Charging Capacitor with one terminal grounded

[gabriel109]

I'm reading the capacity chapter of Serway's book, and I had a question about the charging of a parallel plate capacitor. Let's assume the following situation with a modification of the circuit in the figure: we connect the negative terminal of the battery and one of the capacitor plates to ground. The positive terminal connects directly to the plate as in the figure. I understand that the negative terminal and the plate will be at the same potential (gnd) and there should be no flow of electrons. This is so?

 

[.Scott]

Grounding the negative side will have no effect on the current.
There will be some current as the capacitor is charging. But once charged, there will be no more current.

Please read this next paragraph with caution. It is not part of the text book answer - so don't worry if you don't understand it in detail. But:

It may be useful to comment on the drawing. It is intended to be schematic - not to any real scale.
If a 12-volt lead-acid automobile battery was used and the plates where to scale as pictured, then grounding the circuit might actually have an appreciable effect on the current. This is because those plates are so large and so far apart that they might have noticeable capacitive couplings with other items in the environment. For example, what object are you going to use as the ground you are connecting to? Is there a car nearby connected to that ground?

 

[gabriel109]

Thanks for the reply. I'm ashamed that I still don't understand. We start from the initial situation with the plates discharged, when the cables are connected to the terminals (this means positive terminal to a plate, negative terminal to ground, and the remaining plate to ground), why do electrons flow towards the ground plate? if this plate is at the same potential (0) as the negative terminal.

 

[gabriel109]

To be more graphic:

 

[Gavran]

Although the negative side is grounded the process remains the same.

In the initial situation the negative terminal of the battery connected to the ground is not at the same potential as the plate connected to the ground. The ground plays the same role as the connecting wires. The battery establishes an electric field in the connecting wires and in the ground too. The electric field applies a force on electrons in the wire just outside the plate connected to the ground and this force causes the electrons to move onto the plate. This electric current exists until the negative terminal of the battery, the plate connected to the ground, the wire which connects the negative terminal of the battery and the ground, the wire which connects the ground and the plate, and the ground are all at the same electric potential. A similar process occurs at the other capacitor plate, with electrons moving from the plate to the wire, and in this case the ground does not play the same role as the connecting wires because the positive side is not grounded.

 

[gabriel109]

@Gavran thanks. I had a hard time understanding that while the capacitor is charging the negative terminal and the negative plate are not at the same potential.
I think this happens because the potential drop over a discharged capacitor is 0V, so there cannot be the same potential between the negative terminal and the negative plate. It's right?

 

[DaveE]

I had a hard time understanding that while the capacitor is charging the negative terminal and the negative plate are not at the same potential.

I'm not sure exactly what you mean, but, yes they are at the same potential (assuming the conductivity of the wires is insignificant compared to the rest of the circuit).

Let's consider your drawing with the grounding electrode connected back to the battery cathode (-) through the earth with some resistance.

- When the switch closes, the capacitor voltage is zero and stays at zero.
- This will make the grounding electrode rise to the battery anode potential (+).
- This, in turn, will cause current to flow through the circuit, limited by the resistance of the soil.
- The current flowing through the capacitor will cause it to charge up, perhaps slowly.
- The increased voltage drop across the capacitor will reduce the voltage of the grounding electrode.
- The decreasing voltage of the grounding electrode will cause reduced current flow through the earth according to Ohm's Law.
- The decreasing current flow will still charge the capacitor voltage, but at a slower rate. The result will be an exponential response of both the capacitor voltage and the current.
- Eventually, the grounding electrode will approach the same voltage as the battery cathode (-), and the capacitor voltage will approach the battery voltage.

The capacitor voltage only changes in proportion to the amount of charge that has passed through it. The rest of the circuit is significant in determining what that current flow is. Initially, all of the battery voltage is applied across the soil resistance, later it will all be across the capacitor.

In the drawing from the book, with no earth resistance, the current will be limited by other resistances, like internal to the battery, or the connecting wires. When the resistance gets very small, there are other much more complex things happening with the generated magnetic fields that limit the current (that should be a different thread, really).

 

[gabriel109]

@DaveE really thanks. It's much clearer to me now. Without the resistance of the earth, is the following reasoning correct?

Let's assume a 12V battery. While the capacitor is charging, in the capacitor let's assume a drop of 10V, then I can have a difference of 1V (12V - 11V) between the positive terminal and the positive plate, and 1V (1V - 0V) difference between the negative plate and the negative terminal. This will continue until "...the plate, the wire, and the terminal are all at the same electric potential...". Do you see it well?

 

[DaveE]

@DaveE really thanks. It's much clearer to me now. Without the resistance of the earth, is the following reasoning correct?

Let's assume a 12V battery. While the capacitor is charging, in the capacitor let's assume a drop of 10V, then I can have a difference of 1V (12V - 11V) between the positive terminal and the positive plate, and 1V (1V - 0V) difference between the negative plate and the negative terminal. This will continue until "...the plate, the wire, and the terminal are all at the same electric potential...". Do you see it well?

My assumption in that description was that all wires have zero resistance and therefore have no voltage drop. So there are only two voltages wrt to the battery cathode (-), the battery voltage, and the capacitor voltage. Or, in other terms; three voltage drops: the battery, the capacitor, and the ground resistance.

Yes, in the real world, there is some resistance in every conductor. There is also some inherent resistance within real batteries and capacitors, which is a much more complex discussion. But this sort of problem/description assumes idealized components. This leads into a much more complex discussion of modelling real world components, which is beyond the scope of this thread. But in that modelling process, you will find schematics using ideal components which are tailored to the application (eg. there's no point modelling the behavior at microwave frequencies of a capacitor that's used in a motor drive application). Normally, you would split this up into modelling (where a schematic is created with ideal components to match/model/predict real circuit behavior) versus analysis (where behavior/solutions are derived for the model, using only the model).

For example, this is one model of a real capacitor:

https://wiki.analog.com/university/courses/electronics/electronics-lab-capacitors

 

[gabriel109]

Thanks @DaveE. I really appreciate your help.

 

[hutchphd]

Is it useful to remember that the units of RC is time?? If you assume things occur intantaneously, that usually demands perfect conductors. The world is seldom that cooperative (and much more messy).

 

[sophiecentaur]

why do electrons flow towards the ground plate? if this plate is at the same potential (0) as the negative terminal.

It seems to me that you are making things more difficult for yourself by starting with a ground / floor/earth component. A simple idealised R, connected in series with the charging circuit would make the problem more understandable, I feel.
Current will flow round the circuit as soon as the connection is made. The PD across the C will be proportional to the charge (strictly the charge difference). The current through the R will be the battery volts minus the PD across the C. SO the current will decay exponentially with time.
In the practical example of real earth and two earth spikes, there will intitally be a , PD between the two earth spikes. There is no such thing as a true Zero Earth Voltage; it's always relative to somewhere else.

PS the image in your first post here is nonsensical to analyse correctly because any circuit will actually have a finite series resistance so the C would never charge up instantaneously. You will always have some value of CR even it it's very short.

 

[gabriel109]

It seems to me that you are making things more difficult for yourself by starting with a ground / floor/earth component. A simple idealised R, connected in series with the charging circuit would make the problem more understandable, I feel.
Current will flow round the circuit as soon as the connection is made. The PD across the C will be proportional to the charge (strictly the charge difference). The current through the R will be the battery volts minus the PD across the C. SO the current will decay exponentially with time.
In the practical example of real earth and two earth spikes, there will intitally be a , PD between the two earth spikes. There is no such thing as a true Zero Earth Voltage; it's always relative to somewhere else.

PS the image in your first post here is nonsensical to analyse correctly because any circuit will actually have a finite series resistance so the C would never charge up instantaneously. You will always have some value of CR even it it's very short.

Thanks, it was much clearer to me.
Even though the first image does not make sense, could it be said that at the beginning of the charging of a capacitor there is a potential difference between the positive terminal of the battery (negative) and the positive (negative) plate of the capacitor? And when the capacitor is fully charged, there are no longer those differences?
I want to know if what the book explains makes sense.

 

[sophiecentaur]

I want to know if what the book explains makes sense.

Frankly, I think that's too much to ask. Few PF members will be prepared to read your quote in deep enough detail to identify what it is that gives you a problem. There are hundreds of sources which will give you the correct description of this process. The most fruitfull way into this topic is to read a good text (even wiki is a good source here). When you have understood what you want to know then yhou may or may not wwant to re-read your quote but what would that achieve?

Search "charging a capacitor through a resistor" and select something that is at your level.

Knowing the right answer is far more useful than maybe finding out what's wrong with a poor answer.

 

[DaveE]

I want to know if what the book explains makes sense.

The book is using a shortcut to show the ultimate, steady state, situation of a charged capacitor with a defined potential. They didn't intend this to be a transient response problem and as such it makes no sense, and is best ignored.

A circuit with just a battery, a capacitor, and a switch, all perfect, isn't physically realizable. That transient would be an infinitely large current for an infinitely short time. The actual transient response will be dominated by behavior not shown in the text book's model, primarily parasitic resistance and/or inductance.

 

[DaveE]

BTW, Khan Academy has some excellent tutorials on circuit analysis. Check them out!

 

[gabriel109]

Thank you both for your vocation @sophiecentaur @DaveE