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Exact solution of a 2D thin parallel-plate capacitor (A. E. H. Love, 1924):
C = [\epsilon \cdot K(1-m) \cdot K(m)] \text{ F/m}in which: \epsilon is the material permittivity, m = k^2 is the argument of K(k) , and K(k) = \int_0^{\frac{\pi}{2}} \frac{{\rm{d}}\theta}{\sqrt{1-k^2 \sin^2\theta}} is the complete elliptic integral of the first kind. To determine the value m , first one needs to solve the following equation:\begin{align*} \dfrac{W}{S} = \dfrac{K(1-m)E[\phi,(1-m)]-E(1-m)\cdot F[\phi, (1-m)]}{[E(1-m)-K(1-m)]\cdot K(m)+K(1-m)E(m)} \end{align*}in which: W is the capacitor plate's cross section length, S is the plate spacing, m = k^2 is the argument of K(k) , \phi is the argument of E(\phi, k) , which is calculated according to
\begin{align*} \sin^2 (\phi) &= \dfrac{K'(m)-E'(m)}{(1-m^2)K'(m)} \\ \phi &=\arcsin \left[\sqrt{\dfrac{(K(1-m)-E(1-m)}{(1-m)\cdot K(1-m)}} \right] \end{align*}Finally E(\varphi,k) = \int_0^\varphi \sqrt{1-k^2 \sin^2\theta}\, d\theta is the incomplete elliptic integral of the second kind, and E(k) = E(\frac{\pi}{2},k)=\int_0^{\frac{\pi}{2}}\sqrt {1-k^2 \sin^2\theta}\ {\rm{d}}\theta\! is the complete elliptic integral of the second kind.

To find the capacitance, first guess the value of m and calculate φ using a square root and an arcsin, then evaluate a bunch of elliptic integrals while varying m (and φ), until the second equation is equal to the aspect ratio of the capacitor. Finally plug the value of m into the first equation...

Conclusion: Physics is a mistake for anything other than spherical cows in a vacuum. #electronics

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