Displacement current clarification

[Chriss]

Hy. I am stuck at the displacement current. Here is a picture.

Thanks !

[studiot]

Well it's a very pretty picture, but what is the question?

[Chriss]

First question is that i don't understand what that path P means ?

[studiot]

 

First question is that i don't understand what that path P means ?

 

 

Well P is a circle, but the important thing is it is a closed loop, it could be a square or any shape so long as it is a single closed loop.

 

As a closed loop P is the boundary of surface, S₁.

 

All the current I must pass through this loop or surface.

 

I is the total current and is the sum (integral) of all contributions of curent density J across the whole of S_1.

If the density is uniform then it is just J x Area.

 

We can use Gauss Theorem to relate the flux through the gap across the area to the size of the loop ie path P, if you have studied the necessary vector calculus.

 

Next question?

 

(note this is understanding displacement current the hard way.)

 

Now a question for you.

 

I don't understand why you have posted this in Quantum theory. It is classical physics, due to Maxwell.

Edited December 9, 2013 by studiot

[Chriss]

That path is electric field ?

[studiot]

 

That path is electric field ?

 

 

 

No.

 

This will go much better if you respond to my comments.

 

A path is a route or track in space, just like the footpath along the side of a road, or my route on the surface of the earth.

 

Suppose I went from London to Chicago.

 

That is an open path on the surface of the earth.

 

Now suppose i went London to Chicago to Wellington to Hongkong to Abu Dhabi and back to to London.

 

That is a closed path on the surface of the earth.

 

In your case your path is a circle in a plane so is closed.

[Chriss]

I posted the topic and after that i realized that i posted wrong. If someone can move it...

And what is the easy way to understand displacement current ?

[studiot]

I currently dealing with a server emergency this morning, but will be back to hopefully post this evening that should be a bit easier to digest.

 

Meanwhile please don't let me stop others offering explanation.

[Chriss]

What does S1 and S2 means ? That can only be magnetic field i think.

[imatfaal]

I posted the topic and after that i realized that i posted wrong. If someone can move it...

And what is the easy way to understand displacement current ?

 

I am sure I suggested Hyperphysics and Khan Academy. They really are very good.

[studiot]

 

What does S1 and S2 means ? That can only be magnetic field i think.

 

 

Well you are asking about displacement current, although they are from your diagram which you posted without labelling the axes or any of the variables or any explanation of the diagram.

 

There is no magnetic field involved in displacement current.

 

S1 and S1 are surfaces, but you are still ignoring my comments

 

Why is this?

 

I asked if you have studied vector calculus, because your diagram is from an advanced treatment of displacement current using vector calculus that would be undertaken by someone who already knows what it is.

 

I mentioned Gauss theorem ( which is one of Green's several theorems) which allows a vector surface integral to be converted to a line integral around a closed path and allows us to consider displacement current in free space in total generality.

 

I also said I would post a simpler treatment for you.

 

The first thing to know is that several physical quantities are known by more than one name or have had their names changed over the years.

 

The first important quantity to know this about was originally called the dielectric constant and is now usually known as the 'permittivity' and given the symbol epsilon [math]\varepsilon [/math].

Epsilon is of course a constant, not a variable, and is a property of the space concerned or the substance filling that space.

It is only of general interest for nonconductors since an electric field can exist in the space containing a nonconductor.

 

We use another constant for space containing a conductor called the conductivity, sigma [math]\sigma [/math].

 

So we can have an electric field inside a non conductor or dielectric.

We measure the strenght of the field by the Electric Field Strength, given the symbol E.

I have shown this in bold because E is a vector. That is it has a magnitude and direction at any point - a bit like the strength of the wind at any point.

 

Now we come to the good bit.

 

Years ago a second physical quantity was defined equal to the field strength times the dielectric constant and called the Displacement.

[math]D = \varepsilon E[/math]

 

Not suprisingly it was given the letter D

 

It is also a vector since it is a number times a vector.

 

Nowadays the Displacement is called the Electric Flux.

 

Again originally, experiments showed that if we set up apparatus as in my figure (1) which shows a battery connected to a conductor R in series with a capacitor C and supply a voltage through a switch S then a measurable current flows. This current changes with time as in the equation noted. That is starts off large and gets smaller and smaller, eventually dying away to nothing.

 

To complete the circuit this current must not only flow through the conductor R, but also through the capacitor C.

 

This is very important and was not originally understood. How could a current flow through a non conductor?

 

Well Maxwell suggested that since there is an electric field and therefore a Displacement that this strange effect be called Displacement Current and the name has been kept, even though the name of the quantity D has been changed.

 

We now know that the conducted current is the result of a steady electric field and the displacement current is the result of achanging electric field.

If you wish to know more there is a second figure to explain this.

 

But please let me know how you got on with the part 1 first.

Edited December 12, 2013 by studiot

[Chriss]

I see. I got it.

[Chriss]

Why is the magnetic flux trough a closed surface always zero ?

[imatfaal]

Why is the magnetic flux trough a closed surface always zero ?

 

Magnetic field lines leave and then comeback - the first compensates for the other. You cannot isolate a magnetic monopole - you always end up with two opposite poles and they cancel. It's the basis - from observation - of Gauss' law of magnetism which is also one of Maxwell's famous four.

 

This is completely different to electric charges - you can have an isolated electric charge and thus you can have a closed surface around this charge which has a net electric flux

[Chriss]

I think i understood the first part. You can post the second. Thanks !

[studiot]

OK let us move on from a resistor and capacitor in series to a resistor and capacitor in parallel.

 

I have shown two identical blocks of cross sectional area A and distance between the ends D.

They are connected in parallel to the same (alternating) voltage source.

 

If the resistance is R and the capacitor has capacitance C then

 

i₁ = V/R flows in the resistor
[math]{i_2} = \frac{{dq}}{{dt}} = C\frac{{dV}}{{dt}}[/math] in the capacitor.

 

We call the current through the resistor the conduction current and the current through the capacitor the displacement current.

 

Physical current does not actually flow through the capacitor, it is just as if it did because qharge flows into the end plates and out again as the voltage alternates.

Now let us suppose that the resistor is filled with a conducting medium of conductivity sigma and the capacitor is filled with a dielectric of permittivity (dielectric constant) epsilon.

 

Now for both the capacitor and the resistor the electric field E is the voltage V divided by the separation d

 

E = V/d.

 

Inside the resistor the current density (symbol J₁) equlas the current divided by the crossectional area and also equals the conductivity times the electric field

[math]{J_1} = \frac{{{i_1}}}{A} = E\sigma [/math]

 

The capacitance of a parallel plate capacitor is

 

[math]C = \frac{{\varepsilon A}}{d}[/math] Also V = Ed

 

So

 

[math]{i_2} = C\frac{{dV}}{{dt}} = \frac{{\varepsilon A}}{d}\frac{{d(Ed)}}{{dt}} = \varepsilon A\frac{{dE}}{{dt}}[/math]

 

We can also get an expression for the current density due to the displacement current.

 

[math]{J_2} = \frac{{{i_2}}}{A} = \varepsilon \frac{{dE}}{{dt}} = \frac{{d\varepsilon E}}{{dt}} = \frac{{dD}}{{dt}}[/math]

 

[Chriss]

So the current does not flow between the capacitor plates but it's just creating an electric field, let's say + -, and then when the voltage alternates - +, and that change in electric field creates a magnetic field ?

[studiot]

Studiot post#11

 

There is no magnetic field involved in displacement current.

 

Perhaps I should expand on this.

 

There is no magnetic field associated with the generation of displacement current.

It is a purely electric effect.

 

However a magnetic field appears as a result of the changing electric field.

The displacement current may be input to the usual formula

[math]B = \mu {i_D}[/math]

 

To calculate the value of the magnetic field

 

Where i_D is the displacement currrent.

Edited December 29, 2013 by studiot

[Chriss]

So displacement current is the charging of a capacitor ?

[studiot]

 

So displacement current is the charging of a capacitor ?

 

 

No.

 

The current that charges/discarges a capacitor flows in conductors attached to the capacitor.

It is a conduction current.

This current is external to the capacitor.

 

The displacement current is the current that 'appears to flow' in the region of non conductance.

That is within the capacitor.

It is internal to the capacitor.