Adobe Reader Touch 4) A circuit in Figure consists of paralel-plate capacitors with and without dielectrics. Dielectric constants are K, =2 and K2 = 4. Parallel-plates have the area A = 0.8 m². The distance is a= 0.2 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. (ɛ=8.85×10-12 C²/N×m²) А А K1 d А AV1 AV = 24 V

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4) A circuit in Figure consists of paralel-plate capacitors with and without dielectrics. Dielectric
constants are K, = 2 and K, = 4. Parallel-plates have the area A= 0,8 m². The distance is a = 0.2 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)
d
A
K2
d
A
А
d
d
A
A
AV1
7
AV = 24 V
Transcribed Image Text:Adobe Reader Touch 4) A circuit in Figure consists of paralel-plate capacitors with and without dielectrics. Dielectric constants are K, = 2 and K, = 4. Parallel-plates have the area A= 0,8 m². The distance is a = 0.2 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) d A K2 d A А d d A A AV1 7 AV = 24 V
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