A parallel plate capacitor is made of two circular thin plates of radius 15 cm separated by 2.5 mm of some unknown material.If they are charged to a potential difference of 22 V,what and the charge resident on each plate is 2.0 * 10^-7 C,what is the dielectric constant of the material in the space between the plates?

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Quizplus · Accepted Answer

The capacitance of a parallel plate capacitor is given by C = Q / V, where Q is the charge on each plate and V is the potential difference between the plates. The dielectric constant of the material between the plates is given by k = C * d / A * e0, where d is the distance between the plates, A is the area of each plate, and e0 is the permittivity of free space. Substituting the given values into these equations, we get:C = Q / V = (2.0 * 10^-7 C) / (22 V) = 9.09 * 10^-9 Fk = C * d / A * e0 = (9.09 * 10^-9 F) * (2.5 * 10^-3 m) / (π * (0.15 m)^2 * 8.85 * 10^-12 F/m) = 36

A Parallel Plate Capacitor Is Made of Two Circular Thin