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

[mer584]

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)

 

[mer584]

oops

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?

 

[astrokreng]

In this scenario, the capacitors are connected in series, meaning that they share the same path for current to flow through. This results in the same amount of charge passing through each capacitor. However, the potential difference, or voltage, will be divided between the two capacitors based on their capacitance values.

To calculate the potential difference across each capacitor, we can use the equation V = Q/C, where V is the potential difference, Q is the charge, and C is the capacitance. Since the charge is conserved, we know that the total charge on both capacitors will be equal to the sum of their individual charges. Therefore, we can set up the following equation:

Q1 + Q2 = (2.50uF)(857V) + (6.80uF)(652V)

Solving for Q1 and Q2, we get Q1 = 2.14mC and Q2 = 4.43mC.

To find the potential difference across each capacitor, we can use the equation V = Q/C again. For the 2.50uF capacitor, we get V1 = (2.14mC)/(2.50uF) = 856V. For the 6.80uF capacitor, we get V2 = (4.43mC)/(6.80uF) = 651V.

Therefore, the potential difference across the 2.50uF capacitor will be 856V and the potential difference across the 6.80uF capacitor will be 651V. Both capacitors will have the same charge of 2.14mC and 4.43mC, respectively.

In summary, when capacitors are connected in series, the charge is conserved but the potential difference is divided between the capacitors based on their capacitance values.