Charge on Capacitor C1 & C2: Calculation & Solution

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SUMMARY

The discussion focuses on calculating the charge on two capacitors, C1 and C2, connected in series with a 65-V battery. Capacitor C1, with air as the dielectric, and capacitor C2, filled with porcelain (dielectric constant of 7), both have a plate area of 2.1 cm² and a separation of 0.1 mm. The charge on C1 was initially calculated incorrectly as 1.21e-9 C, while the correct charge is 1.06e-9 C. The participants emphasized the importance of understanding series capacitor behavior and the need to adjust voltage calculations accordingly.

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Homework Statement



Two parallel plate capacitors, C1 and C2, are connected in series with a 65-V battery and a 500-kΩ resistor, as shown in the figure. Both capacitors have plates with an area of 2.1 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.

http://www.webassign.net/bauerphys1/26-p-046.gif

(a) What is the charge on capacitor C1?

(b) What is the charge on capacitor C2?


The Attempt at a Solution



(a) What is the charge on capacitor C1?

C = E0 * A / d

q = C delta V

C = (8.85 * 10^-12 * 2.1 cm^2 / 10000) ( to change it 2.1 cm^2 to m^2) / 0.0001m)

q = C delta V
= 1.8585e-11 * 65V = 1.21 e -9

the answer is 1.06e-9

where did i go wrong?


(b) What is the charge on capacitor C2?

i used C = (E0 L^2 / (2d)) (1+k) and it didn't work
any hints?
 
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The capacitors are in series, so they can't both have the full 65V -- that would total 130 V.

Have you learned how to combine capacitors in series or in parallel to find the effective total capacitance?
 
ya i got it thx
 

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