Simple Capacitor Problem making a Big Problem

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SUMMARY

The discussion centers on the optimal connection of two capacitors to achieve the greatest potential difference. The user, identified as "mr. vodka," explores the concept of connecting capacitors in series versus parallel, ultimately concluding that connecting terminals "2" and "4" yields a total potential difference of 10 V. The conversation delves into the implications of electrostatic equilibrium and equipotential surfaces, highlighting the complexity of charge distribution and the resulting voltage across the capacitors.

PREREQUISITES
  • Understanding of capacitor configurations (series and parallel)
  • Familiarity with electrostatic equilibrium concepts
  • Knowledge of voltage, charge, and capacitance relationships (C = Q/V)
  • Basic principles of equipotential surfaces in electrostatics
NEXT STEPS
  • Study the effects of connecting capacitors in series and parallel configurations
  • Learn about electrostatic equilibrium and charge distribution in capacitors
  • Explore equipotential surfaces and their significance in electrostatics
  • Investigate practical applications of capacitors in electrical circuits
USEFUL FOR

This discussion is beneficial for physics students, electrical engineering students, and anyone interested in understanding capacitor behavior and electrostatic principles.

nonequilibrium
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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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