Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants 2κ1 = 3 and κ2 = 5, as shown in Figure. Parallel-plates have the area A = 0.7 m . The distance is d = 0.05 m. A potential difference of ∆V = 24 ? is applied to the circuit.(a) Find the equivalent capacitance of the system.(b) Find the potential ΔV1. (ε0=8.85×10-12 C2/N×m2) 2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric constantsK¡ = 3 and K2 = 5, as shown in Figure. Parallel-plates have the area A = 0.7 m². The distance isd = 0.05 m. A potential difference of AV = 24 V is applied to the circuit.(a) Find the equivalent capacitance of the system.(b) Find the potential AV. (E=8.85×10-12 C²/N×m²)%3D%3DK1A/2AV,K2A/2AdK2 K1K1A/22d/3 d/3K2A/2dAV = 24 V%3D

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Answered: 2) Two parallel-plate capacitors are…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants 2

κ1 = 3 and κ2 = 5, as shown in Figure. Parallel-plates have the area A = 0.7 m . The distance is d = 0.05 m. A potential difference of ∆V = 24 ? is applied to the circuit.
(a) Find the equivalent capacitance of the system.
(b) Find the potential ΔV1. (ε0=8.85×10-12 C2/N×m2)

2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants
K¡ = 3 and K2 = 5, as shown in Figure. Parallel-plates have the area A = 0.7 m². The distance is
d = 0.05 m. A potential difference of AV = 24 V is applied to the circuit.
(a) Find the equivalent capacitance of the system.
(b) Find the potential AV. (E=8.85×10-12 C²/N×m²)
%3D
%3D
K1
A/2
AV,
K2
A/2
A
d
K2 K1
K1
A/2
2d/3 d/3
K2
A/2
d
AV = 24 V
%3D

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