Two charged parallel square plates with dimensions L × L = 1 m² are separated by a distance of d = 0.1 m. Each side is subdivided into N = 64 equal subareas. The potential of the top plate is + 5 V, and the potential of the bottom plate is - 5 V. (a) Find and plot the charge density distribution

Question

Two charged parallel square plates with dimensions [latex]\mathscr{L} × \mathscr{L} = 1 m^{2}[/latex] are separated by a distance of d = 0.1 m. Each side is subdivided into N = 64 equal subareas. The potential of the top plate is + 5 V, and the potential of the bottom plate is - 5 V.

(a) Find and plot the charge density distribution using the MoM.
(b) Find the capacitance C of this charged conductor system. Compare the limiting case with the known simple classical solution [latex]C / ε_{0} = \mathscr{L}^{2}/d = 10.[/latex]

Accepted Answer

(a) The column vector of the known potential is

[latex][V]= [ +5, +5, .. . , +5; -5, -5, ... , -5]^{T}[/latex]

The matrix equation that must be solved is again (4.101) with a matrix [P] described by ( 4.106) and ( 4.107). The solution of this matrix equation for the surface charge distribution is plotted in the following figure.

[latex][P][Q] = [V] [/latex] (4.101)

[latex]P_{i,j}= \frac{1}{4\pi \varepsilon _{0}}\frac{\Delta s_{j}}{\left|r_{i} - r_{j}\right| } [/latex] (4.106)

[latex]P_{i,i}= \frac{1}{4\pi \varepsilon _{0}}\int_{0}^{2\pi }{\int_{0}^{b}{\frac{\rho ^{\prime } d\rho^{\prime } d\phi^{\prime } }{\rho ^{\prime }} } } = \frac{1}{4\pi \varepsilon _{0}} (2\pi b )= \frac{b}{2\varepsilon _{0}} = \frac{1}{2\varepsilon _{0}}\sqrt{\frac{\Delta s_{i}}{\pi } } [/latex] (4.107)

(b) The normalized capacitance [latex]C_{0} =C / ε_{0}[/latex] is calculated to yield a value of [latex] C_{0} = 13.3811 [/latex] for the value of the number of sections N = 64. This value is larger than that predicted from the elementary formula [latex]C / ε_{0} = \mathscr{L}^{2}/d = 10.[/latex] This formula assumed that the separation distance was significantly less than the area of the plate and all fringing fields at the edges could be neglected. The accuracy can be improved by subdividing the area into smaller subareas. It is also possible to determine the inhomogeneous charge distribution caused by perturbations in the plates or not having the plates exactly parallel as was assumed here.¹

¹ Bai, E. W., and Lonngren, K. E., "Capacitors and the method of moments," Computers and Electrical Engineering, vol. 30, 2004, pp. 223-229.

Question 4.26: Two charged parallel square plates with dimensions L × L = 1...

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