What is the potential difference and charge on capacitors connected in series?

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SUMMARY

The discussion focuses on calculating the potential difference and charge on two capacitors connected in series: a 2.50µF capacitor charged to 857V and a 6.80µF capacitor charged to 652V. The charge on each capacitor is calculated using the formula Q=CV, resulting in Q1 = 0.0021C and Q2 = 0.0044C. The participants emphasize the conservation of charge and suggest using potential energy equations to find the potential difference, although the lack of distance or area data complicates the calculations.

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  • Understanding of capacitor charging and discharging principles
  • Familiarity with the formula Q=CV for calculating charge
  • Knowledge of potential energy equations in electrical circuits
  • Basic concepts of series and parallel capacitor configurations
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mer584
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I was hoping someone could help me approach this problem.

A 2.50uF capacitor is charged to 857V and a 6.80uF capacitor is charged to 652V. These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potential difference across each and the charge on each? (note that charge is conserved)
 
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Sorry, I don't know how to use this, this post should be moved to the homework help section.
What I started to do to attemp the problem was use the formula Q=CV to find the charge that is occurring in each case finding Q1 = .0021C and Q2= .0044 C. I really wasnt sure if this was even helpful or where to go from there. It appears to be a uniform field so I know I can use V=Ed but we don't have a distance or an area in this problem.

In order to get the potential difference I know you need to work with Vb-Va and possibly the potential energy. Should I use PE= V/Q then subtract the potential energies to find the work and then the poential difference?
 

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