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Inductance and Charge Redistribution

  • Thread starter Ayesha02
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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 im not really sure we dont 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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Answers and Replies

  • #2
etotheipi
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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
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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
etotheipi
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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
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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
etotheipi
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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
rude man
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Yep, solving ODE time!
 
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  • #8
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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.
 
  • #9
rude man
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(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
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There can only be mutual inductance when there are conceptually two (or more) inductors (discrete and/or tapped).
Ohh ya true!
 
  • #11
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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
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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:|
A current is defined as change in charge. ;)
 
  • #13
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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
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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
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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
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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
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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
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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
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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.
 
  • #20
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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
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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
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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
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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
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nope
I'm getting so confused rn.
Hm, can you please tell me where things are unclear?
 
  • #25
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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:)
 

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