Black Box, Time-Dependent Current & Potential Rise

In summary: V, there is no way the current could decay to a negligible amount.I had the same initial interpretation, but after some thought I realized that the "one part in a million" refers to the difference between the current entering and leaving the box, not the decay rate of the current itself. So the current is not necessarily exponentially decaying, but it is decreasing by a factor of (1 - 10^-6) over some time interval. As for the assumptions, I believe we are free to make reasonable assumptions for our calculations, as long as they do not significantly alter the final answer. So assuming the current is constant and isotropic is probably a reasonable assumption in this case
  • #1
rbrayana123
44
0

Homework Statement


Consider a black box which is approximately a 10-cm cube with two binding posts. Each of these terminals is connected by a wire to some external circuits. Otherwise, the box is well insulated from everything. A current of approximately 1 amp flows through the circuit element. Suppose now that the current in and the current out differ by one part in a million. About how long would it take, unless something else happens, for the box to rise in potential by 1000 volts.

Homework Equations


Q = CV
V = IR
-dQ/dt = I

The Attempt at a Solution


I'm not exactly sure how to start off this problem. I'm not given the details of the external circuit; nor can I figure out the capacitance of the cube (if that's relevant at all). Here's my attempt though:

Assuming the cube is a conductor, I can calculate out the electric field at the surface using Gauss' Law:
6EA = Q/[itex]\epsilon[/itex]o, where A is the area of one side.

E = Q/[itex]\epsilon[/itex]o6A

However, if I integrate from 0.05 m to infinity, I get an infinite potential. I'm guessing the electric field musn't be constant outside of the cube.

I threw that out of the bag, so I tried calculating the rate at which current changes... except, I don't know how much time it takes to traverse the entire circuit. All I know is that it changes by a factor of (1 - 10^-6) so dI/dt = I(1 - 10^-6), except maybe with some factor of time in there.
 
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  • #2
The potential is a function of position, so to say "rise by 1000 volts" one needs to specify exactly WHERE that is happening.
 
  • #3
I note that the problem statement uses the word "approximately" for both the box size and the current. Presumably that means you are free to make reasonable approximations for your modeling calculations, too. And who's to say that the charge is on the surface of an insulated box rather than on something of simpler geometry inside it?

So, back of the envelope time; Consider a spherical cow*...

-----------------------

*Old joke where a physicist tries to help out his farmer friend with a sick cow.
 
  • #4
Interesting Question:

perhaps you could model it like two parallel plates 10cm apart i.e. a parallel plate capacitor.

Q = CV and C = Area*epsilon/spacing = 0.1*epsilon (for a

So Q = 1000*0.1*epsilon = 100*epsilon

Should be easy turn amount of charge into a time given the amps.
 
  • #5
PhysStudent81 said:
Interesting Question:

perhaps you could model it like two parallel plates 10cm apart i.e. a parallel plate capacitor.

Q = CV and C = Area*epsilon/spacing = 0.1*epsilon (for a

So Q = 1000*0.1*epsilon = 100*epsilon

Should be easy turn amount of charge into a time given the amps.
So running from this line of argument

Q = CV; C = Aε/d; Q = AεV/d

A = (0.1)2 m2
d = 0.1 m
Io = 1 Amp
ε = 8.85 * 10-12 F/m
to is time at which V = 1000 V

Let k = (1 - 10^-6)

dI/dT = -kI
dI/I = -kdt
ln(I) = -kt + c
I = Ioe-kt = dQ/dt

∫Ioe-ktdt, from 0 to to = (Io/k)(1 - e-kto)

(Io/k)(1 - e-kto) = AεV/d
1 - e-kto = AεVk/dIo
e-kto = 1 - AεVk/dIo
-kto = ln(1 - AεVk/dIo)
to = -ln(1 - AεVk/dIo)/k

to = 8.85 * 10-10

Although, I believe the correct answer is 6 ms. I think I messed up somewhere because if the question implies a rise in potential, then the charge should be an increasing function of time.
 
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  • #6
The question states that the charge is building up due to a proposed failure of KCL, wherein the current leaving the box does not balance with the current entering it, to the extent of one part in a million. So the charge will be a linear function of time.

For an approximation, assume that the charge is building up on a spherical conductor at the center of the non-conducting box, and is entirely enclosed by the box.
 
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  • #7
gneill said:
The question states that the charge is building up due to a proposed failure of KCL, wherein the current leaving the box does not balance with the current entering it, to the extent of one part in a million.

For an approximation, assume that the charge is building up on a spherical conductor at the center of the non-conducting box, and is entirely enclosed by the box.

In this case:
Q = CV; C = d/2k; Q = Vd/2k (d/2 would be the radius of the inscribed sphere)
b = 1 - 10^-6
k = 1/4πεo

dI/dT = -bI
dI/I = -bdt
ln(I) = -bt + c
I = Ioe-bt = dQ/dt

∫Ioe-bt, from 0 to to = (Io/b)(1 - e-bto)

(Io/b)(1 - e-bto) = Vd/2k
1 - e-bto = Vdb/2kIo
e-bto = 1 - Vdb/2kIo
-bto = ln(1 - Vdb/2kIo)
to = -ln(1 - Vdb/2kIo)/b

to = 5.56 * 10^-9
 
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  • #8
There won't be any exponential charging of the sphere -- you're given a linear rate at which charge builds up. The charge transported by a current I in time t is I*t. A millionth of the transported charge is being either left on or taken from the device (either implies a charge buildup either positive or negative).
 
  • #9
gneill said:
There won't be any exponential charging of the sphere -- you're given a linear rate at which charge builds up. The charge transported by a current I in time t is I*t. A millionth of the transported charge is being either left on or taken from the device (either implies a charge buildup either positive or negative).

b = (10^-6)

Q = bIot?
Q = Vd/2k
bIot = Vd/2k
t = Vd/2kbIo
t = 5.56 * 10^-3

It appears to be the correct answer but it seems I continually mis-interpreted the "one part in a million" portion of the question. I took it to mean the current must be exponentially decreasing. Is this a completely wrong interpretation or is there one of the following assumptions in play:

a) The current is forced to be 1 Amp (constant) due to some exterior device as much as possible

b) In the time it takes to raise the potential of the sphere to 1000 V, the decay of the current must be negligible (However... having worked out the exact case and missing by orders of magnitude tells me otherwise...)
 
  • #10
rbrayana123 said:
b = (10^-6)

Q = bIot?
Q = Vd/2k
bIot = Vd/2k
t = Vd/2kbIo
t = 5.56 * 10^-3

It appears to be the correct answer but it seems I continually mis-interpreted the "one part in a million" portion of the question. I took it to mean the current must be exponentially decreasing. Is this a completely wrong interpretation or is there one of the following assumptions in play:

a) The current is forced to be 1 Amp (constant) due to some exterior device as much as possible

b) In the time it takes to raise the potential of the sphere to 1000 V, the decay of the current must be negligible (However... having worked out the exact case and missing by orders of magnitude tells me otherwise...)

Your result looks correct to me :approve:

A problem statement open to misinterpretation can be rather vexing :smile:

The way I interpreted it, the current entering is (approximately) a constant 1 Amp (provided by some fixed current source). The current leaving is 1 Amp less some small amount (~1 millionth of an amp less). This leads to a constant rate of charge buildup on the device.

No decay of current or charge is mentioned, so whatever charge is left behind stays on the device.
 
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  • #11
gneill said:
Your result looks correct to me :approve:

A problem statement open to misinterpretation can be rather vexing :smile:

The way I interpreted it, the current entering is (approximately) a constant 1 Amp (provided by some fixed current source). The current leaving is 1 Amp less some small amount (~1 millionth of an amp less). These leads to a constant rate of charge buildup on the device.

No decay of current or charge is mentioned, so whatever charge is left behind stays on the device.

I see. It was simpler than I made it out to be @_@. What exactly would cause an exponential decrease in current, now that I think of it.
 
  • #12
rbrayana123 said:
I see. It was simpler than I made it out to be @_@. What exactly would cause an exponential decrease in current, now that I think of it.

If the box developed a potential across its connections (rather than an overall static charge), and the current was being supplied by a voltage source with an internal resistance, then it would decrease over time. Whether or not it would then "look like" a capacitor and thus create an RC circuit exhibiting exponential current decay would depend upon the details of the device.
 
  • #13
gneill said:
The question states that the charge is building up due to a proposed failure of KCL, wherein the current leaving the box does not balance with the current entering it, to the extent of one part in a million. So the charge will be a linear function of time.

For an approximation, assume that the charge is building up on a spherical conductor at the center of the non-conducting box, and is entirely enclosed by the box.

Why do you make this approximation? What's a binding post anyway?
 
  • #14
PhysStudent81 said:
Why do you make this approximation?
Because it's the simplest model I could think of -- a spherical charge distribution looks like a point charge from any direction (outside the sphere). I justify using a simplified model because,

1. No details were given about the internals of the black box.
2. The problem statement uses the word "approximately" for given values.
3. The problem is a thought experiment with the premise that KCL does not hold.
4. It really, really simplifies the math :wink: and should at the very least give a result with the right order of magnitude.
What's a binding post anyway?
Connectors where wires are attached. Typically a threaded post that you wrap a wire around or pass through a hole in the post, and then tighten down a wire-nut to secure it. You've probably seen this sort of thing on speakers or test equipment. If not, google images will show you many examples.
 
  • #15
gneill said:
Because it's the simplest model I could think of -- a spherical charge distribution looks like a point charge from any direction (outside the sphere). I justify using a simplified model because,

1. No details were given about the internals of the black box.
2. The problem statement uses the word "approximately" for given values.
3. The problem is a thought experiment with the premise that KCL does not hold.
4. It really, really simplifies the math :wink: and should at the very least give a result with the right order of magnitude.

Connectors where wires are attached. Typically a threaded post that you wrap a wire around or pass through a hole in the post, and then tighten down a wire-nut to secure it. You've probably seen this sort of thing on speakers or test equipment. If not, google images will show you many examples.

Thanks - I thought you were envisaging some kind of capacitor and was totally bemused by what you were suggesting.

Totally see what you're saying now. Thanks
 

Related to Black Box, Time-Dependent Current & Potential Rise

1. What is a black box?

A black box refers to a device or system whose internal workings are unknown or not easily understood. It is often used in reference to complex systems, such as electronic circuits or computer programs, where the inputs and outputs can be observed but the processes in between are hidden.

2. How does a black box relate to time-dependent current and potential rise?

In the context of electrical circuits, a black box can represent a component or system whose behavior changes over time. Time-dependent current and potential rise refer to the changes in electric current and voltage that occur over a period of time in a circuit. Therefore, a black box can be used to model and analyze these time-dependent changes in an electrical circuit.

3. What is the significance of studying time-dependent current and potential rise?

Understanding time-dependent current and potential rise is crucial in many fields, such as electronics, telecommunications, and power systems. It allows engineers and scientists to predict and control the behavior of electrical systems, ensuring their safe and efficient operation.

4. How is time-dependent current and potential rise measured?

Time-dependent current and potential rise can be measured using various instruments such as oscilloscopes, voltmeters, and ammeters. These instruments capture and display the changes in current and voltage over time, allowing for accurate measurements and analysis.

5. What are some real-world applications of time-dependent current and potential rise?

Time-dependent current and potential rise play a crucial role in the design and operation of various electrical systems, such as power grids, electronic devices, and communication networks. It is also essential in fields such as medical imaging, where time-dependent changes in electrical signals are used to generate images of the human body.

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