Charge on plates of two capacitors in series are not equal

I have a simple circuit below with two capacitors are initially charged to Vo1 and Vo2 respectively. The two capacitors are in series however the charge on plates are not equal.

Could anyone tell me what is wrong here?

PS:

It seems that no one understands my confusion. Lets me express it again. What I want to check is the charge on plates of two capacitors (with initial charged voltages not zero) are the same. However from the calculation, you can see the expression for Q1 and Q2 are different. So in general, they are not the same. I would like to know why they are not same even the two capacitors are in series. • Which part of the math states that they are unequal? Nov 16 '16 at 7:55
• Nothing is wrong here. If it is given that the capacitors have two different voltages then this is equivalent to saying that they have different charge (provided their capacitances are equal; but I don't see this claim). If that's just given. You don't have to care how this has been accomplished (obviously not by charging them in series; but that's not your problem).
– Curd
Nov 16 '16 at 8:04
• This is such a common question that it's not worth another answer. Did you try google? See: electronics-tutorials.ws/capacitor/cap_7.html
– jonk
Nov 16 '16 at 8:21
• Your assertion that Q1 != Q2 is incorrect. Your calculations do not show that. What they do show is that C1V1 == C2V2. Nov 16 '16 at 12:20
• The charge on capacitors in series is assumed equal if their initial charges are equal (usually because they start with 0 charge). Nov 16 '16 at 16:57

So Q₁ is the total charge on C₁. Q₂ is the total charge on C₂.

This charge will be due to any initial charge ( Q₁(0) = V₀₁·C₁, Q₂(0) = V₀₂·C₂ ), and the charge flowing through the capacitors because of the switch, which we can call Q. So Q₁ = Q + V₀₁·C₁, Q₂ = Q + V₀₂·C₂.

There is no reason why Q₁ and Q₂ should be the same - for example, if V₀₁ = V₀ and V₀₂ = 0, then no current will flow when the switch is closed, Q = 0, Q₁ = C₁·V₀, and Q₂ = C₂·0 = 0.

This one is set up with V₀₁ = 2V, V₀₂ = 0V and V₀ = 4V which ends up with V₁ = 3V and V₂ = 1V, so Q₁ = 3 Q₂: simulate this circuit – Schematic created using CircuitLab

• Thanks. That makes sense. Actually, I was trying to apply charge conservation law for a switched capacitor circuit and stuck with this. The expression should be something like that C1(V1-Vo1) = C2(V2-Vo2) not C1 V1 = C2 V2. Nov 16 '16 at 17:50

One can make any number of separated conductive objects (plates, and in the case of your diagram, the plate pair of C2(negative) and C1(positive) is a single conductive object). One can then apply a voltage difference with a battery, and remove the battery (open the switch).

The C2(negative) charge PLUS the C1(positive) charge, if it started at zero, would still be zero after the battery connection (which doesn't add charge to the plate pair). But while the sum of those charges might be zero, that is not in fact expressed in any of your formulae. Did you wish to add that as an initial condition?

• Please see my update. Nov 16 '16 at 9:45
• Capacitors are NOT in series when the switch is open; they're merely connected at one terminal. There is no signal in a circuit here, this is STATIC ELECTRICITY analysis, not circuit analysis. Nov 16 '16 at 9:47
• Did you read the post? At t = 0, the switch is closed. Nov 16 '16 at 10:18
• Irrelevant. That switch never puts charge onto the plate pair. Those two plates are insulated from the battery and the switch, and ground. They gain no charge, nor lose any. Nov 16 '16 at 10:30

Anhnha, you've made a misteak somewhere, and here's the proof: • This only demonstrates the case with zero initial charge, V₀₁=V₀₂=0. Re-run the simulation with an initial charge such that V₀₁ or V₀₂ are non-zero and you'll get a different result. Nov 16 '16 at 17:22
• Indeed. I misread the question; mea culpa. Thanks for the reality check. :) Nov 16 '16 at 17:42