# Charging of a cpacitor

1. Jul 22, 2007

### pardesi

given an arbitrary circuit when can u say that a given capacitor of course in the circuit has been fully charged

2. Jul 22, 2007

### ranger

Depends on what you mean by arbitrary.

In a simple DC RC circuit, the capacitor will be approx. 99% charged at 5 time constants. Where the time constant in the product of the total resistance of the RC circuit and the capacitance.

Did you have something specific in mind?

Last edited: Jul 22, 2007
3. Jul 22, 2007

### bernhard.rothenstein

charging a capacitor

In a DC circuit it will be charged if no current flows in the circuit.
In an AC circuit the energy stored in the capacitor is a function of time.
Are there more questions?

4. Jul 23, 2007

### pardesi

i want something more specific for a DC circuit as to what decides what the cahrge on the capacitor shoud be. i would like an answer in terms of field concept rather than potential

5. Jul 23, 2007

### bernhard.rothenstein

charging the capacitor

Consider that your capacitor is a plane one. Its capacity is C=$$\epsilon$$(0)S/d ,S representing the surface of its pates and d the distance between them. If the potential difference between the plates is V, the electric field between the plates is E=V/d. The energy stored in the electric field is
W=CV^2/2 and so you can express it as a function of the electric field.
The electric charge on one of the plates is q=CV=CE/d. I think you have now all the elements to answer all your questions.
If$$\Phi$$ represents the electromotive force of the source in the direct cc circuit you will find out that the charging process has ended when the electric field between the plates is E=$$\Phi$$/d.
I hope I have guessed the correct names of the used physical quantities.
If you have more questions...

6. Jul 23, 2007

### pardesi

yes thsi was what i thought (with a slight variation) ...until thsi accorsing to this the charge on a capacitor directly connected to battery of given emf is independent of how it is connected to other circuit elemnts.

but this of course is false.
just consider thsi example u have two capacitances of C connected in series the charge on each of them is $$\frac{CV}{2}$$ but when they are connected individually the charges are $$CV$$

7. Jul 23, 2007

### heafnerj

The capacitor will be "charged" when the *net* electric field everywhere inside the circuit is zero.

8. Jul 23, 2007

### pardesi

i don't think so.even when the charging is taking place it passes through a series of 'equilibrium states'
otherwise consider this case of a simple 1 capacitor being charged which is say connected to one resistor which may be the net resistance of wire itself
any standard book would give a proof like this

$$\frac{q}{C}+\frac{dq}{dt}R - E=0$$
well this done by assuming the potential at the positive terminal of battery is same as that of that in the positive terminal of capacitor which clearly holds only if there is no field 'inside' the wire.

9. Jul 23, 2007

### heafnerj

But they're not actually equilibrium states. It is actually a gradual progression through *steady states* not *equilibrium states*. The net electric field inside a conductor in which the charge carriers are all in static equilibrium is zero and thus there would be no current. If a current exists, the net electric field cannot possibly be zero and a *steady state* exists. In a *steady state*, a uniform current exists and that requires a uniform nonzero electric field. As the capacitor "charges" (and that is really not a good term for describing the process because all that is happening is existing charged particles in the conductor are redistributing themselves) the net electric field inside the circuit decreases in magnitude and thus the current decreases too. Eventually, the net field will be zero and so will the current.

No differential equation is required to understand this. All that is needed is understanding electric fields and how particles respond to them.

10. Jul 23, 2007

### pardesi

if thsi is so then how does the differential equation i wrote above(which is there i suppose in all the books) holds true

11. Jul 23, 2007

### heafnerj

The DE is merely a mathematical description of the physical system in terms of charge, resistance, potential difference, capacitance, and time. It is a description that gives answers against which we can compare measurements. The entire problem can be explained with nothing more than understanding the behavior of charged particles in the presence of an electric field. Similarly, resistance and capacitance are merely macroscopic terms that are redundant in that they include more fundamental microscopic paramaters that explain the behavior.

Consult Matter & Interactions by Chabay and Sherwood if you want a more complete discussion. This is an innovative calculus-based introductory physics textbook that emphasizes micro/macro connections and completely explains all of DC circuit behavior with nothing more than the behavior of charged particles in the presence of an electric field. I've been teaching from this text since 1999.

12. Jul 24, 2007

### pardesi

but without the net fiedl inside the circuit being 0 the above DE is meaningless
also can u post an elink to the book if any

13. Jul 24, 2007

### pardesi

i even don't get why opposite plates get equal charges .most books write that otherwise field will exist in the conducting wires.but this is so if the charge was uniformly distributed but here don't know(probably) how the charge distribute then how can we have such an argument

14. Jul 24, 2007

### heafnerj

The book is published by Wiley and it not available in a digital version. Try an interlibrary loan.

I don't understand your comment about the DE. If a current exists, E cannot be zero.

15. Jul 24, 2007

### pardesi

that's what i am telling if there is a field then the that diffrernetial equation doesn't hold good because it is true by assuming there is no field inside the conducting wires

16. Jul 24, 2007

### heafnerj

That DE is really nothing more than conservation of energy (per charge). It is true as written as far as I can tell. I don't understand why you say that the DE's validity has to do with E being zero inside the wire.

17. Jul 24, 2007

### heafnerj

Okay I see the error in your reasoning (finally). You're confusing the electric field between the plates with the electric field inside the wire, and they are NOT the same! The electric field between the plates can be non-zero while the *net* electric field *inside* the conducting wire is zero. It is the *net* field inside the wire that governs the "charging" (and we really need a better term for that) process. Does this make sense?

Another possible source of confusion is using E to stand for electric field and E to stand for potential difference. In your DE, E is a potential difference and not an electric field.

18. Jul 24, 2007

### pardesi

well can u derive the equation without assuming field in wire is not 0

so upto what does the capacitor gets charged

19. Jul 24, 2007

### heafnerj

Yes. It's nothing more then energy conservation.

20. Jul 24, 2007

### pardesi

well while deriving even by energy conservation i think u have to assume that the potential on one plate of capacitor is same as that of battery....which doesn't hold true when there is an field inside the wire

but anyway my problem is that if why does a capacitor in any given circuit can get charged say only upto $$Q$$ not more or less.what governs the charge distribution. when does it stop charging....
can u please explain them in field concept

21. Jul 24, 2007

### heafnerj

I think you're missing the point, unless I misunderstand what you are saying.

I already did. :-) The "charging" and "discharging" of a capacitor are governed by the *net* electric field inside the wire.

Look here

http://galaxy.cofc.edu/circuits.html

and here

http://galaxy.cofc.edu/rcircuits.html

for simulations that explain what I'm talking about. Chabay and Sherwood is the only text I know of that approaches circuits this way. The traditional way is unnecessarily complicated.

By the way, your textbook probably also says that you should neglect the capacitor's fringe field. If so, your textbook is completely wrong because without a fringe field, a capacitor can neither charge nor discharge. Cheers!

22. Jul 24, 2007

### pardesi

exactly that was what i thought if there were nio field due to capacitor why should charging due to battery ever stop ...
well going like that what i got was given a capacitor when it was connected to a circuit then the field due to it hence charge on it is fixed ...like someone(sorry) did it on the first page of this thread.
this is of course wrong

sir can u tell me is the charge distribution on a capacitor uniform or un-uniform
also why do opposte plates get oppositely charged

23. Jul 24, 2007

### heafnerj

Part of the issue here is that there is more than one *field due to the capacitor.* There's the field inside the plates. There's the field at every point in the wires caused by the charged particles on the plates. There's the field between the two plates. There's the fringe field, the field just outside the plates in the wire.

I can summarize the processes for "charging" using only fields, but I'm afraid this approach will not make much sense to you unless you read Chabay and Sherwood. I'm reaching the limit of how efficiently I can convey this online like this.

Charging:
You have a capacitor, a light bulb, and a battery all in series. The capacitor is uncharged, meaning that the *net* charge on each plate is zero. The *net* electric field inside each plate is zero. The plates of the battery harbor charged particles, effectively making the battery a dipole which creates an electric field *everywhere* in the circuit. This electric field due to the charged particles on the battery cause mobile electrons in the circuit to change their distribution on the surface of the circuit. This changing surface charge distribution establishes a new and different electric field. The *net* field, the combination of the field from the charged particles on the battery and the field from the surface charge distribution, establishes a current. Now, the electric field due to the charged particles on the battery is basically constant in time. But, the electric field due to the changing surface charge distribution does change, so the *net* field changes. The change is such that the *net* field approaches zero. Therefore, the current in the circuit will decreass with time because the electric field driving the mobile charges decreases with time.

Now, the final charge distribution on the capacitor plates will be that particular distribution that makes the final net electric field everywhere in the circuit zero! That distribution happens to make the (identical) plates end up with equal magnitude, but opposite polarity, charges.

I know this probably sounds outlandish, but I assure you it's a correct description. Your assignment is to describe the "discharging" process using only the field concept. Look at the links I posted earlier for some hints. Those web pages also contain references that you might find helpful.

24. Jul 25, 2007

### pardesi

right tahnks i will try to get hold of chabay and sherwood.
that was good as most books describe in terms of potential concept which normally is not that obvious