Connected Capacitors and final charge

[minimario]

Homework Statement

A 25

and a 40

capacitor are charged by being connected across separate

V batteries. The capacitors are then disconnected from their batteries and connected to each other, with each negative plate connected to the other positive plate. What is the final charge on each capacitor, and what is the final potential difference across the 40

capacitor?

Homework Equations

Q = C * V, Parallel: Add Capacitance, Same Voltage; Series: 1/C=1/C1+1/C2, charge equal

The Attempt at a Solution

Cannot tell if they are in series or in parallel...what happens when 2 capacitors are connected without a battery?

 

[ehild]

What is the net charge on the connected plates of capacitors, connected in series? Is it the same in the problem?

 

[minimario]

What is the net charge on the connected plates of capacitors, connected in series? Is it the same in the problem?

The net charges for the 2 capacitors are 1250 and 2000 ##\mu F##, combined, they should be 3250?

 

[ehild]

What is the unit of capacitance?
The capacitor plates are connected positive to negative. If you consider the connected plates like a single plate, what is the charge on it?

 

[minimario]

The unit is ## \mu F## (oops, I meant ## \mu C## in the previous post), it should be 3250 ## \mu C##

 

[ehild]

The unit is ## \mu F## (oops, I meant ## \mu C## in the previous post), it should be 3250 ## \mu C##

It is not 3250 μC. You connect the positive plate of one capacitor to the negative plate of the other. The excess electrons on the negative plate go over to the positive plate and neutralize the positive charges. How much charge remains on the pair of plates?

 

[minimario]

Then, it must be 750 uC. But I still don't understand, why the electrons on negative plate go to positive plate? Then one of the plates has no more charge??

 

[ehild]

Not all go. You can consider the connected plates as one plate of an equivalent capacitor. The charge of that capacitor is 750μC. What is the capacitance of that equivalent capacitor? What is the voltage across the plates?

 

[minimario]

The voltage is 100 V, because the plates are connected in parallel and the previous total voltage was 100 V? And so the capacitance is 7.5 μF? (How can you tell the plates are connected in parallel other than realising the charges are different, so it's not a series connection?)

 

[ehild]

If you connect two capacitors only, they can be considered either series or parallel. But series capacitors have the same charge, positive on one of the connected plates and negative on the other one. It is not the case now, but you can take the two capacitors connected in parallel. What is the resultant capacitance ?

Both capacitors had 50 V voltage across their plates initially, but the positive plates are connected to the negative ones. You can not add the voltages!The voltage becomes the same on parallel connected elements.
But you can calculate the voltage across the equivalent capacitor using the charge and the resultant capacitance.
The image shows the initial situation, the capacitors charged, then connected, electrons move to neutralize the positive charges and the picture below shows the equivalent capacitor.

 

[minimario]

The resulting capacitance is the sum of the original capacitances, or 65 uF. And so the new Voltage is 11.54 V.

 

[ehild]

The resulting capacitance is the sum of the original capacitances, or 65 uF. And so the new Voltage is 11.54 V.

Very good. That is the voltage on both original capacitors. The net charge, 750μC is shared between them. You know the voltage, you know the capacitance of both individual capacitors, so you can calculate the individual charges.

 

[minimario]

If you connect two capacitors only, they can be considered either series or parallel. But series capacitors have the same charge, positive on one of the connected plates and negative on the other one. It is not the case now, but you can take the two capacitors connected in parallel. What is the resultant capacitance ?

Both capacitors had 50 V voltage across their plates initially, but the positive plates are connected to the negative ones. You can not add the voltages!The voltage becomes the same on parallel connected elements.
But you can calculate the voltage across the equivalent capacitor using the charge and the resultant capacitance.
The image shows the initial situation, the capacitors charged, then connected, electrons move to neutralize the positive charges and the picture below shows the equivalent capacitor.

View attachment 80944

I get the idea now, but how do you know the 2 capacitors did not have the same charge at the end? (equalize)

 

[ehild]

I get the idea now, but how do you know the 2 capacitors did not have the same charge at the end? (equalize)

Why should be the charge equal on both capacitors? The connected plates are at the same potential. Therefore, the potential differences between the plates of he individual capacitors are the same.
Imagine two lakes, a big one and a small one, connected by a channel. When both are full, the water level is the same in both lakes. Is the amount of water in in the small lake the same as in the big one?