Two capacitors with opposite terminals connected together

[carlyn medona]

Homework Statement

A capacitor of 5μF is charged to 24V and another capacitor of capacitance 6μF is charged to 12V , the positive plate of first capacitor is now connected to negative plate of second and vice versa, find new charges on the capacitor

Homework Equations

The Attempt at a Solution

I found charge in each capacitor, I have no idea what to do next

 

[Delta2]

what else the exercise asks for? From what you writing in problem statement that's all what we wish to find... I guess you didn't write the complete problem statement?

 

[carlyn medona]

No I mean initial charge, before connection, then I thought charge would flow between the plates till potentials become equal, but answer I got does not match with solution

 

[grzz]

Note that the POSITIVE plate of one capacitor is connected to the NEGATIVE plate of the other.

 

[cnh1995]

No I mean initial charge, before connection, then I thought charge would flow between the plates till potentials become equal, but answer I got does not match with solution

What can you say about the final voltages across the capacitors? What can you say about the polarities of the final voltages?

 

[gneill]

No I mean initial charge, before connection, then I thought charge would flow between the plates till potentials become equal, but answer I got does not match with solution

Can you share more detail about what you actually did for this attempt? What were your thoughts about how the charge would flow? What calculations did you attempt and what were your results?

 

[Delta2]

No I mean initial charge, before connection, then I thought charge would flow between the plates till potentials become equal, but answer I got does not match with solution

Ok I see now. Your mistake is that the equilibrium state will be when the voltages will be opposite $V_1=-V_2$. The condition $V_1=V_2$ is for when you connect the positive plate to the positive plate and the negative to the negative.​

 

[carlyn medona]

Okay guys I solved it, thanks for helping, I got charge on first capacitor = 21.8μC and second capacitor to be equal to 26.2μC

 

[cnh1995]

Okay guys I solved it, thanks for helping, I got charge on first capacitor = 21.8μC and second capacitor to be equal to 26.2μC

That's correct.

 

[Delta2]

That's correct.

Ok I suppose its correct (seems to be an equilibrium state since the new voltages are equal), but someone explains me please why conservation of charge doesn't hold here. Charge before =120+72=192μC, charge after 21.8+26.2=48μC.

 

[gneill]

Ok I suppose its correct, but someone explains me please why conservation of charge doesn't hold here. Charge before =120+72=192μC, charge after 21.8+26.2=48μC.

Conservation of charge holds.

When the plates are connected, one of the paired plates has a positive charge while the other a negative charge. This is true for both of the pairs that are connected. Unlike charges will meet and annihilate via the available conduction paths leaving the net sum the same, hence conservation of the net charge.

Note that (usually) on a charged capacitor the two plates carry equal but opposite charges. Hence the net charge on the device as a whole is zero. When we talk about the charge on a capacitor, we typically refer to the magnitude of the charge on one plate, since both plates have the same charge but of opposite sign. When charged capacitors are connected in parallel, the charges on the connected plates sum. If the charge signs are opposite, they will combine and cancel until no differently signed charges are left. Whatever un-combined charges are left represents the net remaining charge on that plate.

 

[Delta2]

Conservation of charge holds.

When the plates are connected, one of the paired plates has a positive charge while the other a negative charge. This is true for both of the pairs that are connected. Unlike charges will meet and annihilate via the available conduction paths leaving the net sum the same, hence conservation of the net charge.

Note that (usually) on a charged capacitor the two plates carry equal but opposite charges. Hence the net charge on the device as a whole is zero. When we talk about the charge on a capacitor, we typically refer to the magnitude of the charge on one plate, since both plates have the same charge but of opposite sign. When charged capacitors are connected in parallel, the charges on the connected plates sum. If the charge signs are opposite, they will combine and cancel until no differently signed charges are left. Whatever un-combined charges are left represents the net remaining charge on that plate.

Sorry but you got me confused. Only definite conclusion I can make from your statements is that net (positive+negative) charge is zero before and is zero after in both capacitors. Ok in that sense, conservation of charge holds.

Ok I see now conservation of charge holds for the net charge of the plates that are connected (don't know if you actually was trying to say that). Since we connect the positive to negative and vice versa, charge before=120-72=48 (or 72-120=-48), charge after=48 (or -48 if we sum the charges on the other pair of connected plates that carry negative charges) as well.

 

[gneill]

Ok I see now conservation of charge holds for the net charge of the plates that are connected (don't know if you actually was trying to say that). Since we connect the positive to negative and vice versa, charge before=120-72=48 (or 72-120=-48), charge after=48 (or -48 if we sum the charges on the other pair of connected plates that carry negative charges) as well.

Right.