plz help this problem show me the work and equation Two parallel plate capacitor

Question

plz help this problem show me the work and equation

Two parallel plate capacitors. C_1 and C_2, are connected in series with a 53.8-V battery and a 130.-kOhm resistor, as shown in the figure. Both capacitors have plates with an area of 5.42 cm^2 and a separation of 0.292 mm. Capacitor C_1 has air between its plates, and capacitor C_2 has the gap filled with a certain porcelain (dielectric constant of 7.00 and dielectric strength of 5.70 kV/mm). The switch is dosed, and a long time passes. What is the charge on capacitor C_1?

Explanation / Answer

Here ,

for the capacitance C1

C1 = epsilon_0 * Area/d

C1 = 8.854 *10^-12 * 5.42 *10^-4/(0.292 *10^-3)

C1 = 1.643 *10^-11 F

for C2

C2 = k *epsilon_0 * Area/d

C2 = 7 * 8.854 *10^-12 * 5.42 *10^-4/(0.292 *10^-3)

C2 = 11.501 *10^-11 F

for the capacitors in series

1/Cs = 1/(1.643 *10^-11) + 1/(11.501 *10^-11 )

Cs = 1.44 *10^-11 F

Charge on capacitors = Cs * V

Charge on capacitors = 1.44 *10^-11 * 53.8 C

Charge on capacitors = 7.734 *10^-10 C

the Charge on capacitor C1 is 7.734 *10^-10 C

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