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(a) Find the equivalent capacitance of the system. (b) Find the potential AV1. (E,=8.85×10-12 C²/N×m²)

(a) Find the equivalent capacitance of the system. (b) Find the potential AV1. (E,=8.85×10-12 C²/N×m²)

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2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants K 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. (8=8.85×10-12 C²/N×m²) 2 and K, = 4, as shown in Figure. Parallel-plates have the area A = 0.8 m². The %3D A/2 AV1 A/2 A K1 A/2 2d/3 d/3 K2 A/2 AV = 24 V ------
Transcribed Image Text:2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants K 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. (8=8.85×10-12 C²/N×m²) 2 and K, = 4, as shown in Figure. Parallel-plates have the area A = 0.8 m². The %3D A/2 AV1 A/2 A K1 A/2 2d/3 d/3 K2 A/2 AV = 24 V ------
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