Help ASAP 1) In the figure how much charge is stored on the parallel-plate capac

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Help ASAP

1) In the figure how much charge is stored on the parallel-plate capacitors by the 18.0 V battery? One is filled with air, and the other is filled with a dielectric for which ? = 4.10; both capacitors have a plate area of 2.90 × 10–3 m2 and a plate separation of 3.10 mm.

2)The figure shows a parallel-plate capacitor with a plate area A = 4.27 cm2 and plate separation d = 4.64 mm. The top half of the gap is filled with material of dielectric constant ?1 = 12.8; the bottom half is filled with material of dielectric constant ?2 = 11.6. What is the capacitance?

Explanation / Answer

part 1 )

C2 = eo*A/d

d = 3.10 mm = 3.10 x 10^-3 m

C2 = 8.279 x 10^-12 F

C1 = k*C2 = 3.39 x 10^-11 f

Ceq = C1 + C2

Ceq = 4.22 x 10^-11 F

Q = Ceq*V

Q = 7.6 x 10^-10 V

Q1 = C1V = 6.102 x 10^-10 C

Q2 = C2V = 1.49 x 10^-10 C

part 2 )

this is series combination

C1 = k1*eoA/d/2

C2 = k2*eoA/d/2

1/Ceq = 1/C1 + 1/C2

Ceq = (2eo*A/d )*(k1*k2/(k1+k2))

Ceq = 9.912 x 10^-12 F

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