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

AI Thread Summary
The discussion revolves around calculating the potential difference and charge on capacitors connected in series after being charged to different voltages. Two capacitors, 2.50uF at 857V and 6.80uF at 652V, are connected with their positive and negative plates joined. The charge on each capacitor is calculated using the formula Q=CV, resulting in charges of 0.0021C and 0.0044C. Participants express uncertainty about the next steps, including whether to use potential energy formulas to find the potential difference. The conversation highlights the need for clarity on charge conservation and the relationships between voltage, charge, and capacitance in series circuits.
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 forumla 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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