Question

dielectric constant of the paper is 3.5

The plates of a parallel plate capacitor are 4 cm wide and 6 cm long. The plates are separated by a 1.8-mm-thick layer of paper (a) Calculate the capacitance of the device (in pF) using the dielectric constant of paper from this table. 0.0836 X pF (b) Any dielectric material other than a vacuum has a maximum electric field that can be generated in the dielectric material before it physically or chemically breaks down and begins to conduct. This maximum electric field is called dielectric strength. The dielectric strength for paper is reached at a value of 15 x 106 V/m. Calculate the maximum charge (in HC) that can be placed on the capacitor at this dielectric strength. HC

Answer 1

Width w = 4 cm length l = 5 cm separation d = 1.8 mm = 1.8 times 10^-3 m Area of plate = width times length = 4 times 6 cm^2 24 cm^2 = 24 times 10^-4 m^2 capacitance C = A epsilon/d = A epision_0 k/d k = dielectric constant of paper (24 times 10^-4) times (8.85 times 10^-12) times 3.5/1.8 times 10^-3 = 41.3 times 10^-12 F = 41.3 pF (b) Dielectric strength of paper = 15 times 10^6 V/m E = 15 times 10^6 V/m field is given by 5/eisilon_0 k = Q/A epsilon_0 k Q/A epsilon k = 15 times 10^6 rightarrow Q = (15 times 10^6) A epsilon_0 k = 15 times 10^6 times 24 times 10^4 times 8.85 times 10^-12 times 3.5 = 0.126 times 10^-6 C = 0.126 mu C