Asking some Capacitor questions to help my understanding

[Chenkel]

TL;DR Summary

Explaining positive charge on one side of capacitor

I have watched a number of tutorials on how a capacitor works and I'm still confused about a couple things. Suppose we have a battery who's terminals are connected to the battery. It make sense to me that negative electrons accumulate on one side of the dialectric, because they're attracted to the otherside because there's a voltage difference between the terminals of the battery, as electrons flow from the negative terminal to the positive terminal they collide with the dialectric in the capacitor, preventing them from going any further, creating a negative charge on one side of the capacitor, but I don't understand how a positive charge is created on the otherside of the capacitor. How does one side of the capacitor get a net positive charge unless electrons have left it, what explains the absence of electrons on one side of the terminal?

 

[anorlunda]

There are many animated explanations available via this Youtube search.
https://www.youtube.com/results?search_query=capacitor+charge+animation

 

[Chenkel]

I watched this tutorial from your YouTube search, it seems that as negative electrons accumulate on one plate they create a pushing force on the electrons of the other plate, that's very cool! I wonder if these electrons disperse to create negative ions later on in the circuit, are materials just willing to accept negative electrons, is there a limit to how many electrons an atom is willing to accept?

 

[Baluncore]

I wonder if these electrons disperse to create negative ions later on in the circuit, are materials just willing to accept negative electrons, is there a limit to how many electrons an atom is willing to accept?

Metal conductors have huge quantities of electrons that can move like a mist between the atoms. Capacitance involves upsetting the density or balance of the electron mist. The number of electrons, or the charge that is pushed out of place, is proportional to the voltage applied; Q = V * C. That is the definition of capacitance; C = Q / V.

The limit comes when the voltage is high enough to cause electrons to break down the insulation of the dielectric.

 

[Chenkel]

Metal conductors have huge quantities of electrons that can move like a mist between the atoms. Capacitance involves upsetting the density or balance of the electron mist. The number of electrons, or the charge that is pushed out of place, is proportional to the voltage applied; Q = V * C. That is the definition of capacitance; C = Q / V.

The limit comes when the voltage is high enough to cause electrons to break down the insulation of the dielectric.

I've heard one volt is one joule per coulomb, I assume C in your example is coulombs, does that means Q is in joules?

 

[Baluncore]

Q is the quantity of charge in coulombs.
C is the capacitance in farads.
V is the voltage difference.
E is the energy; E = ½ C V² joules.

 

[Chenkel]

Q is the quantity of charge in coulombs.
C is the capacitance in farads.
V is the voltage difference.
E is the energy; E = ½ C V² joules.

Does this mean that if the voltage difference between the two plates is 12 volts, and the quantity of charge stored is 24 coulombs then the amount of farads required for the capacitor is 2 farads? Does this also mean, based on the definition of voltage being joules per coulomb, that the amount of energy stored would be 12*24 or 288 joules of energy.

Does this mean you could run a 1 kilowatt device for .288 seconds with this capacitor?

 

[Baluncore]

Not quite.
Your “definition of voltage being joules per coulomb” is misleading you.
To drop one volt, charge must flow through a one ohm resistor, at the rate of one coulomb per second. One coulomb per second, is one ampere. The energy dissipated in the resistor will be one watt. One watt, is one joule per second.

When charge flows into a capacitor, the voltage rises as the charge flows. The average voltage is therefore half the final voltage, so the capacitor actually stores 144 joules, NOT 288 J.
Use; E = ½ C V²;
For your example; 0.5 * 2 * 12² = 144 J .

 

[Chenkel]

Q is the quantity of charge in coulombs.
C is the capacitance in farads.
V is the voltage difference.
E is the energy; E = ½ C V² joules.

I'm a little confused with how you came to the equation of energy, so as electrons are flowing into the capacitor the voltage of the capacitor increases, it's like pushing water into a tank, the pressure will increase, the current to the capacitor is constant if I'm not mistaken. If you take the volts and multiply it by the coulombs wouldn't you expect that to give you the joules? Also E looks a lot like an anti derivative of some equation, that may be coincidence, but because some quantity is changing overtime i.e voltage, it leads me to believe there might be some hidden calculus that is key to understanding.

 

[anorlunda]

If the voltage ramps from 0 to V volts in a straight line, what was the average voltage during the ramp?

 

[Chenkel]

If the voltage ramps from 0 to V volts in a straight line, what was the average voltage during the ramp?

The area under the line f(v) = v is (1/2)v^2 so the average from 0 to V is ((1/2)V^2)/V that's V/2. Average of a function confusing to me because I don't see the direct analogy to statistics which do make sense to me, for example if you have 3 dogs with heights 2ft, 2ft, 3ft respectively, then the average height is (3 + 2 + 3) / 3 = 2. Let me know if my average function math is correct, thank you!

 

[anorlunda]

Yes your average is correct. I make no sense out of trying to apply statistics to this. It is more like filling a glass with water.

 

[Baluncore]

For a constant current I;
In a resistor R, the voltage V, will also be constant, proportional to the current.
For a capacitor C, the voltage rises in proportion to charge Q, the integral of the current.
For an inductor L, the voltage is zero, the derivative of the current.

Power W, is the rate of energy flow.
Energy is dissipated as heat in a resistor, at the rate of; W = I²·R = V·I watts.
Energy is stored in the electric field of a capacitor; E = ½·C·V² joules.
Energy is stored in the magnetic field of an inductor; E = ½·L·I² joules.

 

[Chenkel]

I watched a YouTube video on the subject and learned some things, one of them is the derivation of the formula E = (1/2)CV^2



The amount of charge a capacitor stores is Q=CV

A little bit of work is dU = VdQ where V is the voltage between the plates, and dQ is a small amount of charge, this indicates that the bigger Q is the more work is required to move a small charge dQ, this makes sense because the capacitor is charging and as more electrons are being pushed onto one of the plates, an increase in the density of electrons occurs, creating a larger repeling force.

Using the two equations above we can write

dU = (Q/C)dQ

Integrating both sides we get the total change in charge

U = (1/2C)Q^2

Plugging the first equation in we get

U = (1/2C)(C^2)(V^2) = (1/2)C(V^2)

The only part of the video I was confused about was when he was talking about the movement of positive charges, the protons stand still so I don't see how a positive charge moves around.

 

[Baluncore]

The only part of the video I was confused about was when he was talking about the movement of positive charges, the protons stand still so I don't see how a positive charge moves around.

When an electron moves away, the hole it was in, and the proton next to it, remains. The electron no longer cancels that proton's charge.

If you have a sheet full of holes, and plug all but one of them, you can move that hole by moving one plug.

 

[Chenkel]

When an electron moves away, the hole it was in, and the proton next to it, remains. The electron no longer cancels that proton's charge.

If you have a sheet full of holes, and plug all but one of them, you can move that hole by moving one plug.

What you said makes sense, but I feel the when people say the positive charge is moving it's a harder to comprehend than saying a negative charge is moving because a negative charge moves with the electron it's associated with, where as a positive charge doesn't move with the proton it's associated with (this is true for copper wires and materials where the atoms don't move, but not true for all situations). If you have atoms A B next to each other consecutively, if A loses an electron, then it will gain one from B, at first the positive charge is associated with A, but then the positive charge is associated with B. So a positive charge is moving around but the charged particle is changing.

 

[Baluncore]

It often makes little difference if the charge carrier is negative or positive.
Conventional current is positive.
Later, it turned out that the mobile component was negative, the electron.

 

[Tom.G]

For the Positive/Negative current flow, this post (or the whole thread) may help:
https://www.physicsforums.com/posts/6010519

Cheers,
Tom

 

[Chenkel]

So convention is to say current is in direction from positive to negative, that is, when current is one amp that means every second one coulomb of negative charge is moving from negative to positive and one coulomb of positive charge is moving from positive to negative. It's like those puzzle games where there is one missing piece and you slide the blocks around by moving the empty space around. This time the empty space is analogous to a positive charge.

Here's an example of what's happening in a copper wire when one electron is moving

Net	+1	0	0	0
Electrons	28	29	29	29
Protons	29	29	29	29

Net	0	0	0	+1
Electrons	29	29	29	28
Protons	29	29	29	29

Net is the net charge and we see in that as the electrons move from negative to positive the positive charge 'flows' in the other direction

So when we see the convention of a arrow pointing from + to - for current it means the direction the positive charges are moving.

 

[Chenkel]

I did some more research and found out that in the case of battery and a copper wire, the copper wire has a net charge of 0 and is connected from + to - where the - side of the battery contains a compound which wants to lose an electron and the + side of the battery is connected to a compound that wants to gain an electron. So the last post I made is almost correct but not quite, there are no copper ions and the coppers outer electrons move freely to allow a flow of electrons from - side of the battery to + side of the battery, the copper wire is gaining electrons from the negative terminal and at the same time the wire is losing electrons to the positive terminal, since the wire is gaining the same amount of electrons as it's losing the wires net charge is 0. Since we are 'removing' positive charges from the positive compound (by adding electrons) and we are 'adding' positive charges to the negative compound (by removing negative electrons) we can say there is a movement of positive charge from positive to negative.