Two parallel plate capacitors, C 1 and C 2 , are connected in series with a 70-V

Question

Two parallel plate capacitors, C1 and C2, are connected in series with a 70-V battery and a 440-k? resistor, as shown in the figure. Both capacitors have plates with an area of 1.5 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?
                                                  

8.13e-10 C

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I used C = (E_0 * A)/d to get capacitance and then plugged into C= q/dV to get q but I'm not getting the right answer....

Two parallel plate capacitors, C1 and C2, are connected in series with a 70-V battery and a 440-k? resistor, as shown in the figure. Both capacitors have plates with an area of 1.5 cm2and 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. What is the charge on capacitor C1? What is the charge on capacitor C2? What is the total energy stored in the two capacitors? What is the electric field inside capacitor C2? I used C = (E_0 * A)/d to get capacitance and then plugged into C= q/dV to get q but I'm not getting the right answer....

Explanation / Answer

v = 70 volts

R = 440*10^3 ohms

C = C1 = C2 = A*epsilon/d = 13.281*10^-12 F


a) Cnet = C/2 = 6.64*10^-12 F

Q = cnet*v = 4.65*10^-10 C

b) U = 0.5*Cnet*v^2 = 1.627*10^-8 J

c)V2 = 35 volts

E2 = V2/d = 35/0.1*10^-3 = 3.5*10^5 N/C

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