Two like capacitors in parallel

[zengodspeed]

I'm having a little trouble understanding capacitors in parallel and series.

I understand that if there are two capacitors in a circuit with a potential difference V_ab then the potential difference across the capacitors is also V_ab.
Every example I have seen have been for two capacitors with different capacitance, therefore the charge on each capacitor will also differ.

But what about if there are two capacitors in parallel with equal capacitance? What consequence will this have for the resulting charge on each capacitor?

Thanks

 

[Borek]

a. one will be greater
b. other will be greater
c. they will be identical

Apply logic, choose one.

 

[zengodspeed]

a. one will be greater
b. other will be greater
c. they will be identical

Apply logic, choose one.

Given that the potential across the capacitors is the same and the capacitance is the same it seems nonsensical to assume that the charges will be anything other than identical. However, I understand that capacitors in parallel can be seen as equivalent to a single capacitor with plates of area equal to the sum of the areas of the parallel capacitor plates. This single capacitor would therefore have a higher capacitance, and given the electric potential v_ab is constant, the charge must also be higher.
So although the charge on each capacitor in parallel is identical, it is not equal in magnitude to the charge if there was just a single capacitor in the circuit, right?

 

[Borek]

So although the charge on each capacitor in parallel is identical, it is not equal in magnitude to the charge if there was just a single capacitor in the circuit, right

Twice the capacity, twice the charge (assuming no changes in the voltage applied).

 

[zengodspeed]

Twice the capacity, twice the charge (assuming no changes in the voltage applied).

So is there not some limit to how much charge can be achieved through this process?

 

[Borek]

So is there not some limit to how much charge can be achieved through this process?

The only limit is the one you can easily calculate from the voltage and capacitance, just by rearranging the definition C = q/V.

(That is, for huge capacitors you may need a lot of energy to charge them, so if you are using some small battery it won't work - but it doesn't falsify the general idea).

 

[NascentOxygen]

But what about if there are two capacitors in parallel with equal capacitance? What consequence will this have for the resulting charge on each capacitor?

When connected in parallel, each capacitor is unaware of others in parallel to it. So each behaves the same as it would if it were alone.

 

[berkeman]

When connected in parallel, each capacitor is unaware of others in parallel to it.

Please never anthropomorphize capacitors. They hate it when you do that.

 

[phinds]

I understand that if there are two capacitors in a circuit with a potential difference V_ab then the potential difference across the capacitors is also V_ab.

Uh ... you think maybe that might depend on the WAY they are in the circuit? As a blanket statement, what you said makes no sense.

 

[Tom.G]

@zengodspeed, Let us try this.

Step 1:

• Charge a capacitor to a known voltage.
• Disconnect the capacitor and measure its charge and write down the result. (maybe by discharging with a constant current and measuring the time it takes to decay to a given voltage)

Step 2:

• Two capacitors, each of which is ½ the capacitance of Step 1, wired in parallel.
• Charge the capacitors.
• Disconnect them.
• Measure the charge on one of the capacitors, and write down the result.
• Measure the charge on the other capacitor, and write down the result.

How does the measured charge from one of the capacitors in Step 2 compare to the charge from Step 1?
How much total charge from the capacitors was measured in Step 2?

That handles the parallel case; same general approach applies in the series case... but a bit harder to think about.
(You may not have to actually do this experiment in the lab. Just calculating the answers should suffice. Or even as a mental exercise )

Additional Thought: as pointed out by @phinds

capacitors in parallel can be seen as equivalent to a single capacitor with plates of area equal to the sum of the areas of the parallel capacitor plates.

So although the charge on each capacitor in parallel is identical, it is not equal in magnitude to the charge if there was just a single capacitor in the circuit, right?

But you haven't stated the capacitance of each capacitor, so that question can't be answered as posed.

 

[CWatters]

The amount of charge is given by Q=CV.

So a capacitor with twice the capacitance will hold twice the charge if connected to the same voltage.

If you differentiate both sides you get

dQ/dt = CdV/dt

But dQ/dt is current. So to charge two capacitors in parallel at the same rate will take twice the current.

It takes twice a much water to fill two buckets than it does one. If you want to fill two buckets in the same time as one you need a higher flow rate.

 

[phinds]

But you haven't stated the capacitance of each capacitor, so that question can't be answered as posed.

He hasn't even stated whether the caps are in parallel or in series, are separated by a resistor, etc. In short, he seems to have some circuit in mind and thinks we can all read his mind as to what it look like.

 

[CWatters]

I was going by the thread title.

 

[phinds]

I was going by the thread title.

As I would have been if I had been careful enough to read it