Question

Problem 2 Find the maximal charge that can be stored by the parallel plate capacitors of the equal plate area (by one of the plates) if the dielectric material between the plates are characterized by the following dielectric constant and breakdown fields: (a) Air: €r-1, EBD 3 MV/m. (b) Barium titanate: Er = 1200, EBD-3 MV/m. (c) Silicon dioxide: Er = 3.78, EBD = 16 MV/m. (d) Polyethylene: Er = 2.26, EBD = 4.7 MV/mm

Please show all work. Thank you!

Answer 1

Parallel plate capacitor electric field strength between plates is given by therefore E = Q/er epsilon_0 A where, E_r = dielectric material permittivity Q = charge in palte A = arm of plate charge density is given by sigma = Q/A c/m^2 therefore E = sigma/Epsilon_0 E_R ......... (1) e_r = 1, epsilon_0 = 8.85 times 10^- 12, e_BD = 3 times 10^6 N/m 3 times 10^6 = sigma_1 times 1/1 times 8.85 times 10^- 12 sigma_1 = 3 times 10^6 times 8.85 times 10^- 12 = 2.65 times 10^- 5 c/m^2 maximum charge will be sigma_1 = 2.65 times 10^- 5 theta/m^2 \r\n Borium E_r = 1200, E_BD = 3 times 10^6 v/m 3 times 10^6 = sigma_2/8.85 times 10^- 12 times 1200 sigma_2 = 3 times 10^5 times 1200 time s8.85 times 10^- 12 = 3.186 times 10^- 2 c/m^2