# Understanding nature of current in LC circuit

Alright, I have an LC circuit.

Using conservation of energy, I get $\frac{dI}{dt}= \frac{-q}{CL}$ which gives $\frac{d^2q}{dt^2}= \frac{-q}{CL}$, therefore, $q=q_o sin(\omega t +\phi)$, where $\omega= \frac{1}{\sqrt {LC}}$, giving $q=q_o cos(\omega t)$ which implies $I= -q_o \omega sin(\omega t)$.

Now the equation clearly tells that there is an alternating current when the capacitor is given some initial charge. But I am finding it difficult to physically imagine whatever's going on. I think it is because of some fundamental flaw in my understanding of how capacitors work. Can someone explain me in small steps what exactly happens in one cycle of current flowing from one end to the other? I am being told that as soon as the charge on one plate is reduced by suppose dq, the charge on the other is also reduced by the same amount (in magnitude). But I really don't understand how exactly.

Thank You

## Answers and Replies

*bump*

First don't look at capacitor as 2 plates, but look at it as one component, that can store electrostatic energy. The analogy and whats happening is very simple.

Essentially what is happening is an oscillation, transfer of energy between capacitor(electrostatic energy) and inductor(electromagnetic energy). You can write a differential equation for that, but I won't bug u with that because I see you are in high school.

Best analogy is a spring, loaded with some mass m, oscillating on the frictionless surface.

[PLAIN]http://pokit.org/get/bce67f9a1ac811f658633291f8ef8809.jpg [Broken]

Figure (a)

Capacitor is fully charged. And there exist only voltage v across the capacitor. Current is 0.

Figure (b)

When switch S is closed, current rushes from the positive side of the capacitor. But nature in inductor is that it doesn't like big changes in current. As current slowly rises to Imax, inductor stores electromagnetic energy.

Figure (c)

Current starts to drop from Imax back to 0, by that charging the capacitor again, but now with opposite polarity.(voltage would be -v, with respect to the original voltage across the capacitor) Electrostatic energy is stored.

Figure (d)

Again voltage slowly drops to 0 across the capacitor, slowly building up the current again to I max and storing electromagnetic energy.

Figure (e)

Full period. Again current slowly dropping down to 0, and voltage across the capacitor builds up to v ergo charging it to its original state.

I think you can figure out from the picture, the analogy of the spring.

A nice thing to say is that the solution for the differential equation of this oscillation is a sinusoid. Both for voltage and current. And this, in Theory will oscillate into infinity.

Real case is a RLC circuit, all in series which leaves you with damped oscillation.

R comes from internal resistance of the wire, capacitor and inductor.( Which can be equivalented as series resistor with inductor capacitor)

[PLAIN]http://beltoforion.de/pendulum_revisited/Damped_oscillation_graph2.png [Broken]

Hope I helped

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Thanks for the replies. I understood the process that you people explained. What I don't understand is why the charge on one plate of the capacitor is reduced automatically if the the charge on the other is reduced as well. I know that I need to consider the capacitor as a single unit, but I still need to understand how something happens the way it does.

Thanks for the replies. I understood the process that you people explained. What I don't understand is why the charge on one plate of the capacitor is reduced automatically if the the charge on the other is reduced as well. I know that I need to consider the capacitor as a single unit, but I still need to understand how something happens the way it does.

Why that happens, and why charge is uniformly released has to do with property of the dielectric. Its much too complicated to be explained in one thread but loosely saying:

When your capacitor is fully charged, Its charge is being held in place by the dielectric, assuming you have one in your capacitor(not counting air as dielectric, although it is). To be more precise, charge on one plate is held by the dielectric and the charge on the other plate. You may say that the charges on the plates are coupled(through electric field).

[PLAIN]http://climatetechwiki.org/sites/default/files/images/extra/430px-Capacitor_schematic_with_dielectric.svg_.png [Broken]

When capacitor is charged, dielectric is polarized, meaning it has its own Electric field, that is opposing the original field applied across the dielectric.

When the capacitor is short circuited, electrons from the negatively charged plate, want to go to the positive plate. This results, in current, by convention that goes from + to -.

If you reduce the charge on one plate, electric field weakens. And when electric field weakens, polarization weakens too, ergo dielectric can hold less charge on the other plate.

(Negative charge in the dielectric, holds positive charge of the plate next to it, in place.)

Basically the electrons get back to positive plate, returning the neutrality, meaning when 1 electron gets back to positive plate, you get 1C less charge in the capacitor. So one may say that the charge is uniformly distributed and released.

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I hope I didn't confuse you, I don't have my textbooks here, this was clearly out my intuition and head.

Thanks for the replies. I understood the process that you people explained. What I don't understand is why the charge on one plate of the capacitor is reduced automatically if the the charge on the other is reduced as well. I know that I need to consider the capacitor as a single unit, but I still need to understand how something happens the way it does.

But you must know that one pole of a magnet cannot be stronger or weaker than the other pole. As the magnet is unseparable so is capacitor. In LC circuit the electric field of the capacitor is turning into the magnetic field of the iductance and vice versa.

If you reduce the charge on one plate, electric field weakens. And when electric field weakens, polarization weakens too, ergo dielectric can hold less charge on the other plate.

I don't think I clearly understand. The charge on one plate doesn't depend only on the dielectric does it? In fact why should it depend on anything at all? By conservation of charge, nothing should happen to the amount of charge on the negative plate for example plate even if a little charge from the positive plate enters the wires in form of current.

(Negative charge in the dielectric, holds positive charge of the plate next to it, in place.)
Does that mean a capacitor cannot store charge in place if there is vacuum between the plates?

Basically the electrons get back to positive plate, returning the neutrality, meaning when 1 electron gets back to positive plate, you get 1C less charge in the capacitor. So one may say that the charge is uniformly distributed and released.
Exactly that is what I imagined as well. But the decrease in charge on the positive plate for example isn't because of a few electrons filling the space but because of decrease in charge on the negative plate.

I don't think I clearly understand. The charge on one plate doesn't depend only on the dielectric does it? In fact why should it depend on anything at all? By conservation of charge, nothing should happen to the amount of charge on the negative plate for example plate even if a little charge from the positive plate enters the wires in form of current.

Does that mean a capacitor cannot store charge in place if there is vacuum between the plates?

Exactly that is what I imagined as well. But the decrease in charge on the positive plate for example isn't because of a few electrons filling the space but because of decrease in charge on the negative plate.

If you are talking about vacuum, positive charge on the plate is being held by the negative charge on the other plate(what else would hold them?). I was giving you a general example, where mostly all capacitors have dielectric. And in fact it can be shown that it can store LESS energy if no dielectric exists between the plates.

And why shouldn't nothing happen if you reduce charge on one plate?

Electric field, holds the charges in place. Positive and negative plate form that field. But really you are going very deep here.

I think the problem here is the concept of charge. If you have negative charge on the negative plate, and it leaves it, and goes to positive plate, they cancel out. That is the main idea of the capacitor in the first place.

Negative charge "wants" to go on the positive plate. Attraction exists, if they would arrive there, no charge would exist, as they would cancel out, positive+negative.

And why shouldn't nothing happen if you reduce charge on one plate?

Electric field, holds the charges in place. Positive and negative plate form that field. But really you are going very deep here.

Lets suppose you have two plates A and B having charges +q and -q respectively just like in a capacitor. If you suppose remove x amount of charge from A, the charge on B would still be -q wouldn't it; since it doesn't have anywhere to go.

I think the problem here is the concept of charge. If you have negative charge on the negative plate, and it leaves it, and goes to positive plate, they cancel out. That is the main idea of the capacitor in the first place.

Negative charge "wants" to go on the positive plate. Attraction exists, if they would arrive there, no charge would exist, as they would cancel out, positive+negative.

I know, that is what I thought too. But it seems that the charges don't cancel each other out. I am told that when current in the wire is Imax, and charge on the capacitor is 0, all the charge that left the plates is in the wire.

Had it been the simple case of charge on one plate cancelling each other out, why would the capacitor start charging again after coming down to 0. Current only develops because of a potential difference. But that won't exist when there is 0 charge.

You got me confused here too a bit. Studiot will probably come by and clear this misunderstanding we both have. In the mean time I will do some more research to try clarify this myself.

@mishrashubham

The capacitor starts charging again because once the switch in a circuit is closed, a potential difference across the capacitor exists and hence that allows charge to build up on the parallel plates once again. For example, you can treat the capacitor element as an open circuit.

You got me confused here too a bit. Studiot will probably come by and clear this misunderstanding we both have. In the mean time I will do some more research to try clarify this myself.
Hope he does.

@mishrashubham

The capacitor starts charging again because once the switch in a circuit is closed, a potential difference across the capacitor exists...

Could you explain how a potential difference exists without any charge difference?

Lets suppose you have two plates A and B having charges +q and -q respectively just like in a capacitor. If you suppose remove x amount of charge from A, the charge on B would still be -q wouldn't it; since it doesn't have anywhere to go.

You are correct... The charge wouldn't just fall off the plate. It would have to stay there in this scenario that you have laid out here.

However, this situation is unlike your original question, which involved a complete circuit. There is a place for it to go. With the circuit, if one electron leaves plate A then that means that one electron also arrived at plate B.

Could you explain how a potential difference exists without any charge difference?

Why are you saying there is no charge difference? The potential was still there, it just had no place to go with the switch open.

You are correct... The charge wouldn't just fall off the plate. It would have to stay there in this scenario that you have laid out here.

However, this situation is unlike your original question, which involved a complete circuit. There is a place for it to go. With the circuit, if one electron leaves plate A then that means that one electron also arrived at plate B.

I know, that is what I thought too. But it seems that the charges don't cancel each other out. I am told that when current in the wire is Imax, and charge on the capacitor is 0, all the charge that left the plates is in the wire.

Had it been the simple case of charge on one plate cancelling each other out, why would the capacitor start charging again after coming down to 0. Current only develops because of a potential difference. But that won't exist when there is 0 charge.

Why are you saying there is no charge difference? The potential was still there, it just had no place to go with the switch open.

Here I am talking about the instant when charge on the capacitor becomes 0 and current becomes maximum, after which the capacitor starts charging again but with opposite polarity. Now in that instant since charge is 0 potential difference between the ends should also be 0 (please correct me if I am wrong).

Check out http://www.allaboutcircuits.com/vol_2/chpt_6/1.html" [Broken]... It gives very detailed descriptions of what is happening in an LC circuit. I think this is the answer you're searching for:
The inductor will maintain current flow even with no voltage applied. In fact, it will generate a voltage (like a battery) in order to keep current in the same direction. The capacitor, being the recipient of this current, will begin to accumulate a charge in the opposite polarity as before

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Thanks for the input people. I still don't get it, but I think I will have to do some re-analyzing. I'll be back in a day or two after I try to figure it out myself.

jim hardy
Gold Member
Dearly Missed
i think you fellows have touched on a real basic point here, namely absolute potential..

Mechanical engineers have an advantage over us electricals because their "absolute zero " pressure is something tangible and perceptible every day; you feel an approach to it with an inordinately long soda straw in your soft drink. Hook two straws together and take a swig....

For us unfortunate electricals, "absolute potential" might only be mentioned briefly in the introductory course - i vaguely remember mention of some thiings called "statvolts" and "statcoulombs".

Your local "Absolute potential" would be the work required to bring a unit of charge from infinity to where ever you are. That's impractical to say the least...and something your senses won't feel in an ordinary day.
So we work instead in terms of "potential difference" between two points, at least one of the points usually nearby.
Both of them may well be at substantial "absolute potential" but we only work with the local differences of that absolute potential, pick one of them as our reference and call it zero.
But unlike those lucky mechanicals our reference is not absolute zero.

So your two capacitor plates might both be carrying an excess or a shortfall of charge
what determines the field between them is the difference between their respective excess/shortfalls.
If you connect them with a wire, charge will move from one to the other so that they come to same absolute potential. That's because of simple electrostatic repulsion and attraction, q1*q2/r^2.

maybe i just muddied things further, sorry if so.
but that's an important little detail IMHO.
remember - electric circuits like thermodynamics is usually studied in context of closed systems.

mental exercise - ask yourself what is potential difference between 'earth ground' and Alpha Centauri.
....or even the moon.

old jim

Thanks for the replies. I understood the process that you people explained. What I don't understand is why the charge on one plate of the capacitor is reduced automatically if the the charge on the other is reduced as well. I know that I need to consider the capacitor as a single unit, but I still need to understand how something happens the way it does.

Such metaphor comes to my mind: a plate is being pressed by many point force from both sides alternatively. As soon as there appears a hole on one side of the plate there appears a bulge on the opposite side. Then application of the force changes direction and pressing the holes/bulges changes direction as well. And a bulge on one side is unimaginable without a hole on the other side, and vice versa (so long as quantum mechanics is not concerned at least).