Discussion in 'General Electronics Chat' started by activee, May 16, 2014.

1. ### activee Thread Starter Member

Jan 16, 2014
39
0
So when a capacitor is charging shouldn't in theory the electron that are travelling towards the negative plate of the capacitor being repealed by this plate since it is itself negative or am I missing something ?

2. ### crutschow Expert

Mar 14, 2008
16,186
4,324
You are correct, they are repelled. You are missing the applied voltage across the plates that causes them to become negative and positive. This voltage pulls the electrons from the positive side and forces them to the negative side until the repelling voltage equals the applied voltage (the capacitor is charged). If you remove the voltage source and connect the two plates together, than the electrons on the negative side will be repelled and rush back to the positive side until both sides have neutral net charge (the capacitor becomes discharged).

3. ### activee Thread Starter Member

Jan 16, 2014
39
0
Thanks that's what I thought but I still got a question and it's hard to put into a sentence. It seem to me that the closer the electron is to the negative plate the bigger the force to make it leave it will be. Since that electron will eventually be always closer to the negative plate it sounds intuitive to me to say that the force making it leave it will be bigger than the force making it go towards the positive plate. In other words : the positive plate has an electric field ( a + voltage relative to the e-) making the electrons travelling towards the negative plate. The negative plate has an electric field ( a - voltage relative to the e-) that makes the electron leave it. Since the electrons are closer to the negative plate I intuitively think that the electrons should be repealed. It's not the case so there is a problem in how I see things.

Last edited: May 17, 2014
4. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
519
It doesn't work anything like that.

It is said that

"Current leads voltage in a capacitor"
alternatively
"Voltage lags current in a capacitor"

This means that if you start with an uncharged capacitor and supply electrons (charge) to one of the plates from some attached circuitry the voltage between the plates will start from zero and increase as the electrons are added, after a short time delay.

The first electron added takes no work at all, but creates a small negative charge on the plate.

The second electron is harder to add since we have to work against the mutual repulsion of like charges. This means we have to do work (add energy) to place the second electron.
This energy appears as electrical potential energy stored in the capacitor and manifests itself as a voltage between the plates.
As we add each successive electron it becomes harder and harder as we need to do more work against the increasing repulsion.

Now if the negative charge on one plate is increasing because we are adding electrons, then this negative charge is repelling electrons in the other plate.
That is it is pushing them away somewhere.
So the other plate becomes positive, due to a deficiency of electrons.
We say that there is an induced positive charge on the other plate.
As this positive charge builds up it attracts the electrons back making each electron successively harder to push away.
This goes on until the push from the negative exactly balances the pull from the positive plate.

At this point the capacitor is charged and no further changes take place.

We can add them mechanically you should read the thread in the Physics section on the Leyden Jar.
We can add them by electrically connecting something more negative than the capacitor terminal eg a battery. Then charge will flow from the more negative source to the capacitor plate.

Last edited: May 17, 2014
5. ### activee Thread Starter Member

Jan 16, 2014
39
0
Thank you so much !!! You should be a teacher.

I understood everything you said but I still got a question.

The energy comes from the voltage of the battery ? Thus the more voltage a battery has the more we could load a capacitor ? Furthermore does that mean that a circuit with a battery of 1 Volt in which there is a capacitor with a capacitance of 1 Farad will store 1 Cb of charge (both plates considered I presume)?

really thank you !

Last edited: May 17, 2014
6. ### toffee_pie Active Member

Oct 31, 2009
165
7
another thing,

I had a circuit recently and was debugging, now it had power going into it, (<4volts). I used the buzzer on the meter looking for shorts and a few capacitors buzzed out as short (zero ohms too with the resistance measure).

How come only a few of them did this? i know doing these readings on a live board is not good.

I guess you get inaccurate readings as long as the circuit is live.

7. ### wayneh Expert

Sep 9, 2010
13,426
4,264
You've almost answered your own question. The other part of the answer is that is would be unlikely for ALL the capacitors in a circuit to have failed, unless the device is ancient.

8. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
519
Yes, a capacitance of 1 Farad (1F) will have 1 Volt (1V) across it when charged with 1 Coulomb (1C) of charge. Note the standard symbols for each quantity.

In equations

Q = VC

The energy may be chemical energy from a battery, or it may be mechancial energy from a static generator as in the linked video.

Note that capacitors are different from batteries. Batteries have the (chemical) means to replenish the charge or energy taken out (up to the point of exhaustion) whilst capacitors have no such means. They are limited to the charge (energy) put in. Rechargable batteries are somewhere between the two.