# Final potential difference of a 2 capacitor system

• Zack K

## Homework Statement

An isolated parallel-plate capacitor of area ##A_1## with an air gap of length ##s_1## is charged up to a potential difference ##\Delta V_1## A second parallel-plate capacitor, initially uncharged, has an area ##A_2## and a gap of length ##s_2## filled with plastic whose dielectric constant is ##K##. You connect a wire from the positive plate of the first capacitor to one of the plates of the second capacitor, and you connect another wire from the negative plate of the first capacitor to the other plate of the second capacitor. What is the final potential difference across the first capacitor?

## Homework Equations

##\Delta V=V_B - V_A##
##\Delta V=-\vec E \Delta r##
##U_{electric}=Q\Delta V##

## The Attempt at a Solution

My 2 main problems are:
1) I'm not sure what the potential of the second capacitor is compared to the first one when connected by a wire. I know that since we have a conductors, charge will transfer on to the second capacitor. I'm assuming that the charge capacitor 1 and 2 will equal the total charge of the system. I think that the charge on the second capacitor will be less than of the first, since everything in the universe likes to travel in the path of least resistance (not sure if that's the case for this type of problem though). I know that ##\Delta V_2## is less than ##\Delta V_1## since we have a dielectric in the middle which would decrease it's net electric field, and therefor it's ##\Delta V##.
2) Given the above, I'm not sure how to relate these 2 capacitors in terms of an equation.

• yohanes vianei

Conductors are equipotentials. They'll have the same potential difference when the charge has rearranged.

And you're correct that charge is conserved, so the charge on each capacitor adds up to the original charge.

• Zack K
Conductors are equipotentials. They'll have the same potential difference when the charge has rearranged.

And you're correct that charge is conserved, so the charge on each capacitor adds up to the original charge.
So the sum of the new potential difference of the new connected system will be the same as the original value?

So the sum of the new potential difference of the new connected system will be the same as the original value?

I don't think so, but haven't checked. What you know is that V = Q/C is the same for both capacitors, with their appropriate values of Q and C.

Thinking about it, I really don't think so. The voltage across the two capacitors is the same, and there's no reason to think that each new voltage is half of the original voltage unless each charge is half of the original charge. And that would only be true if the two capacitors were identical, which they're not.