2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants Kj = 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 AV, K2 A/2 A d K1 A/2 2d/3 d/3 K2 A/2 d AV = 24 V
2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants Kj = 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 AV, K2 A/2 A d K1 A/2 2d/3 d/3 K2 A/2 d AV = 24 V
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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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)
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