Charge on Capacitor C1 & C2: Calculation & Solution

  • Thread starter Thread starter yjk91
  • Start date Start date
  • Tags Tags
    Capacitor Charge
Click For 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.
yjk91
Messages
73
Reaction score
0

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?
 
Physics news on Phys.org
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
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
6
Views
3K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K