You have three parallel plate capacitors, each consisting of two parallel circul
ID: 1571058 • Letter: Y
Question
You have three parallel plate capacitors, each consisting of two parallel circular disks of metal separated by a gap filled with some material. Rank the capacitors by their capacitances from highest to lowest, based on the provided plate radii (r), plate separations (d), and filler materials. Dielectric constants for each material can be found in the table to the right. Capacitor 1 (C_1): r = 8.74 times 10^-3 m, d = 2.85 times 10^-6 m, filled with paper. Capacitor 2 (C_2): r = 7.50 times 10^-3 m, d = 27.0 times 10^-6 m, filled with porcelain. Capacitor 3 (C_3): r = 9.38 times 10^-3 m, d = 21.4 times 10^-6 m, filled with Gallium Arsenide (GaAs)Explanation / Answer
as capacitance C = kA0/d = k(r2)0/d
hence C1 = (3.50){(3.14)(8.74x10-3)2}(8.85x10-12)/(2.85x10-6)
= (3.14)(8.85)(10-12){(3.50)(8.74)2/2.85} = (27.789)(10-12)(93.809333) = 2606.8675 pF
Similarly
C2 = (3.14)(8.85)(10-12){(7.00)(7.50)2/27} = (27.789)(10-12)(14.583333) = 405.25625 pF
C3 = (3.14)(8.85)(10-12){(13.2)(9.38)2/21.4} = (27.789)(10-12)(54.27075) = 1508.129911 pF
So C1, C3, C2.