Inductance and Charge Redistribution

  • Thread starter Ayesha02
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  • #26
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various thoughts crossing my mind-
1. I have four unknowns ( ##Q_0## ,##Q_A## ,##Q_B## and i ) and only 3 equations
2. Even if manage to solve the equations and find ##Q_A##(t)/ whatever, how will i get the time in which charge flow occurs

could you give me a birds eye view of the sum once again and then let's proceed with equation solving:)
At this point, ignore ##Q_0##. I think that, up until now, we are clear on$$\frac{kQ_A(t)}{r} - \frac{k(Q_0-Q_A(t))}{r} = L\frac{di}{dt}$$
We need to find the appropriate expression for ##i(t)##; either ##Q'_B(t)##, or ##Q'_A(t)##.
Before losing half of its initial charge, it should be clear that ##Q_A(t)\geq Q_B(t)##, since ##Q_B(t)## starts from being zero coulombs and grows by taking the charge lost from A's.
Thus
$$\frac{kQ_A(t)}{r} - \frac{k(Q_0-Q_A(t))}{r}\geq0$$
as long as the charge in A is greater than half of its initial value.
Now, we have$$\frac{kQ_A(t)}{r} - \frac{k(Q_0-Q_A(t))}{r} = L\frac{di}{dt}$$
or, in other terms, ##\alpha=\beta##. If ##\alpha>0##, then what can you infer about ##\beta##?
 
  • #27
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If ββ is the derivative of some function, then what does its sign tell you about that function?
 
  • #28
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If ##\beta## is the derivative of some function, then what does its sign tell you about that function?
Oh so I know that ##i## is an increasing function isn't it?

what next then?
 
  • #29
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Ok, we know that ##i(t)## is growing, and since ##i(t)=Q'(t)##, then ##Q'(t)## should also be growing, or, in mathematics, ##Q''(t)\geq0##.
I haven't foreseen this. We definitely know that ##Q'_A\leq0## and that ##Q'_B\geq0##, but I don't see how we can say anything about their second derivatives without knowing the actual functions.
Since it seems that we have no idea, I suggest we choose ##Q=cQ_B(t)=c(Q_0-Q_A(t))##, where ##c## is either ##1## or ##-1##, and is to be determined later.
This should work as ##cQ''_B=-cQ''_A=ci'(t)## (substitute with the expression you have found for ##Q_B## to check).
 
  • #30
rude man
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Ohh ya true!
Actually, for a given coil of N turns, each turn has mutual inductance with every other turn, which is why coil inductance basically goes as the square of the number of turns. So you weren't completely off base after all! :smile:
 
  • #31
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Ok, we know that ##i(t)## is growing, and since ##i(t)=Q'(t)##, then ##Q'(t)## should also be growing, or, in mathematics, ##Q''(t)\geq0##.
I haven't foreseen this. We definitely know that ##Q'_A\leq0## and that ##Q'_B\geq0##, but I don't see how we can say anything about their second derivatives without knowing the actual functions.
Since it seems that we have no idea, I suggest we choose ##Q=cQ_B(t)=c(Q_0-Q_A(t))##, where ##c## is either ##1## or ##-1##, and is to be determined later.
This should work as ##cQ''_B=-cQ''_A=ci'(t)## (substitute with the expression you have found for ##Q_B## to check).
oh man!
why do I need second derivative here now:(
 
  • #32
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oh man!
why do I need second derivative here now:(
could you do me a favor and send a pic of your working for this sum-- I'm going terribly wrong somewhere but i cant pinpoint it:(
 
  • #33
rude man
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So
I'm not sure I'm quite getting how to eliminate one of ##Q_A## or ##Q_B## there buddy
well, where does the care from q1 go to? Is any charge lost between the two?
="Ayesha02, post: 6336768, member: 677354"]
See
various thoughts crossing my mind-
1. I have four unknowns ( ##Q_0## ,##Q_A## ,##Q_B## and i ) and only 3 equations
you have 3 unknowns but can easily eliminate one of them, leaving you with 2 unknowns but related by a derivative..
 
  • #34
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Instead,
Can't we consider a loop at infinity and just write the loop law simply?
 
  • #35
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What an awful problem!
It's too tempting to neglect the mutual capacitance between A and B and consider only the self capacitance. However (https://en.wikipedia.org/wiki/Capacitance) points out the mutual capacitance equals the self capacitance. We end up with to capacitors "connected to infinity" and another between points A and B.
dude can u please elaborate on this?
 
  • #36
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So
My textbook's solution says that :

angular frequency of the LC oscillation is w=sqrt( 2*pi*##E_0##*R*L)
And that required time is one fourth of time period

Can you explain this please @rude man @archaic
 
  • #37
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Instead of my other suggestion, you should take ##Q(t)=cQ_A(t)##.
$$\frac{kQ_A(t)}{r} - \frac{k(Q_0-Q_A(t))}{r} = L\frac{di}{dt}=L\frac{d^2}{dt^2}\left(cQ_A(t)\right)\\
\implies\frac{k}{r}\left(2Q_A(t)-Q_0\right)=cL\frac{d^2Q_A(t)}{dt^2}$$
Take ##f(t)=2Q_A(t)-Q_0##, this gives you ##f''(t)=2Q_A''(t)##, thus
$$\frac{k}{r}f(t)=\frac{cL}{2}\frac{d^2f(t)}{dt^2}$$
 
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  • #38
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So
My textbook's solution says that :

angular frequency of the LC oscillation is w=sqrt( 2*pi*##E_0##*R*L)
And that required time is one fourth of time period

Can you explain this please @rude man @archaic
Can you please explain this @archaic
 
  • #39
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Can you please explain this @archaic
You have$$\frac{k}{r}f(t)=\frac{cL}{2}\frac{d^2f(t)}{dt^2}$$where ##f(t)=2Q_A(t)-Q_0##. You want to find the time at which ##Q_A(t)=\frac{Q_0}{2}##.
Start by solving the differential equation, and remember what I said about ##c##!
 
  • #40
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You have$$\frac{k}{r}f(t)=\frac{cL}{2}\frac{d^2f(t)}{dt^2}$$where ##f(t)=2Q_A(t)-Q_0##. You want to find the time at which ##Q_A(t)=\frac{Q_0}{2}##.
Start by solving the differential equation, and remember what I said about ##c##!
Hint: what are ##Q_A(0)## and ##i(0)##?
 
  • #41
Delta2
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I am not sure I agree with all the posts here, but I believe the final equation first presented at post #37 should have a minus sign in front. coming from the fact that the voltage in an inductor is ##V=-L\frac{dI}{dt}##.

Also because the "circuit" is not so local the current will vary spatially across the two ends of the inductor, so i believe we should explicitly state as a vital assumption that the current does not vary spatially but only temporally.
 
  • #42
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I am not sure I agree with all the posts here, but I believe the final equation first presented at post #37 should have a minus sign in front. coming from the fact that the voltage in an inductor is ##V=-L\frac{dI}{dt}##.

Also because the "circuit" is not so local the current will vary spatially across the two ends of the inductor, so i believe we should explicitly state as a vital assumption that the current does not vary spatially but only temporally.
I have multiplied by ##c## for lack of knowledge abour the sign, and yes, it turned out to be ##-1##!
 
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  • #43
etotheipi
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I am not sure I agree with all the posts here, but I believe the final equation first presented at post #37 should have a minus sign in front. coming from the fact that the voltage in an inductor is ##V=-L\frac{dI}{dt}##.
I think the ##\frac{kQ_A(t)}{r} - \frac{k(Q_0-Q_A(t))}{r} = L\frac{di}{dt}## is okay because that's just passive sign convention, OP just needs to be careful when substituting in ##i = -\frac{dQ_A}{dt} = \frac{dQ_B}{dt}## since you should get a negative sign out the front. From then on, it's just a mass on the end of a spring problem :wink:.
 
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  • #44
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I think the ##\frac{kQ_A(t)}{r} - \frac{k(Q_0-Q_A(t))}{r} = L\frac{di}{dt}## is okay because that's just passive sign convention, OP just needs to be careful when substituting in ##i = -\frac{dQ_A}{dt} = \frac{dQ_B}{dt}## since you should get a negative sign out the front. From then on, it's just a mass on the end of a spring problem :wink:.
I have multiplied by ##c## for lack of knowledge abour the sign, and yes, it turned out to be ##-1##!
Oh, a sign convention? So we don't care at all about the actual sign by analyzing the charge functions?
 
  • #45
etotheipi
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You can work it all out logically, but it's often quicker to do anything circuit related with the so-called passive sign convention
1588662353095.png

Essentially, if the reference direction for the voltage is in the opposite direction to the reference direction for current, then the formulae for passive components hold without negative signs, i.e. ##v=ir##, ##i = c\frac{dv}{dt}##, ##v = l\frac{di}{dt}##. A consequence is that passive components have positive power. If you define the reference directions the other way around (the so-called active sign convention, which I hear isn't usually done in electrical engineering), then all of those defining equations have negative signs in them.

It just means that here we can define the reference voltage direction as from B to A (i.e. V(A) - V(B)) and the reference current direction from A to B, and not worry about getting the signs wrong.
 
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  • #46
Delta2
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I think the ##\frac{kQ_A(t)}{r} - \frac{k(Q_0-Q_A(t))}{r} = L\frac{di}{dt}## is okay because that's just passive sign convention, OP just needs to be careful when substituting in ##i = -\frac{dQ_A}{dt} = \frac{dQ_B}{dt}## since you should get a negative sign out the front. From then on, it's just a mass on the end of a spring problem :wink:.
Thanks for explaining the sign convention used in circuits at post #45 but one more thing I want to ask, why do you take current to be ##I=-\frac{dQ_A}{dt}=\frac{dQ_B}{dt} ## and not ##I=\frac{dQ_A}{dt}=-\frac{dQ_B}{dt}##
 
  • #47
etotheipi
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Thanks for explaining the sign convention used in circuits at post #45 but one more thing I want to ask, why do you take current to be ##I=-\frac{dQ_A}{dt}=\frac{dQ_B}{dt} ## and not ##I=\frac{dQ_A}{dt}=-\frac{dQ_B}{dt}##
I defined the reference direction of ##i## to be in the direction from A to B, so positive current results in B gaining charge and A losing charge.

Sorry for not making that clear; I have the feeling that we defined everything in exactly the exact opposite way!
 
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  • #48
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I think the ##\frac{kQ_A(t)}{r} - \frac{k(Q_0-Q_A(t))}{r} = L\frac{di}{dt}## is okay because that's just passive sign convention, OP just needs to be careful when substituting in ##i = -\frac{dQ_A}{dt} = \frac{dQ_B}{dt}## since you should get a negative sign out the front. From then on, it's just a mass on the end of a spring problem :wink:.
@etotheipi
Can you explain how do i solve the equation after substituting ,##i## ?
 
  • #49
etotheipi
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@Ayesha02 Try what @archaic suggested (I have substituted for ##c## for clarity):
Instead of my other suggestion, you should take ##Q(t)=-Q_A(t)##.
$$\frac{kQ_A(t)}{r} - \frac{k(Q_0-Q_A(t))}{r} = L\frac{di}{dt}=L\frac{d^2}{dt^2}\left(-Q_A(t)\right)\\
\implies\frac{k}{r}\left(2Q_A(t)-Q_0\right)=-L\frac{d^2Q_A(t)}{dt^2}$$
Take ##f(t)=2Q_A(t)-Q_0##, this gives you ##f''(t)=2Q_A''(t)##, thus
$$\frac{k}{r}f(t)=\frac{-L}{2}\frac{d^2f(t)}{dt^2}$$
That last equation is the same as ##\frac{-2k}{Lr} f(t) = f''(t)##. That reminds me of another physical phenomenon, i.e. a mass on a spring which behaves like ##a _x= -\omega^2 x##. Can you then work out the solution of the differential equation?
 
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  • #50
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@Ayesha02 if you haven't solved differential equations before, then suppose that ##f (t)=A\cos (\omega t-B)## and use the initial conditions I hinted at in a previous post to find the unknowns.
 
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