A parallel-plate air capacitor of area A = 21.9 cm2 and plate separation of d =

Question

A parallel-plate air capacitor of area A = 21.9 cm2 and plate separation of d = 3.60 mm is charged by a battery to a voltage of 64.0 V. What is the charge on the capacitor? If a dielectric material with ? = 4.90 is inserted so that it fills the volume between the plates (with the capacitor still connected to the battery), how much additional charge will flow from the battery onto the positive plate?

Explanation / Answer

Capacitance of parallel plate air capacitor, C = eo A/d = 8.9 x 10^-12 * 21.9 x 10^-4 / 3.60 x 10^-3 farad = 5.41 x 10^-12 => Charge accumulated on the plates = CV = 5.41 x 10^-12 * 64.0 C = 3.465 x 10^-10 C When the volume between the plates is completely filled with dielectric material with k = 4.9, total charge on the plates will be 4.9 * 3.465 x 10^-10 C => additional charge that will flow = 3.9 * 3.465 x 10^-10 C = 1.35 x 10^-9 C.

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