Discharging a charged capacitor to another neutral capacitor

[Hijaz Aslam]

Homework Statement

After a capacitor of capacitance $C_1$ is charged up to a voltage $V_o$, the Battery is removed from the circuit. Another capacitor of capacitance $C_2$ is connected in series with the capaciotr $C_1$. Find the charge on each capacitor after equilibrium is attained. ($C_1=5microF$, $C_2=5.3microF$, $V_o=6.50V$).

Homework Equations

$q=CV$

The Attempt at a Solution

I tried cracking the problem as follows:

Initially $C_1$ has a voltage of $V_o$. After connecting $C_2$ voltages of both the capacitors will become equal. So we have $V_1=V_2$ . Now as the initial potential was $V_o$, and as the capacitors are connected in series $V_1+V_2=V_o$. Therefore $V_1=V_2=V_o/2$. Hence $q_1=C_1V/2$ and $q_2=C_2V/2$. But after substituting the values, the final figure seems wrong!

My textbook attempts this question by taking: $V_1=V_2$ and assuming that $C_1$ initially had charge $q_o$. Therefore as $V_1=V_2$ we have $q_1/C_1=q_2/C_2$ and taking $q_1+q_2=q_o$ and simplifying they arrive at the answer.
Textbook method seems fine. But what is wrong with my method? Am I missing any concepts? (I think I am likely to be wrong at $V_1+V_2=V_o$).

 

[mfb]

Why do you think the sum of voltages stays the same?
The charge stays the same, the voltage does not have to.

 

[Hijaz Aslam]

Well my text mentions that in case of a Battery the voltage is maintained, whereas in case of Capacitors as it lacks the electro-chemical reaction (as in case of a cell) the voltage is not conserved. Why is the charge conserved but not the voltage. Can I have an explanation?

 

[vela]

Why would voltage be conserved?

 

[BvU]

Dear Hijaz,
If you make a little drawing you will see that, as you already expected, $V_1+V_2=V_o$ can not be correct.
Fortunately, you have enough equations left over to determine q₁ and q₁

 

[Hijaz Aslam]

I thought about it and arrived at the following explanation : The moment $C_2$ is connected to the circuit, nature urges the voltage of both the capacitors to be the same. At the same time $C_1$ pushes charges into $C_2$ to attain the state represented by $q_1/C_1=q_2/C_2$. And at some point when the required charge $q_2$ is consumed by $C_2$, $C_1$ in the hurry to attain equilibrium sheds any left over voltage somehow (or any similar method).
Am I right? Or is my explanation coincidental and a blunder?

But still we can argue: why can't $q_1/C_1=q_2/C_2$ be still attained at the same time following the condition $V_1+V_2=q_1/C_1+q_2/C_2=V_o$.
And I think I've found a possible explanation for that too. Isn't it because the charge supplied by $C_1$ is limited to $q_o$ (which $C_1$ attained during it's initial charging by the cell)?

 

[BvU]

Thinking is good. Making a little drawing is better. If you follow V₁ plus V₂, then you are back at the same point (*). But you traversed C₂ in the opposite direction, meaning you have V₁ - V₂ = 0, in other words V₁ = V₂ -- which you already had (and correctly so).

"The moment C₂ is connected to the circuit, nature urges the voltage of both the capacitors to be the same" yes ( but again: *)
"C₁ pushes charge into C ₂ to attain the state represented by q₁ /C₁ =q₂ /C₂" yes ( but again: *)

There is no "hurry", but it goes fast, yes. Lightning speed, so to say (there is no mention of resistance to limit the rate of discharge).
"Shedding of voltage" is the same thing as unloading charge: q = CV. Driving force is voltage difference. Unloading stops when voltage difference is zero, well before q₀ is supplied.

- - - - - - -
(*)
I suddenly realize there may be a misunderstanding: how exactly do you interpret

Another capacitor of capacitance C₂ is connected in series with the capacitor C₁

-- again a good reason for a drawing!