(a) A capacitor has square plates of side a, except that the top plate is inclin
ID: 2164895 • Letter: #
Question
(a) A capacitor has square plates of side a, except that the top plate is inclined so that the spacing between the plates increases from d to 2d as shown below. Assuming that d<<a, calculate the capacitance by using strips of width dx and length a to approximate differential capacitors of area a*dx and separation s=d+(d/a)x connected in parallel.
(b) By what numerical factor (<1) does this capacitance differ from the capacitance obtained if the plates were parallel and separated by d?
(c) Since the angle of inclination is so small, you might reasonably think that the capacitance of the wedged capacitor would equal the capacitance of a parallel plate capacitor whose spacing was equal to the average separation of (3/2)d. But capacitance is a bit bigger than this. Can you offer a physical explanation for this result?
(Hint: Is the charge density uniform over the surface of the plates? Why?)
Explanation / Answer
Area of each plate (A) = 6.5^2 cm^2 = 42.25 cm^2 = 0.004225 m^2 Dielectric constant of quartz (e) = 4.2 Permittivity of free space (e0) = 8.85 x 10^-12 F/m Distance between plates (d) = 2.0 mm = 0.002 m Capacitance = (e)(e0)(A/d) = (4.2)(8.85 x 10^-12)(0.004225/0.002) = 7.8522 x 10^ -11 F = (7.8522 x 10^-11)(10^12) pF = 78.522 pF
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