An air-filled capacitor consists of two parallel plates, each with an area of 7.

Question

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 2.00 mm. If a 16.4-V potential difference is applied to these plates, calculate the following. (a) the electric field between the plates magnitude kV/m (b) the capacitance pF (c) the charge on each plate pC

Explanation / Answer

(a) By +/-s = Q/A & E = s/? =>E = Q/A? =>E = V/d =>E = 25/[1.60 x 10^-3] =>E = 15625 Volt/m =>E = 15.63 kV/m (b) By C = ?A/d =>C = [8.85 x 10^-12 x 7.60 x 10^-4]/[1.60 x 10^-3] =>C = 4.2 x 10^-12 F =>C = 4.2 pF (c) By V = Qd/?A =>Q = V?A/d =>Q = [25 x 8.85 x 10^-12 x 7.60 x 10^-4]/[1.60 x 10^-3] =>Q = 105.09 x 10^-12 C =>Q = 105.09 pC

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