A circuit in Figure consists of paralel-plate capacitors with and without dielectrics. Dielectric constants are K¡ = 3 and K2 = 5. Parallel-plates have the area A = 0.7 m². The distance is d= 0.1 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×m?) %3D d А K2 K1 d. HHE А AV1 Δν 24 V

Principles of Physics: A Calculus-Based Text
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Chapter20: Electric Potential And Capacitance
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A circuit in Figure consists of paralel-plate capacitors with and without dielectrics. Dielectric
constants are K, = 3 and K, = 5. Parallel-plates have the area A = 0.7 m². The distance is d= 0.1 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?)
%3D
K2
A
K1
А
AV1
AV = 24 V
Transcribed Image Text:A circuit in Figure consists of paralel-plate capacitors with and without dielectrics. Dielectric constants are K, = 3 and K, = 5. Parallel-plates have the area A = 0.7 m². The distance is d= 0.1 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?) %3D K2 A K1 А AV1 AV = 24 V
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