Find the area of the plates of a parallel plate capacitor having a capacitance of 0.491 uF (micro-Farads), with the plates separated by 0.2 mm of paper. The dielectric constant for paper is "k = 3.7". answer choices a)5.0 m^2 b)4.0 m^2 c)2.0 m^2 d)3.0 m^2

A parallel plate capacitor has a capacitance of 0.20 μF. The capacitor is notconnected to a battery or a power supply. There is a charge of 24 μC on thecapacitor. How much work must be done on the capacitor to double the separation between the plates?

a) A capacitor consists of two parallel metal plates immersed in a high-dielectric liquid, chloro- cyclohexane (e = 30). When the separation of the plates is 1mm, the capacitance is 0.06 F. Initially it is charged, at 500 V. Assume that 0 = 8.85 x 10-12 Fm-1. i) == Use this information to calculate the area of the plates, and the charge and the energy stored. ii) The plates are now pulled apart, to a gap of 1cm. Calculate the voltage it is now at, the energy now stored, and the force required to pull the plates apart.

Parallel plates are spaced apart by the thickness of a sheet of paper placed between them. The area of the plates is 0.850m2 and the thickness of the paper is 1.2mm. The dielectric constant of the paper is 3.7 . What is the capacitance of this arrangement?