# Battery and capacitor in series

1. Jul 24, 2013

### Qroid

In 1. (see attached), a is attached to ground, so has a voltage of 0 V. The voltage at b is -1 V, so the voltage across the capacitor (Vb-Va) is -1 V.

My understanding is so rusty... What happens when a is now wired to the positive end of a 1.5 V battery, and the negative end of the battery is wired to the ground, as in 2? Specifically, what are the new voltages at a and b, and what is the new voltage drop across the capacitor? Is it still -1 V, or has it changed?

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2. Jul 25, 2013

### ehild

The voltage (potential) is referred to the ground, where the potential is taken zero. The voltage (potential) difference is 1.5 V across the terminals of the battery. One terminal is connected to the ground. What is the voltage (electric potential) of the other terminal connected to a?
The voltage (potential) of point b is kept -1 V with respect to the ground. Does it change if something is connected to an other point ?

ehild

3. Jul 25, 2013

### Qroid

Hi ehild, thanks so much for your help. I'm embarrassed how rusty I am at this!

In the second configuration, a is now at 1.5 V (relative to ground). Doesn't that mean that b is now 1.5-1 = 0.5 V relative to ground? So the voltage difference across the capacitor is still -1 V (b-a)? Is it correct to say that changing configuration from 1. to 2. had no effect on the voltage across the capacitor?

4. Jul 25, 2013

### ehild

NO. B is still -1V with respect to the ground:Vb=-1 V. You would be right if the potential difference across the capacitor would be unchanged. But that is the question. The potential of b is unchanged, and the potential difference across the capacitor is Vb-Va.

ehild

5. Jul 25, 2013

### Qroid

Hi ehild - sorry to be pedantic, but I just want to make sure I have this right.

The voltage at b does not change. Vb=-1V in both situations.
The voltage at a does change. Va=0v in situation 1., and Va=1.5V in situation 2.
Therefore the voltage across the capacitor, Vb-Va, changes. Vb-Va=-1V in situation 1. to Vb-Va=-2.5V in situation 2.

Is this correct? Many thanks.

6. Jul 25, 2013

### ehild

It is correct.

ehild

7. Jul 25, 2013

### Qroid

There's something I don't understand...

If the potential across the capacitor changes, then the charge on the capacitor would have to change. But b is floating, so there's no source of charge for b (no way to get current to it). So how does this work? Specifically, how does the charge on the b side of the capacitor change?

Again many thanks.

8. Jul 25, 2013

### ehild

The text of the problem is not clear. You can understand it in the way that b is floating, (it was not stated in the OP), in that case the charge of the capacitor and the potential difference across the capacitor remain the same, and the potential of b increases by 1.5 V, so it becomes 0.5 V.

The other possibility is that the potential of b originates from an external source, for example, it is connected to the negative pole of a battery, with the other pole connected to the ground. That makes a closed circuit, current can flow to recharge the capacitor causing a different voltage across it.

I understood the problem in the second way. Did you copy the problem text exactly as it was written?

ehild

9. Jul 25, 2013

### Qroid

Hi - sorry for the confusion! b is indeed floating. I (think) I set this in the circuit diagram, but I didn't explicitly state it in the question. The question is worded by me, so all confusion is my fault!

I understand what happens now. Thanks very much - super useful!

10. Jul 25, 2013

### ehild

If you said that a charged capacitor is connected to point a it would have been more clear. Anyway, I am pleased that you understood the concept of potential and potential difference.

ehild