# If capacitors in series reach the same charge and all have the same ca

1. May 8, 2014

### jaredvert

Doesn't this imply their voltages are the same???? Why does my textbook say v= v1+ v2 .... Am I missing something here??? Someone elucidate please .. And one more question...... Is there any reason when using gauss's law why you use an upright cylinder in an infinite area and a knocked over cylinder in an infinite line (of lambda) to calculate the electric field??? ...thanks

2. May 8, 2014

### 256bits

Parallel capacitors have the same voltage, but can have different charge accumulation if the capacitances are not equal.
Series capacitors have the same charge accumulation, but can have different voltages if the capacitances are not equal.

3. May 8, 2014

### jaredvert

But if they are equal capacitance in series, then doesn't that imply they are all the same voltage across?? Since they all accumulate the same charges??? And so emf= v1=v2= emf

4. May 8, 2014

### Staff: Mentor

When they're in series, they add. The high-voltage side of one capacitor is connected to the low-voltage side of the next one, so the high-voltage side of the next capacitor must be even higher - it's like a stairway, where each step starts at the level of the previous step and takes you one step higher.

5. May 9, 2014

### jaredvert

"The high voltage side of one cap is connected to the low voltage side of another" what does this even mean? So you're saying the first capacitor in this series does not reach the full emf? How can that be since there is nothing else to deter it or no other path for the electrons to go. By logic wouldn't all the voltage appear across that first capacitor and hence all the charge and then since the charge is equal to the other ones, across the whole series?

6. May 9, 2014

### 256bits

With a series of capacitors in series all the charge has to flow around the circuit. So whatever charge exits the emf is the same for for all series capacitors. ( This is the same thing as for series resistors - all the current flowing through each resistor is the same. ) this is a principle of continuity.

As you know whenever a capacitor accumulates a charge, there is a resulting voltage across its terminals.
Since each capacitor is accumulating charge there has to be a voltage across each capacitor terminal. The sum of all the capacitor voltages in series adds up to the emf.

If the series capacitors were all equal, then each capacitor would have the same voltage, and all of them add up to the emf.

7. May 9, 2014

### Staff: Mentor

yes. That's what those plus signs in the formula in your textbook are saying.

The easiest way to understand what's going on is to think through the simplest case: two capacitors connected in series. We have three wires in the circuit: one from the battery plus terminal to one plate of the first capacitor; one from the other plate of that capacitor to a plate in the second capacitor; and one from the other plate in the second capacitor to the battery negative terminal.

Now consider the behavior of the charge carriers in that middle segment of wire. The electrons move towards the first capacitor leading to an accumulation of negative charge there. But this leaves the other end with a positive charge, which leads to an accumulation of negative charge on the plate connected to the negative terminal of the battery.

8. May 9, 2014

### enorbet

Simply put Voltage is a potential relative from one point to another. Commonly the reference point is ground but it can simply be across a component, irrespective of system ground. If a 20v battery is placed across two capacitors in series, each will display roughly 10v from it's anode to it's cathode, but across the system of two (anode of one cap to the cathode of the other) you will see 20v.

9. May 9, 2014

### dauto

If a 30 m high building has 10 stories does that mean each story is 30 m high?

10. May 9, 2014

### jaredvert

Oh ok so what (in the two capacitor case you described) keeps all the charge from accumulating on that first capacitor??? I know there is repulsion from what has already been placed there but it should be enough to counter up to 20 v, where is Thai extra repulsion or is this just how it is shown experimentally?

11. May 9, 2014

### jaredvert

Ok kind of random but say you have a 10 ohm resistor before a branch with two parallel resistors, the voltage that goes through each branch is the same but it is equal to the emf - I * 10 ohms right??? So there is a potential drop before the current reaches the branch?

12. May 9, 2014

### Staff: Mentor

If you're going to ask a totally new and totally unrelated question, start a new thread... You'll get better answers that way.

13. May 9, 2014

### Staff: Mentor

Consider that middle length of wire, the one that connects one plate of one capacitor to one plate of the other, before we've made the connections between the other plates and the battery terminals. The total charge in that wire is zero, right? And it's going to stay that way, because it's electrically isolated - the electrons from the battery cannot jump across the gap between plates to get to that wire, and the electrons already in the wire cannot jump across the gap to leave the wire.

So if a negative charge builds up on the plate connected to one end of that wire, then to keep the total charge zero an equal positive charge has to appear at the plate in the other capacitor that is connected to the other end of that wire. Then the battery will deliver charge to the two plates connected to its two terminals... And we end up with two charged capacitors.

14. May 9, 2014

### jaredvert

Yeah I get that, I was just curious why their the voltage isn't v=v1= v2 f their capacitances are the same

15. May 9, 2014

### Staff: Mentor

Let's call the voltage between the two terminals of the battery $V_b$, and let $v_1$ and $v_2$ be the voltages across the two capacitors.

You are right that if the capacitances are equal, then $v_1$=$v_2$

What is the voltage between the plate connected to the negative terminal of the battery and the Playe connected to the positive terminal of the battery?

What is the voltage between the two plates connected to the center segment of wire?

Answer those two questions and you'll be able to get to $V_b=v_1+v_2$, or (for the case of equal capacitances) $v_1=v_2=V_b/2$. This is dauto's building from #9 with two stories instead of ten, and my stairway from #4 with two equal-height steps.