# LC Circuits Qmax

1. Jul 27, 2010

Hey everyone,

So I'm working with this simple LC circuit where the capacitor has been charged to some Qmax. I want to derive the relationship for Q(t) (aside from using the book). But I'm stuck.

The book's derivation uses kirchoff's law of voltages around a loop. But it seems to me like you can't use that law in this case because the closed loop integral of E dot dl is not zero around the loop. Rather, it now equals the change in magnetic flux in time.

Is this a true statement?

If so, then I now have the closed loop integral of E dot dl = - d/dt(magnetic flux).

And now I want to evaluate the integral on the left side. And the only place where there is an electric field is in the capacitor. But I don't know how to proceed from there to getting the standard second order diff eq. that allows us to solve for Q(t) as a S.H.O. solution.

If I have my circuit set up with the polarities as shown in the attached image, then it seems like I get - Q/C = - LdI/dt. And this will ultimately give me wrong diff eq. I get the -Q/C because the electric field is pointing in the opposite direction of the path I've chosen.

Sorry for the long post. If anyone understood this and can offer some help, that would be greatly appreciated. Thanks.

2. Jul 27, 2010

I didn't post the image!

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3. Jul 27, 2010

### hikaru1221

Good point!

Can you go a bit into details on this? Maybe reading your work may help me understand why you arrive at that equation.

4. Jul 27, 2010

Hey thanks a lot for the response!

So, using the attached image as a reference, there is an electric field that points from the positive terminal to the negative terminal of the capacitor. And this electric field is E = sigma/epsilon.

And i'm going to choose my path for this closed loop integral to be clockwise (which is the direction we assume positive charge moves). So I move this positive charge from the negative terminal (lower potential) to the positive terminal (higher potential) and so now I have my E-field pointing in the opposite direction to my chosen path, dl. This will bring about a negative sign. And if I say the distance between the capacitor plates is d, then E dot dl becomes -sigma/epsilon * d, but this simplifies to Q/C (since sigma is Q/A). So basically I have -Q/C and this is just the voltage across the capacitor and this equals -L*dI/dt.

So the negatives from both sides cancel out and I am left with Q/C = L*dI/dt, which is off by a negative sign for the diff eq. that gives us the proper solution of S.H.O. motion.

I tried to be clear! I hope my train of though made sense. Let me know anywhere where I'm wrong in my logic! Thanks again.

5. Jul 27, 2010

### hikaru1221

First, I have to say that you shouldn't write the signs (+) and (-) next to the plates, as you know, the sign of each plate changes periodically, especially when you found a subtle point about using Kirchhoff's rule. When you want to do it right, you should do it right from the start to the end. Moreover, when you mention the invalidity of using Kirchhoff's rule, you should forget the term "potential difference" or "electric potential". I'm not sure if electric potential is defined in this case, but leaving it out won't do any harm.
Just my 2 cents

The integral will be: $$\oint\vec{E}\vec{dl}=-Ldi/dt$$ if you set the current i moving in the same direction of the path chosen for computing the integral. You are still right at this point.

Then you will have: $$\oint\vec{E}\vec{dl}=-Q/C$$ only if you choose Q as the charge of the upper plate. The sign would change (i.e. $$\oint\vec{E}\vec{dl}=Q/C$$) if you chose Q as the charge of the lower plate. Note that the direction to take the integral is still clockwise. If it were anti-clockwise, then Q in $$\oint\vec{E}\vec{dl}=-Q/C$$ would be the charge of the lower plate.

Until now, you have: $$-Q/C = -Ldi/dt$$. Since Q is the charge of the upper plate, and the direction of current i is clockwise, i.e. the current i is "going out of" the upper plate or Q, we have: $$i=-dQ/dt$$. This equation should make sense, since the current i is going out of the plate Q, i.e. the charges are going out of the plate Q, then Q should decrease, which corresponds to the minus sign in the equation.

Now it goes all the way back to the good old solution, right?

6. Jul 28, 2010

### K^2

When you apply Kirchoff laws to an LC circuit, you shouldn't be trying to apply it to inductor wire. You consider an inductor as a simple element in the circuit, whose voltage drop is known.

Kirchoff current law applied to capacitor tells you this.
$$\dot{q_C} = I_C$$

Voltage across the two elements is easy.

$$V_C=\frac{q_C}{C}$$
$$V_I=L\frac{dI_L}{dt}$$

Kirchoff current law applied to the whole circuit.
$$I_L=I_C=\dot{q_C}$$

And the voltage law.
$$V_C+V_L = 0$$

Or if you expand it in terms of qC.

$$\frac{q_C}{C}+L\ddot{q_C}=0$$

And that's a differential equation with a simple solution.

$$q_C=Acos(\omega t) + Bsin(\omega t)$$

Where

$$\omega=\sqrt{\frac{1}{LC}}$$

There is no need to invent anything more complicated. You could in principle integrate over the field inside the inductor and still get the same result, but you really shouldn't be using Kirchoff's laws at that point anymore. These were invented specifically so that you don't have to do any of this.

7. Jul 28, 2010