary conditions reflecting the continuity of tangential electric field at the three surfaces and that Gauss’ law be satisfied through the sheet are
Bulk relations are given by Table 2.16.1. In the upper region
and in the lower
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In view of Eq. 6, Eq. 5 becomes
so what is now required is the amplitude The surface charge, given by Eq. 6, as the Difference follows in terms of the potentials from taking the difference of Eqs. 7 (second row) and 8 (first row). The resulting expression is solved for
Substituted into Eq. 9 (where the self terms in are imaginary and can therefore be dropped) the force is expressed in terms of the given excitations
b) Translation of the given excitations into complex amplitudes gives
Thus, with the even excitation, where liveexamhelper.com
and with the odd excitation, < fz >z = 0. c) This is a specific case from part (b) with ω = 0 and δ = λ/4. Thus,
The sign is consistent with the sketch of charge distribution on the sheet and electric field due to the potentials on the walls sketched.
Figure 2: Potentials = −V0 cos kz shown with surface charge σf = σ0 sin kz (Image by MIT OpenCourseWare.) liveexamhelper.com
Problem 3 a) Cartesian hx = 1 hy = 1 hz = 1
Cylindrical hr = 1 hφ = r hz = 1
Spherical hr = 1 hθ = r hφ = r sin θ
b)
c)
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d) Divergence
Figure 3: Contour for determining
(Image by MIT OpenCourseWare.) liveexamhelper.com
Curl
Similarly
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