- #1
eprparadox
- 138
- 2
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 kirchhoffs 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.
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 kirchhoffs 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.