How Do Potential Differences Distribute Across Series Capacitors?

[Renaldo]

Homework Statement

A 2600-nF capacitor is disconnected from the 15-V battery and used to charge three uncharged capacitors, a 100-nF capacitor, a 200-nF capacitor, and a 300-nF capacitor, connected in series.

After charging, what is the potential difference across each of the four capacitors?

Homework Equations

Qo = CV = 3.9e-5 C

Q_o = Q₂₃₄C₁[(1/100)+(1/200)+(1/300)] + Q₂₃₄

The Attempt at a Solution

Q_o = 5.33Q₂₃₄

Q_o = 3.9e-5 C

Q₂₃₄ = 7.32e-6 C

Q₁/C₁ = (Q_o-Q₂₃₄)/C₁ = [(3.9e-5 C) - (7.32e-6 C)]/2600 nF = 12.2 V

V₂ = (7.32e-6 C)/100 nF = 7.32 V

V₃ = (7.32e-6 C)/200 nF = 3.65 V

V₄ = (7.32e-6 C)/300 nF = 2.43 V

This is incorrect, and I haven't been able to figure out what's wrong.

 

[NascentOxygen]

In charging the others, the big capacitor loses a charge Q and drops its voltage from 15V to V.

 

[Renaldo]

I express that in this equation,

Q_o = Q₂₃₄C1[(1/100)+(1/200)+(1/300)] + Q₂₃₄

which is derived thusly:

(Q_o-Q₂₃₄)/C₁ = V₁

Q₂₃₄[(1/100)+(1/200)+(1/300)] = V₂₃₄

V₁ = V₂₃₄

(Q_o-Q₂₃₄)/C₁ = Q₂₃₄[(1/100)+(1/200)+(1/300)]

Q_o = Q₂₃₄C1[(1/100)+(1/200)+(1/300)] + Q₂₃₄

 

[NascentOxygen]

So, you can finish this now?

 

[Renaldo]

No, I cannot.

Correct answers are:
V₁ = 14.7 V
V₂ = 8.01 V
V₃ = 4.01 V
V₄ = 2.67 V

These differ from my answers in the op.

 

[NascentOxygen]

Your method looks right. When I finish it I get 14.692V, 8.0137V, etc.

So it looks like you may be making an arithmetic error.

 

[Renaldo]

Yes, I figured out what I was doing wrong. instead of adding [(1/100) + (1/200) + (1/300)]
I was doing 1/600. lol. Gotta be careful. Thanks for your help.

 

[Equilibrium]

In charging the others, the big capacitor loses a charge Q and drops its voltage from 15V to V.

So is it like this:
the previous charge of the energized capacitor is used to charge other capacitors. The previous energized charge is distributed to new charge of C1 and the equal charge of the series capacitors.
$$Q_{\mbox{energized C1}}=Q_{\mbox{new charge of C1}}+Q_{234}$$
Hence the new charge is
$$Q_{\mbox{energized C1}}-Q_{234}=Q_{\mbox{new charge of C1}}=V_{1}C_{1}$$
is my analogy correct?
forgive me for hijacking but this also came up in the exam but the difference is a capacitor is charging two parallel capacitors.

I already also solved the problem in the op
Q_234 is 8.01x10^-7Coloumb btw

 

[NascentOxygen]

So is it like this:

That looks right.

 

[NascentOxygen]

Yes, I figured out what I was doing wrong. instead of adding [(1/100) + (1/200) + (1/300)]
I was doing 1/600. lol. Gotta be careful. Thanks for your help.

As a check, you could consider charging one capacitor of 600/11 nF, since that's the equivalent capacitance of those 3 capacitors in series.