Two parallel plate capacitors, C1 and C2, are connected in series with a 65.0-V

Question

Two parallel plate capacitors, C1 and C2, are connected in series with a 65.0-V battery and a 320-k? resistor, as shown in the figure. Both capacitors have plates with an area of 1.79 cm2 and a separation of 0.250 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 5.70 kV/mm). The switch is closed, and a long time passes.

a) What is the charge on capacitor C1?

b) What is the charge on capacitor C2?

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

d) What is the electric field inside capacitor C2?

Explanation / Answer

C1 = eA/d = (8.85 x 10^-12)(1.79 x 10^-4)/2.5 x 10^-4 = 6.33 pf C2 = 7C1 = 44.31 pf Ceqv = 44.31*6.33/50.64 = 5.53 pf q=Ceqv*65 = 3.6nC or 3.6x 10^-9 Coulombs This is the charge on both capacitors V C1 = 3.6 x 10^-9/0.63 x 10^-11 = 57.1 volts V C2 = 65 -57.1 = 7.9 Volts, or by V = q/C = 3.6 x 10^-9/0.44 x 10^-10 = 8.18 volts Energy in C1 = 1/2 C*V^2 = (3.16 x 10^ -12)(57.1^2) = 1.03 v 10^-8 joules Energy in C2 = (1/2)44.31 pf)(2.76^2) = 1.69 x 10^-9 joules The E field in C2 = 8.18/1 x 10^-4 = 8.18 x 10^4 Volts per meter

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