Find the capacitance of a parallel-plate capacitor consisting of two metal surfaces of area A held a distance d apart (Fig. 2.52).

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If we put +Q on the top and −Q on the bottom, they will spread out uniformlyover the two surfaces, provided the area is reasonably large and the separationsmall.[latex]^{13}[/latex] The surface charge density, then, is [latex]\sigma =Q/A[/latex] on the top plate, and so thefield, according to Ex. 2.6, is [latex](1/\epsilon _{0})Q/A[/latex]. The potential difference between theplates is therefore

[latex]V=\frac{Q}{A\epsilon _{0}}d, [/latex]

and hence

[latex]C=\frac{A\epsilon _{0}}{d}, [/latex] (2.54)

If, for instance, the plates are square with sides 1 cm long, and they are held 1 mm apart, then the capacitance is [latex]9 imes 10^{-13}F[/latex].

[latex]^{13}[/latex]The exact solution is not easy—even for the simpler case of circular plates. See G. T. Carlson and B. L. Illman, Am. J. Phys. 62, 1099 (1994).

Question 2.11: Find the capacitance of a parallel-plate capacitor consistin...

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