# About capacitor -- how can current flow "through" it?

Usually people get kind of frustrated about how can a capacitor charge up when there's no current flowing through, including me.
Can I interpret the picture better by considering the electric potential set up by the battery across the circuit? Is this interpretation a correct one?
Sorry for asking this question which has been asked for a thousand times.

NascentOxygen
Staff Emeritus
Usually people get kind of frustrated about how can a capacitor charge up when there's no current flowing through, including me.
A capacitor has two wires emerging from it. While it's charging, current flows "in" on one wire and flows "out" by the other wire. To most people, that's as good as flowing "through" the capacitor.

For those who are reluctant to accept this, the concept of displacement current was introduced to show a current passing through the dielectric.

For those who are reluctant to accept this, the concept of displacement current was introduced to show a current passing through the dielectric.

So you mean it's nothing to do with the electric potential difference set up by the battery or charger?

Drakkith
Staff Emeritus
So you mean it's nothing to do with the electric potential difference set up by the battery or charger?

It's not that it has nothing to do with those concepts, they are related after all, it's that charging a capacitor adds charge to one plate while removing charge from the other plate. Positive charge if you're using conventional current. Negative charges if you want to talk about electrons. It's like pumping water from one cup to another cup. The water doesn't pass through the cup itself, it has to go through a hose, with some sort of pump providing the energy and force to move the water.

NascentOxygen
Staff Emeritus
So you mean it's nothing to do with the electric potential difference set up by the battery or charger?
The charge accumulating on the plates is pushed along by the battery potential, but if you are considering a circuit of ideal elements then you must include a resistance in series with the capacitor and battery to account for the behaviour we observe for this arrangement using practical devices.

jim hardy
Gold Member
Dearly Missed
Here's an explanation written for beginners

http://amasci.com/emotor/cap1.html
When we "charge" a conventional metal-plate capacitor, the power supply pushes electrons into one plate, and the fields from these extra electrons reach across the gap between the plates, forcing an equal number of electrons to simultaneously flow out of the other plate and into the power supply. This creates opposite areas of imbalanced charge: one plate has less electrons and excess protons, and the other plate has more electrons than protons. Each individual plate does store charge.

However, if we consider the capacitor as a whole, no electrons have been put into the capacitor. None have been removed. The same number of electrons are in a "charged" capacitor as in a capacitor which has been totally "discharged." Yes, a certain amount of charge has been forced to flow momentarily during "charging," and a rising potential difference has appeared. But the current is directed through the capacitor, and the incoming electrons force other electrons to leave at the same time. Every bit of charge that's injected into one terminal *must* be forced out of the other terminal at the same time. The amount of charge inside the capacitor never changes. The net charge on each plate is cancelled by the opposite charge on the other plate. Capacitors are never "charged" with electric charge!

just like with a wire,
for every charge that goes in one end a charge comes out the other end.
But they are not the same exact Tom Dick and Harry charges.
Push Tom in one side of a capacitor and Harry or Frank or 'an electron named Sue' might come out the other side.
So it appears as if current is flowing through the capacitor.

(Apologies to Johnny Cash and Shel Silverstein... old jim )

RotatingUniverse and davenn
davenn
Gold Member
2021 Award
nice one Jim

jim hardy
This humorous picture explain how AC current flow through capacitor.

CWatters
Homework Helper
Gold Member
When we "charge" a conventional metal-plate capacitor, the power supply pushes electrons into one plate, and the fields from these extra electrons reach across the gap between the plates, forcing an equal number of electrons to simultaneously flow out of the other plate and into the power supply.

What happens when all the free electrons have been repelled from the "other" plate :-)

RotatingUniverse
Just wanting to check whether I get the correct concept.

Is it that theoretically, when I connect a circuit like the one shown in below, at the two end of the switch would accumulate some charges, though small, and the ammeter shall read a small current passing through. However this effect can be ignored as the end of switch is so small and narrow that only a very small amount of charge can be stayed at equilibrium state.

Is this guess correct?

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What happens when all the free electrons have been repelled from the "other" plate :-)

This seems an interesting question to think about.
Perhaps the charge on one side would stop accumulating? Due to repulsion?

CWatters
Homework Helper
Gold Member
Just wanting to check whether I get the correct concept.

Is it that theoretically, when I connect a circuit like the one shown in below, at the two end of the switch would accumulate some charges, though small, and the ammeter shall read a small current passing through. However this effect can be ignored as the end of switch is so small and narrow that only a very small amount of charge can be stayed at equilibrium state.

Is this guess correct?

View attachment 101193
There might be a small capacitance between the contacts in a switch but for something like a lighting circuit this can usually be ignored.

Sometimes there is more capacitance between the wires on each side of the switch, especially if they are cores in the same cable. If the circuit is AC this capacitance can allow enough power to bypass the switch to light LED lamps. I have this problem in my bathroom. When I changed from halogen to LED lamps I found they were glowing even with the switch off.

jim hardy
Gold Member
Dearly Missed
What happens when all the free electrons have been repelled from the "other" plate :-)
Hmmm saturation, like iron in a magnetic medium ?
i guess if there's enough voltage it'd start stripping them from next shell ?

I've long wondered
since polar molecules in the dielectric are rotating into alignment with the field
is there a saturation effect where Q and V are no longer proportional ?

analogdesign
What happens when all the free electrons have been repelled from the "other" plate :-)

You'll have dielectric breakdown LONG before you can repel all the free electrons from the other plate. Then you'll have a short circuit and the plates will equalize (and the capacitor will probably never work again, but that depends on what it's made of).

To see why, think of the density of states of a conductor (e.g. Aluminum), then compare that to say the charge on a 1uF cap at 10V (remember an electron has charge 1.6e-19 C). You'll see there are many, many orders of magnitude more conduction electrons than there is charge on the capacitor (as in there are many trillion times more free electrons in the plates).

If you're playing along at home, what voltage would you need on a 1 uF cap made from two plates each with a volume of 1cm^3? If you can invent a dielectric that can take that voltage in that dimension you might get a Nobel Prize.

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NascentOxygen
Staff Emeritus
you might get a Noble Prize.
a legacy of Swedish inventor, Alfred Nobel

jim hardy
Gold Member
Dearly Missed
what voltage would you need on a 1 uF cap made from two plates each with a volume of 1cm^3?

?? volume of plates ? Am i missing something ?
V = Q/C

do you mean voltage to oops- edit: hold i mean remove from one plate all the conduction electrons , and double their number on the other plate ?

"Density of States ".. the depth of my ignorance is unfathomable.... i need a tutorial

If there's on the order of 10^22 free charges per cm^3
10^22charges X 1.6X10^-19C/charge = 1600 Coulombs

V = 1600 / 1E10^-6 = 1.6 gigavolts ? Now, there's a Flux Capacitior !

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dlgoff
analogdesign
?? volume of plates ? Am i missing something ?
V = Q/C

do you mean voltage to oops- edit: hold i mean remove from one plate all the conduction electrons , and double their number on the other plate ?

Yes, exactly. You're not missing anything. The question was what would happen if the capacitor plate ran out of free electrons to remove. My little puzzle was what voltage would be required to do that in a real capacitor.

"Density of States ".. the depth of my ignorance is unfathomable.... i need a tutorial
If there's on the order of 10^22 free charges per cm^3
10^22charges X 1.6X10^-19C/charge = 1600 Coulombs

You don't need a tutorial. You got the concept, just not the term. Density of States is just the number of states at each energy level that are available to be filled by a charge carrier. If you multiply it by the probability a given state is filled (Fermi-Dirac statistics) you get the number of charge carriers at each energy. In the quantum theory of solids, if a charge carrier is free, it is said to be in the conduction band. So, I'm saying density of states as a shorthand for the number conduction band electrons that are filled at thermal equilibrium (I'm assuming a metal room temperature). In a metal, virtually all available states in the conduction band are filled so my shorthand works for a metal. In fact, some people use that as the very definition of a metal (from a quantum standpoint). So, if virtually all available conduction band states are filled --> metal. If virtually no available conduction band states are filled at thermal equilibrium --> insultator. If most states are unfilled but a meaningful number ARE --> semiconductor. It is really as simple as that.

You got the value right. I picked a volume because if you have plates with atomic-level thickness there wouldn't be that many free electrons.
V = 1600 / 1E10^-6 = 1.6 gigavolts ? Now, there's a Flux Capacitior !
Indeed!

jim hardy
jim hardy
Gold Member
Dearly Missed
Density of States is just the number of states at each energy level that are available to be filled by a charge carrier. If you multiply it by the probability a given state is filled (Fermi-Dirac statistics) you get the number of charge carriers at each energy. In the quantum theory of solids, if a charge carrier is free, it is said to be in the conduction band. So, I'm saying density of states as a shorthand for the number conduction band electrons that are filled at thermal equilibrium (I'm assuming a metal room temperature).

I THANK YOU for that wonderful explanation. The Wiki one was , well, out of my league.
I learned a new term today and that makes it a good day.

In fact, some people use that as the very definition of a metal (from a quantum standpoint). So, if virtually all available conduction band states are filled --> metal. If virtually no available conduction band states are filled at thermal equilibrium --> insultator. If most states are unfilled but a meaningful number ARE --> semiconductor. It is really as simple as that.

I've run across that description of metal before, and i envision metal as an amalgam of atoms awash in a sea of loosely bound electrons .
My mental crutch for the concept is liquid mercury and i envision the electron sea as a shiny fluid engulfing the metal's crystal lattice..
But when demonstrating "conventional" versus "electron" current in Kirchoff's laws i draw "electron flow" with Green arrows.

dlgoff