# Black Box, Time-Dependent Current & Potential Rise

1. Jan 13, 2013

### rbrayana123

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Q = CV
V = IR
-dQ/dt = I

3. 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/$\epsilon$o, where A is the area of one side.

E = Q/$\epsilon$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.

2. Jan 14, 2013

### voko

The potential is a function of position, so to say "rise by 1000 volts" one needs to specify exactly WHERE that is happening.

3. Jan 14, 2013

### Staff: Mentor

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*...

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*Old joke where a physicist tries to help out his farmer friend with a sick cow.

4. Jan 14, 2013

### PhysStudent81

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. Jan 14, 2013

### rbrayana123

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.

Last edited: Jan 14, 2013
6. Jan 14, 2013

### Staff: Mentor

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.

Last edited: Jan 14, 2013
7. Jan 14, 2013

### rbrayana123

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

Last edited: Jan 14, 2013
8. Jan 14, 2013

### Staff: Mentor

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. Jan 14, 2013

### rbrayana123

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. Jan 14, 2013

### Staff: Mentor

Your result looks correct to me

A problem statement open to misinterpretation can be rather vexing

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.

Last edited: Jan 15, 2013
11. Jan 14, 2013

### rbrayana123

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. Jan 15, 2013

### Staff: Mentor

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. Jan 15, 2013

### PhysStudent81

Why do you make this approximation? What's a binding post anyway?

14. Jan 15, 2013

### Staff: Mentor

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 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.

15. Jan 15, 2013

### PhysStudent81

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