Inductance and Charge Redistribution

[Ayesha02]

Homework Statement

Two fixed identical metallic spheres A and B of radius R=50cm each are placed on a non-conducting plane at a very large distance from each other and they are connected by a coil of inductance L=9mH as shown in figure. One of the spheres (say A) is imparted an initial charge and the other is kept uncharged. The switch S is closed at t=0. After what minimum time t does the charge on the first sphere decrease to half of its initial value ?

Relevant Equations

Loop law: kq/r- L di/dt -kq/r=0
where L is inductance, q is initial charge on spheres.

I tried applying loop law but I am not really sure we don't really have a closed loop here.

I guess they're testing some concept here that I'm not very good at (Why do i keep coming back to mutual inductance for some reason:|)
Any help will be appreciated:)

 

[etotheipi]

Kirchhoff's loop rule has to do with adding up changes in potential around a closed loop (so there are actually two oddities with the 'loop law' in your relevant equations: $\frac{kq}{r}$ is a potential and not a potential difference ); though I guess you could consider a closed loop to infinity and back in which case your construction is nearly correct.

Instead, suppose a current $i$ is flowing from sphere A to sphere B through the inductor, and that the charges of the spheres at any given time $t$ are $Q_A(t)$ and $Q_B(t)$. What is the potential difference across the inductor: can you find two different ways of writing this, and equate them? How do you relate $i$ to $Q_A(t)$ and/or $Q_B(t)$? How do you relate $Q_A(t)$ and $Q_B(t)$ to the total charge in the system, in order to eliminate one of them?

A few things to think about there...

 

[Ayesha02]

Kirchhoff's loop rule is not going to help you here; that has to do with adding up changes in potential around a closed loop (so there are actually two problems with the 'loop law' in your relevant equations: $\frac{kq}{r}$ is a potential and not a potential difference ).

Instead, suppose a current $i$ is flowing from sphere A to sphere B through the inductor, and that the charges of the spheres at any given time $t$ are $Q_A(t)$ and $Q_B(t)$. What is the potential difference across the inductor: can you find two different ways of writing this, and equate them? How do you relate $i$ to $Q_A(t)$ and/or $Q_B(t)$? How do you relate $Q_A(t)$ and $Q_B(t)$ to the total charge in the system, in order to eliminate one of them?

A few things to think about there...

So
Potential difference across the inductor is -L di/dt.
and (I'm not really sure of this-)
$Q_A(t)$ should be $Q_0$ - $i$ t
$Q_B(t)$ should be $Q_0$ + $i$ t

Is that right?

 

[etotheipi]

So
Potential difference across the inductor is -L di/dt.
and (I'm not really sure of this-)
$Q_A(t)$ should be $Q_0$ - $i$ t
$Q_B(t)$ should be $Q_0$ + $i$ t

Is that right?

A few things. $Q_A(t) = Q_0 - it$ would be correct if the current were constant, however we have no grounds to assume it here. Your second equation also assumes constant current, but $Q_B(0) = 0$ in the problem statement so even if the current were constant there should be no $Q_0$.

You're right about the PD across the inductor; that is the change in potential in the direction of the current.

Think about it like this; if sphere $A$ has a charge of $Q_A(t)$ and a radius of $r$, what is its potential? Likewise, what is the potential of sphere $B$? What is the potential difference from sphere A to sphere B?

 

[Ayesha02]

A few things. $Q_A(t) = Q_0 - it$ would be correct if the current were constant, however we have no grounds to assume it here. Your second equation also assumes constant current, but $Q_B(0) = 0$ in the problem statement so even if the current were constant there should be no $Q_0$.

You're right about the PD across the inductor; that is the change in potential in the direction of the current.

Think about it like this; if sphere $A$ has a charge of $Q_A(t)$ and a radius of $r$, what is its potential? Likewise, what is the potential of sphere $B$? What is the potential difference from sphere A to sphere B?

Okay so if sphere $A$ has a charge of $Q_A(t)$ and a radius of $r$, its potential should be k$Q_A(t)$ /r ; likewise for B.
the potential difference between the two then becomes k$Q_A(t)$ /r - k$Q_B(t)$ /r .

is it okay?

 

[etotheipi]

Sure, so that gives you (being careful with the signs)
$$\frac{kQ_A(t)}{r} - \frac{kQ_B(t)}{r} = L\frac{di}{dt}$$ There are a few different things you could do; I would suggest eliminating either $Q_A$ or $Q_B$, and rewriting $\frac{di}{dt} = \frac{d}{dt}(i)$ in terms of either $Q_A$ or $Q_B$ (depending on which one you have left).

 

[rude man]

Yep, solving ODE time!

 

[Gordianus]

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.

 

[rude man]

(Why do i keep coming back to mutual inductance for some reason:|)

There can only be mutual inductance when there are conceptually two (or more) inductors (discrete and/or tapped).

 

[Ayesha02]

There can only be mutual inductance when there are conceptually two (or more) inductors (discrete and/or tapped).

Ohh you true!

 

[Ayesha02]

Sure, so that gives you (being careful with the signs)
$$\frac{kQ_A(t)}{r} - \frac{kQ_B(t)}{r} = L\frac{di}{dt}$$ There are a few different things you could do; I would suggest eliminating either $Q_A$ or $Q_B$, and rewriting $\frac{di}{dt} = \frac{d}{dt}(i)$ in terms of either $Q_A$ or $Q_B$ (depending on which one you have left).

So
I'm not sure I'm quite getting how to eliminate one of $Q_A$ or $Q_B$ there buddy:|

 

[archaic]

So
I'm not sure I'm quite getting how to eliminate one of $Q_A$ or $Q_B$ there buddy:|

A current is defined as change in charge. ;)

 

[Ayesha02]

A current is defined as change in charge. ;)

Fine so i=d $Q_A$/dt or d$Q_B$/dt. yet, how to eliminate?

 

[archaic]

Fine so i=d $Q_A$/dt or d$Q_B$/dt. yet, how to eliminate?

Well, what do you think of the sign of the left hand side of your equation?

 

[archaic]

Moreover, where do you think the charge lost by A will go? If the initial charge of A is $Q_0$, how can you relate it to that of B as time varies forward?

 

[archaic]

Moreover, where do you think the charge lost by A will go? If the initial charge of A is $Q_0$, how can you relate it to that of B as time varies forward?

If I am drinking from a bottle of water of initial volume $V_0$, and the volume that I drank as a function of time is $V(t)$, then what is the bottle's volume as time goes on?

 

[Ayesha02]

Moreover, where do you think the charge lost by A will go? If the initial charge of A is $Q_0$, how can you relate it to that of B as time varies forward?

So the charge lost by A goes to B.
which means considering $Q_0$ as the initial charge given to A, $Q_B$ should eventually become $Q_0$ - $Q_A$
so far so good?

 

[Ayesha02]

Well, what do you think of the sign of the left hand side of your equation?

Though i didnt quite get you on this..

 

[archaic]

So the charge lost by A goes to B.
which means considering $Q_0$ as the initial charge given to A, $Q_B$ should eventually become $Q_0$ - $Q_A$
so far so good?

Perfect. Gj.

Though i didnt quite get you on this..

The sign of the left hand side of your equation is equal to the sign of its right hand side, so you should choose the change in the current accordingly.

 

[archaic]

Before A's charge becoming lower than the half of its initial charge, the sign of the difference is positive. This tells you that the derivative of the current is postive, thus the current should be growing. Feel free to add a minus aign if necessary depending on which charge you have chosen to eliminate.

 

[Ayesha02]

Perfect. Gj.

The sign of the left hand side of your equation is equal to the sign of its right hand side, so you should choose the change in the current accordingly.

Alright so what i can gather so far:

1. k $Q_A$/r - k $Q_B$/r =L di/dt
2. i= ($Q_0$ - $Q_A$) /t
3. i= ($Q_B$) /t

I still don't see how are we solving this:|

 

[archaic]

You have found that
$$\frac{kQ_A(t)}{r} - \frac{kQ_B(t)}{r} = L\frac{di}{dt}$$
and have chosen $Q_B(t)=Q_0-Q_A(t)$, which give you
$$\frac{kQ_A(t)}{r} - \frac{k(Q_0-Q_A(t))}{r} = L\frac{di}{dt}$$
You know that $Q_A(t)-(Q_0-Q_A(t))=2Q_A(t)-Q_0\geq0$, thus for for $Q_A(t)\geq\frac{Q_0}{2}=\frac{Q_A(0)}{2}$, we have $\boxed{Li'(t)\geq0\implies i(t)=\frac{dQ}{dt}\text{ is growing.}}$
Since you have chosen to eliminate $Q_B(t)$, you need to express $i(t)$ using the expression you have found for $Q_B(t)$, and it needs to satisfy the condition above, and then solve the differential equation.

 

[Ayesha02]

You have found that
$$\frac{kQ_A(t)}{r} - \frac{kQ_B(t)}{r} = L\frac{di}{dt}$$
and have chosen $Q_B(t)=Q_0-Q_A(t)$, which give you
$$\frac{kQ_A(t)}{r} - \frac{k(Q_0-Q_A(t))}{r} = L\frac{di}{dt}$$
You know that $Q_A(t)-(Q_0-Q_A(t))=2Q_A(t)-Q_0\geq0$, thus for for $Q_A(t)\geq\frac{Q_0}{2}=\frac{Q_A(0)}{2}$, we have $Li'(t)\geq0$.
Since you have chosen to eliminate $Q_B(t)$, you need to express $i(t)$ using the expression you have found for $Q_B(t)$, and it needs to satisfy the condition above, and then solve the differential equation.

nope
I'm getting so confused rn.

 

[archaic]

nope
I'm getting so confused rn.

Hm, can you please tell me where things are unclear?

 

[Ayesha02]

Hm, can you please tell me where things are unclear?

See
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:)

 

[archaic]

See
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 $\beta$?

 

[archaic]

If ββ is the derivative of some function, then what does its sign tell you about that function?

 

[Ayesha02]

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?

 

[archaic]

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).

 

[rude man]

Ohh you 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!

 

[Ayesha02]

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:(

 

[Ayesha02]

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 can't pinpoint it:(

 

[rude man]

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..

 

[Ayesha02]

Instead,
Can't we consider a loop at infinity and just write the loop law simply?

 

[Ayesha02]

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?

 

[Ayesha02]

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

 

[archaic]

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
$$\frac{k}{r}f(t)=\frac{cL}{2}\frac{d^2f(t)}{dt^2}$$

 

[Ayesha02]

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

 

[archaic]

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 $c$!

 

[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 $c$!

Hint: what are $Q_A(0)$ and $i(0)$?

 

[Delta2]

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.

 

[archaic]

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 $-1$!

 

[etotheipi]

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 .

 

[archaic]

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 .

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?

 

[etotheipi]

You can work it all out logically, but it's often quicker to do anything circuit related with the so-called passive sign convention

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.

 

[Delta2]

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 .

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 $I=\frac{dQ_A}{dt}=-\frac{dQ_B}{dt}$

 

[etotheipi]

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!

 

[Ayesha02]

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 .

@etotheipi
Can you explain how do i solve the equation after substituting ,$i$ ?

 

[etotheipi]

@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?

 

[archaic]

@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.