Simple Capacitor Problem making a Big Problem

  • Thread starter nonequilibrium
  • Start date
  • Tags Tags
    Capacitor
1 reply · 2K views · last reply
nonequilibrium
Messages
1,412
Reaction score
2

Homework Statement


Circuit diagram of two capacitors in series-style layout: C1 with terminals 1 and 2, V1 = 10V and V2 = 15 V; C2 with terminals 3 and 4, V3 = -5V and V4 = -10V.

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
 
Attachments
  • Circuit diagram of two capacitors in series-style layout: C1 has terminals 1 and 2 with V1 = 10V and V2 = 15 V; C2 has terminals 3 and 4 with V3 = -5V and V4 = -10V.
    capacitor.gif
    2.2 KB · Views: 545
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...