You have three parallel plate capacitors, each consisting of two parallel circul

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.

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.