Gauss law in electrodynamics


by quietrain
Tags: electrodynamics, gauss
quietrain
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#1
Apr21-11, 09:57 AM
P: 651
if i have a current carrying straight long wire, I = I0sin(wt)

why is gauss law ∇.E = 0?

i thought only for steady currents , then the charges reside on surface, thats why 0 charge enclosed, and hence gauss law gives 0 right?

so now since i have a time varying current, what should gauss law give?

thanks!
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G01
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Apr21-11, 10:32 AM
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A (long) straight wire can carry a current (steady or oscillating) and still have a zero net charge distribution throughout.

A Caveat:

If you consider the end points of the wire, this may not be the case. Charge will build up at the ends, and in the case of an harmonically varying current, the charge on the end points will also vary harmonically, but out of phase with the current by [itex]\pi /2[/itex].

In this case, your charge distribution will consist of two delta functions at the ends of the wire, whose amplitudes oscillate in time.

So, if you're wire ends are at +-z:

[tex]\rho(r)= q_o \cos(\omega t) \delta(z) - q_o \cos(\omega t) \delta(-z)[/tex]
quietrain
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#3
Apr21-11, 10:33 AM
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Quote Quote by G01 View Post
A wire can carry a current and still have a zero net charge distribution throughout.
issn't current the flow of charges?

how can the charged enclosed be 0?

erm, what do you mean by net charge distribution?

quietrain
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#4
Apr21-11, 10:38 AM
P: 651

Gauss law in electrodynamics


i am getting very confused :(

if i apply a gaussian cylindrical surface to the wire, then the flux is 0?

so my net charge is 0? so i assume its the electrons and +ve charges that do the cancelling?
quietrain
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#5
Apr21-11, 10:41 AM
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Quote Quote by G01 View Post
A (long) straight wire can carry a current (steady or oscillating) and still have a zero net charge distribution throughout.

A Caveat:

If you consider the end points of the wire, this may not be the case. Charge will build up at the ends, and in the case of an harmonically varying current, the charge on the end points will also vary harmonically, but out of phase with the current by [itex]\pi /2[/itex].

In this case, your charge distribution will consist of two delta functions at the ends of the wire, whose amplitudes oscillate in time.


i think the caveat part is out of my scope at the moment :(
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Apr21-11, 10:42 AM
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Quote Quote by quietrain View Post
issn't current the flow of charges?

how can the charged enclosed be 0?

erm, what do you mean by net charge distribution?


Consider a length of the ideal wire we are considering carrying no current.

There are the same number of positive and negative charges in it. Thus it is neutral. Agreed?

Now apply a voltage so we get a current. At any one time, the protons and electrons are still equal in number and the wire has no net (i.e. total) charge. However, the electrons are moving, but the protons are not, resulting in a net (total/overall) current. Thus, a wire can have no net charge, but still carry current. Does this make sense?
quietrain
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#7
Apr21-11, 11:11 AM
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Quote Quote by G01 View Post
Now apply a voltage so we get a current. At any one time, the protons and electrons are still equal in number and the wire has no net (i.e. total) charge. However, the electrons are moving, but the protons are not, resulting in a net (total/overall) current. Thus, a wire can have no net charge, but still carry current. Does this make sense?
ah yes i see, but in the case of a varying magnitude current as in the original question, wouldn't this mean that the electrons moving in that unit length be not equal to the number of protons which were not moving in that unit length?

for example, say the unit length has 5 protons, since now my current magnitude is varying, i will have 5,4,3,4,5 electrons at different instances of time.
so the net charge will be different?
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Apr21-11, 11:27 AM
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Quote Quote by quietrain View Post
ah yes i see, but in the case of a varying magnitude current as in the original question, wouldn't this mean that the electrons moving in that unit length be not equal to the number of protons which were not moving in that unit length?

for example, say the unit length has 5 protons, since now my current magnitude is varying, i will have 5,4,3,4,5 electrons at different instances of time.
so the net charge will be different?

No. You are picturing the current varying in space. This is not what we are talking about here. In this case, the current is varying in time only!

Consider one instant in time. Call it time 0. The value of the current in the wire is [itex]I_o[/itex]. All along the wire, the current value is [itex]I_o[/itex].

Now, at a later time, say time=5s, the current value is different, [itex]I_o sin(5\omega)[/itex], but yet it is the same at every point in the wire.

Thus, in both cases, the net charge distribution is still uniform and zero, the speed of the electrons in the only thing to have changed.


As an aside,

If the current was varying in space, and you had "pockets" of greater current and lesser current along the wire, then, the divergence of the current would not be zero, and by the continuity equation, we would indeed have a changing charge distribution...
quietrain
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#9
Apr21-11, 11:35 AM
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Quote Quote by G01 View Post
No. You are picturing the current varying in space. This is not what we are talking about here. In this case, the current is varying in time only!

Consider one instant in time. Call it time 0. The value of the current in the wire is [itex]I_o[/itex]. All along the wire, the current value is [itex]I_o[/itex].

Now, at a later time, say time=5s, the current value is different, [itex]I_o sin(5\omega)[/itex], but yet it is the same at every point in the wire.

Thus, in both cases, the net charge distribution is still uniform and zero, the speed of the electrons in the only thing to have changed.


As an aside,

If the current was varying in space, and you had "pockets" of greater current and lesser current along the wire, then, the divergence of the current would not be zero, and by the continuity equation, we would indeed have a changing charge distribution...
ah i see thank you!


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