Question

Consider a round parallel plate capacitor consisting of two circular metal plates of radius, R = 1.0 cm, separated by d = 150 μm of air. The capacitor is connected to a potential source, as shown in the figure below. Part a Calculate the capacitance of this capacitor in picoFarads? (1pF 10-12F) pF Enter answer here

Answer 1

a) Capacitance of a capacitor C = k epsilon_0 A/d dielectric constant of air = k = 1 A is the are of circular plate A = pi R^2 = 0.0003142 m^2 and plate separation d = 150 mu m epsilon_0 = 8.854 times 10^-12 C^2/N m^2 substituting all the know values given C = 1.8542 - 011 = 18.54 pF b) potential source = V_0 = 335 V we plug the values in hand while using the relation give us energy stored in capacitor = U = 1/2 CV^2 = 1.041 e = 06 J = 1041 nJ rightdoublearrow U = 1041 nJ c) if the radius is decreased by half and the new radius = r' = r/2 new are of cross section = A' = pi r^2/4 = A/4 and new separation between plates = d' = d/2 new capacitance of capacitor = C' = k epsilon_0 A'/d' = C/2 and the new energy = U' = 1/2 C' V^2 = 1/2 (1/2 C V^2) = U/2 THE ENERGY STORED IN THE CAPACITOR IS CUT IN HALF