Two parallel-plate capacitors are constructed using dielectric materials with dielectric constantsκ1 = 3 and κ2 = 5, as shown in Figure. Parallel-plates have the area A = 0.7 m2. The distance isd = 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 K, = 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,. (&,=8.85×10-12 C²/N×m²)A/2AV1K,A/2AdK1A/22d/3 d/3A/2d3-5/5AV = 24 V

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Answered: 2) Two parallel-plate capacitors are…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants κ1 = 3 and κ2 = 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 ∆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 K, = 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,. (&,=8.85×10-12 C²/N×m²)
A/2
AV1
K,
A/2
A
d
K1
A/2
2d/3 d/3
A/2
d
3-5/5
AV = 24 V

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