# Homework Help: Simple Capacitor Problem making a Big Problem

1. Apr 11, 2010

### nonequilibrium

1. The problem statement, all variables and given/known data

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

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

3. 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

#### Attached Files:

• ###### capacitor.gif
File size:
2.2 KB
Views:
188
2. Apr 11, 2010

### nonequilibrium

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...