# A simple capacitor problem

1. Nov 28, 2015

### gracy

1. The problem statement, all variables and given/known data
Consider two capacitors $C_1$=10μf and $C_2$=40μFif the two are connected in series to a 9.0 volt battery What are the charges $Q_1$ and $Q_2$ stored on the two capacitors?

2. Relevant equations

$Q$-$C$$V$
3. The attempt at a solution
The capacitors are connected in series hence equivaIent capacitance would be 8μF (by solving I got this)I want to know shall I put 9.0 in place of V in the formula above,just because they are connected to a 9.0 volt battery ,potential difference between the two capacitors would be same as battery emf?is it always the case that potential difference between components is same as emf of connected battery?or is just happens in case of series connection?

2. Nov 28, 2015

### nrqed

Whenever a *single* circuit element is connected to a battery, the voltage across it will be equal to the emf. If several elements are connected in parallel to a battery, they will each have a voltage equal to the emf. In your example, the *equivalent* capacitor has a voltage equal to the emf across it, but the individual capacitor will each have a smaller voltage than 9.0 volts.

3. Nov 28, 2015

### gracy

If several elements are connected in series to a battery then?

4. Nov 28, 2015

### gleem

How much charge is provided by the battery to the equivalent capacitor. and thus what charge is on each capacitor.

5. Nov 28, 2015

### SammyS

Staff Emeritus
Draw the circuit.
Apply what you have learned in the last two days.

You tell us: How much charge would be on each plate of the equivalent capacitor.

Then, converting that back to the two resistors capacitors in series, how much charge is on each of their plates, and finally, how much voltage is across each capacitor?

Above correction done in Edit: .

Last edited: Nov 30, 2015
6. Nov 29, 2015

### gracy

Means all three (2 capacitors and one battery)are in series?

7. Nov 29, 2015

### cnh1995

I don't know if the battery is said to be in "series" with the capacitors. But the charge will be same on both the caps since they are in series with each other. I would say series combination of caps is connected "across" the battery.

8. Nov 30, 2015

### gracy

9. Nov 30, 2015

### Staff: Mentor

Where elements are connected so that their current is common (i.e., all of the current through each goes through each other), then those elements are in series.

Note: even though a capacitor has an insulator right in its centre, we still can think of current passing through a capacitor for the purpose of circuit analysis, because of the way capacitors work.

10. Nov 30, 2015

### gracy

11. Nov 30, 2015

### SammyS

Staff Emeritus
What was that question? ... Oh, yes !

That's a rather strange question, but wait! It's in answer to another question by you.

Well, technically, that's not a question, but it's part of a run-on sentence in the OP.

A little more of the OP is:
Well, by now you know that I'm not going to simply give an answer, but I'll try to lead you to it.

You correctly found that the two capacitors in series have an equivalent capacitance of 8μF.

So replace the two capacitors with a single capacitor having capacitance of 8μF.

(That reminds me: You never answered my questions in post #5 of your thread https://www.physicsforums.com/posts/5303965/) I can play that game too.​

How much charge is stored in that 8μF capacitor ? What is the voltage across it?

(Please: Try to answer more than one question in each one of your posts. I.e., organize them more than you have been doing.)

12. Nov 30, 2015

### gracy

The equivalent capacitor has the same charge as it's individual component I am sure it is true for series connection but what to know is it applicable in parallel connection as well?

13. Nov 30, 2015

### gracy

When we replace the two capacitors with equivalent capacitor,it becomes *single* circuit element .That means equivalent capacitor will have same voltage across it as emf of battery.Hence,I can put 9.0 volt in place of V in the formula below
$Q$=$C$$V$
Here Q=Charge on equivalent capacitor
C=Capacitance of equivalent capacitor =8μF
V=Voltage across equivalent capacitor=9.0 volt
using formula mentioned
Q=72μC
But we are asked to find $Q_1$ and $Q_2$
Hence
$Q$=$Q_1$=$Q_2$=72μC
Right ?

14. Nov 30, 2015

### gracy

Thanks @nrqed for addressing the question in OP.

15. Nov 30, 2015

### SammyS

Staff Emeritus
Amazingly, nrqed did this is post #2 of this thread.

16. Dec 1, 2015

### gracy

Yes,but I took time to understand that .I know it was crystal clear,nothing to understand so I should say I took time to consider that .But what's important is I solved the problem on my own.
Still I am waiting for the answer of my post #12.

17. Dec 1, 2015

### ehild

NO.

18. Dec 1, 2015

### cnh1995

In case of parallel connection, the equivalent capacitor will have charge equal to the sum of the charges on individual capacitors.