A parallel-plate capacitor is constructed by filling the space between two square plates with a block of three dielectric materials, as shown in the figure below. You may assume that { >> d. d/2 (a) Find an expression for the capacitance of the device in terms of the plate area A and d, x₁, K2, and K. (Use the following as necessary: K₁, K2, K3, 80, A and d.) C = (b) Calculate the capacitance using the values A = 1.10 cm2, d = 2.00 mm, k₁ = 4.90, x2 = 5.60, and K3 = 2.40. F

A parallel-plate capacitor is constructed by filling the space between two square plates with a block of three dielectric materials, as shown in the figure below. You may assume that { >> d. d/2 (a) Find an expression for the capacitance of the device in terms of the plate area A and d, x₁, K2, and K. (Use the following as necessary: K₁, K2, K3, 80, A and d.) C = (b) Calculate the capacitance using the values A = 1.10 cm2, d = 2.00 mm, k₁ = 4.90, x2 = 5.60, and K3 = 2.40. F

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter20: Electric Potential And Capacitance

Section: Chapter Questions

Problem 81P

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