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,. (&-8.85×10-12 C²/N×m²) P A K2 d A d A A AV1 AV = 24 V

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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,. (&-8.85×10-12 C²/N×m²)
d
K2
d
A
K1
THE
d
d
A
А
AV,
Ду — 24 V
Transcribed Image Text: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,. (&-8.85×10-12 C²/N×m²) d K2 d A K1 THE d d A А AV, Ду — 24 V
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