Two parallel plate capacitors, C1 and C2, are connected in series with a 55-V ba

Question

Two parallel plate capacitors, C1 and C2, are connected in series with a 55-V battery and a 580-k resistor, as shown in the figure. Both capacitors have plates with an area of 1.7 cm2 and a separation of 0.1 mm. Capacitor C1 has air between its plates, and capacitor C2 has the gap filled with porcelain (dielectric constant of 7 and dielectric strength of 5.7 kV/mm). The switch is closed, and a long time passes.

(a) What is the charge on capacitor C1?
1

(b) What is the charge on capacitor C2?
2

(c) What is the total energy stored in the two capacitors?
3

(d) What is the electric field inside capacitor C2?

Explanation / Answer

C1 = eA/d = 8.85e-12 * 1.7e-4 / 0.1e-3 = 1.5e-11 F C2 = keA/d = 7 * 8.85e-12 * 1.7e-4 / 0.1e-3 = 1.05e-10 F Ceq = 1/(1/C1+1/C2) = 1/(1/1.5e-11 + 1/1.05e-10) = 1.31e-11 F Q = Emf * Ceq = 55 * 1.31e-11 = 7.22e-10 C a) Charge on C1 is Q = 7.22e-10 C b) Charge on C2 is Q = 7.22e-10 C c) U = 1/2*Q^2/C = 1/2*(7.22e-10)^2/1.31e-11 = 2e-8 J d) V2 = Q/C2 = 7.22e-10 / 1.05e-10 = 6.88 V E2 = V2/d = 6.88V / 0.1e-3m = 6.88e4 V/m

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