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 AV,. (E=8.85×10-12 C²/N×m²) A/2 AV1 K2 A/2 A d K2 KI K1 A/2 2d/3 d/3 K2 A/2 d. AV = 24 V -------

Delmar's Standard Textbook Of Electricity
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ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter19: Capacitors
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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 AV,. (E=8.85×10-12 C²/N×m²)
A/2
AV1
K2
A/2
A
d
K2 KI
K1
A/2
2d/3 d/3
K2
A/2
d.
AV = 24 V
-------
Transcribed Image Text: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 AV,. (E=8.85×10-12 C²/N×m²) A/2 AV1 K2 A/2 A d K2 KI K1 A/2 2d/3 d/3 K2 A/2 d. AV = 24 V -------
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