Voltage of Capacitors in Series, a widely held assumption

[Hirams_bro]

Say we have a simple circuit with just a battery and two capacitors. Why must the voltage of each capacitor in series add up to the total voltage of the battery? It seems intuitively plausible, I admit. But can someone show how to go about proving this using the laws of electrostatics ( please, not Kirchoff's Laws )?
I've already consulted 3 physics texts and none of them shed any light on this question. In one book, it says "clearly the voltage Vac = Vab + Vbc." That's it. Just because they wrote it in the form of an equation and said its clear, I guess they think they proved it. But that doesn't really prove much.
In Jewett's Physics for Scientists, their attempt at explaining this is to simply refer to a diagram and say "as can be seen, the diagram shows that these voltages add up..." Well drawing a diagram and labeling it doesn't it make it so. And this fact is used to prove the Law of Series Capacitance.

It seems to be a widely held assumption that individual voltages add up to the circuit voltage for series capacitors, but how can we assume this? I'm not aware of any conservation of voltage. Can someone please explain it?

 

[atyy]

http://en.wikipedia.org/wiki/Maxwell's_equations

Look at the integral form of the "Maxwell-Faraday equation", set the RHS to zero for electrostatics. This gives Kirchoff's loop law that the voltage in a loop sums to zero.

 

[Hirams_bro]

Thank you. I'm sure after this semester, when I finish Calc 3, I'll be able to make sense of Maxwell's Equations, gradient and curl, etc., and derive it from those. Until then is there any other way of showing this relationship, one that might be more familiar for a Physics 2 student?

 

[atyy]

OK, maybe try this. In electrostatics, we can use an electric potential instead of an electric field. Work done by a force usually depends on the path taken, but if the force can be represented by a potential, the work done only depends on the start and end points of the path. In a circuit, the charge starts out and ends up at the same place, so the total work done (charge times potential difference) in a loop is zero.