Physics II / Conductors and Capacitance
Practice question · Numerical answer

A parallel-plate capacitor has plate area 0.02 m2^2, separation 1.0 mm, and vacuum between the plates. Compute its capacitance in pF, using ε0=8.854×10−12\varepsilon_0 = 8.854 \times 10^{-12} F/m. Give the answer to the nearest whole number.

Hints
  1. Use the parallel-plate formula: permittivity times area over separation.
  2. Compute 8.854e-12 x 0.02 / 1.0e-3.
Show the answer

177 (answers within ±3 count)

Why

C=ε0A/d=8.854×10−12×0.02/10−3=1.77×10−10C = \varepsilon_0 A/d = 8.854 \times 10^{-12} \times 0.02/10^{-3} = 1.77 \times 10^{-10} F =177= 177 pF. Note how large an area is needed for even this modest capacitance.

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