Inductance and Charge Redistribution

In summary: I'm not sure what point B is supposed to be!In summary, 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 :wink:); though I guess you could consider a closed loop to infinity and back in which case your construction is nearly correct. However, if we ignore the mutual capacitance between A and B, we end up with two capacitors "connected to infinity".
  • #1
Ayesha02
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5
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:)
 

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  • #2
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 :wink:); 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...
 
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  • #3
etotheipi said:
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 :wink:).

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?
 
  • #4
Ayesha02 said:
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?
 
  • #5
etotheipi said:
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?
 
  • #6
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).
 
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  • #7
Yep, solving ODE time!
 
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  • #8
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.
 
  • #9
Ayesha02 said:
(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).
 
  • #10
rude man said:
There can only be mutual inductance when there are conceptually two (or more) inductors (discrete and/or tapped).

Ohh you true!
 
  • #11
etotheipi said:
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:|
 
  • #12
Ayesha02 said:
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. ;)
 
  • #13
archaic said:
A current is defined as change in charge. ;)

Fine so i=d ##Q_A##/dt or d##Q_B##/dt. yet, how to eliminate?
 
  • #14
Ayesha02 said:
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?
 
  • #15
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?
 
  • #16
archaic said:
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?
 
  • #17
archaic said:
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?
 
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  • #18
archaic said:
Well, what do you think of the sign of the left hand side of your equation?

Though i didnt quite get you on this..
 
  • #19
Ayesha02 said:
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.
Ayesha02 said:
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.
 
  • #20
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.
 
  • #21
archaic said:
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:|
 
  • #22
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.
 
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  • #23
archaic said:
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.
 
  • #24
Ayesha02 said:
nope
I'm getting so confused rn.
Hm, can you please tell me where things are unclear?
 
  • #25
archaic said:
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:)
 
  • #26
Ayesha02 said:
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##?
 
  • #27
If ββ is the derivative of some function, then what does its sign tell you about that function?
 
  • #28
archaic said:
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
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
Ayesha02 said:
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! :smile:
 
  • #31
archaic said:
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
Ayesha02 said:
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:(
 
  • #33
Ayesha02 said:
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
Instead,
Can't we consider a loop at infinity and just write the loop law simply?
 
  • #35
Gordianus said:
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?
 
<h2>1. What is inductance?</h2><p>Inductance is a property of an electrical circuit that describes the ability of the circuit to generate an electromotive force (EMF) in response to a changing current. It is measured in units of Henrys (H).</p><h2>2. How is inductance related to charge redistribution?</h2><p>Inductance is related to charge redistribution because when a current flows through a circuit, it creates a magnetic field. This magnetic field can induce a current in nearby conductors, causing charge redistribution. The amount of charge redistribution is directly proportional to the inductance of the circuit.</p><h2>3. What factors affect inductance?</h2><p>The factors that affect inductance include the number of turns in a coil, the cross-sectional area of the coil, the material of the coil, and the presence of any ferromagnetic materials in the circuit. Increasing any of these factors will increase the inductance of the circuit.</p><h2>4. How does inductance impact circuit behavior?</h2><p>Inductance impacts circuit behavior by causing a delay in the current flow when a voltage is applied. This delay is known as reactance and it opposes the flow of current. Inductance also affects the frequency response of a circuit, causing it to behave differently at different frequencies.</p><h2>5. How can inductance be controlled or manipulated in a circuit?</h2><p>Inductance can be controlled or manipulated in a circuit by using different materials for the conductors, changing the number of turns in a coil, or adding inductors or capacitors to the circuit. Additionally, the shape and size of the circuit can also affect the inductance. By understanding the factors that affect inductance, engineers can design circuits with specific inductance values to meet their desired specifications.</p>

1. What is inductance?

Inductance is a property of an electrical circuit that describes the ability of the circuit to generate an electromotive force (EMF) in response to a changing current. It is measured in units of Henrys (H).

2. How is inductance related to charge redistribution?

Inductance is related to charge redistribution because when a current flows through a circuit, it creates a magnetic field. This magnetic field can induce a current in nearby conductors, causing charge redistribution. The amount of charge redistribution is directly proportional to the inductance of the circuit.

3. What factors affect inductance?

The factors that affect inductance include the number of turns in a coil, the cross-sectional area of the coil, the material of the coil, and the presence of any ferromagnetic materials in the circuit. Increasing any of these factors will increase the inductance of the circuit.

4. How does inductance impact circuit behavior?

Inductance impacts circuit behavior by causing a delay in the current flow when a voltage is applied. This delay is known as reactance and it opposes the flow of current. Inductance also affects the frequency response of a circuit, causing it to behave differently at different frequencies.

5. How can inductance be controlled or manipulated in a circuit?

Inductance can be controlled or manipulated in a circuit by using different materials for the conductors, changing the number of turns in a coil, or adding inductors or capacitors to the circuit. Additionally, the shape and size of the circuit can also affect the inductance. By understanding the factors that affect inductance, engineers can design circuits with specific inductance values to meet their desired specifications.

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