2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants K, = 2 and K, = 4, as shown in Figure. Parallel-plates have the area A = 0.8 m². The distance is d = 0.04 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 AV1. (E=8.85×10-12 C²/N×m²) %3D %3D A/2 AV1 K, A/2 K2 K1 A/2 2d/3 d/3 A/2 d AV = 24 V -----

2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants K, = 2 and K, = 4, as shown in Figure. Parallel-plates have the area A = 0.8 m². The distance is d = 0.04 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 AV1. (E=8.85×10-12 C²/N×m²) %3D %3D A/2 AV1 K, A/2 K2 K1 A/2 2d/3 d/3 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 55P

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Can you solve a and b parts? Thanks in advance.

2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric
constants K, = 2 and K, = 4, as shown in Figure. Parallel-plates have the area A = 0.8 m². The
distance is d = 0.04 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 AV1. (E,=8.85×10-12 C²/N×m?)
K1
A/2
AV1
K2
A/2
A
d
K2 | K1
K1
A/2
2d/3 d/3
K2
A/2
d
AV = 24 V

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