Simple Capacitor Problem making a Big Problem

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To achieve the greatest potential difference with two capacitors, they should be connected in series, allowing their voltages to add up. The confusion arises from the concept of electrostatic equilibrium, where connecting capacitors can lead to a redistribution of charges, affecting their individual voltages. While the charge distribution remains largely unchanged, the creation of an equipotential surface during connection is a subtle effect that may not significantly impact the overall potential difference. The key takeaway is that connecting capacitors in series maximizes the voltage across the entire setup. Understanding these principles clarifies the behavior of capacitors in different configurations.
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Homework Statement


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Given the two capacitors, in what way should you connect them to get the greatest potential difference over the whole construction?

Homework Equations


More a concept question, no relevant equations (except definition C = Q/V)

The Attempt at a Solution


Okay I know I'm supposed to think that if you connect "2" to "4" (or "1" to "3") you get a total potential difference of 5 + 5 = 10 V. But I'm having trouble truly seeing that they just add. When you connect "2" to "4", they're clearly on a different potential, and because it's a conductor, it'll want to go into electrostatic equilibrium and make the two connected pieces the same voltage, resulting in a shift of charges.

If you say "the uttermost left and uttermost right halves keep the charges in the middle where they are so they don't mix and change things", I understand, but still, then what is the deal with the (in this explanation non-existing) equipotential surface. So my clear question is: on one hand I understand that the capacitor-charge distribution (of each) doesn't change upon connection of "2" and "4", but on the other hand I don't understand, because there should be an equipotential surface (created) in between. What's the big picture?

I thank you,
mr. vodka
 

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Is it maybe that there IS a shift of charges to make it an equipotential surface, but that it's pretty much negligable? That does sound pretty qualitative and even vague...
 
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