Charge on Capacitor C1 & C2: Calculation & Solution

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AI Thread Summary
The discussion revolves around calculating the charge on two parallel plate capacitors, C1 and C2, connected in series with a 65-V battery. The user initially attempts to calculate the charge on C1 using the formula for capacitance but arrives at an incorrect answer. For C2, the user struggles with the correct approach to account for the dielectric and the series configuration, realizing that the total voltage must be divided between the two capacitors. The conversation emphasizes the importance of understanding how to combine capacitors in series to find the effective capacitance. Accurate calculations are crucial for determining the correct charge on each capacitor.
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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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