Displacement current inside a wire

**bakshi** · Dec9-05, 06:02 PM

The example of the charging capacitor is often used to show that Ampère's law cannot work in electrodynamics, since the value of the integral (I'm talking about Ampère in integral form here) would depend on the choice of surface delimited by the Amperian loop. It is then shown that since the displacement current between the plates of the capacitor is equal to the conduction current in the wire of the circuit, Ampère-Maxwell's law gives the same result for the integral.

My problem is the following: why isn't there also a displacement current inside the wire of the circuit? The electric field obvously changes inside the wire, since the voltage changes while the capacitor is charging. Since there is a changing electric field, there should be a displacement current, no?

Thank you,

Bakshi

**marlon** · Dec10-05, 11:12 AM

You need to look at how displacement current is defined. If the electrical flux, in between two capacitor plates, changes with respect to time, this change is equal to a "quantity" that must have the dimension of a current. This is predicted by the Ampere-Maxwell Law. However this is ofcourse a fictitious current since there is no real current present between the two plates. Hence the name displacement current, although "displacement" is rather poorly chosen because nothing is "being displaced".

Look at two capacitor plates inside a closed loop. The actual (ie socalled encapsulated by the closed loop when applying Ampère's Law) current will encounter the first capacitor plate, which gets charges. depending on the dielectric in between the plates, the second plate will get a certain opposite (equal in magnitude) charge. In between the two plates there is an E-field. Then the (real) current starts from the second plate and goes to the electrode. Now, it seems as if the real current has passed through the material (or vacuum) between the two capacitor plates. One looks at this as if current has flown in that specific region and it is this current that we call displacement current. You see ?

In a wire, you always have real current (ie moving electrons) and what you are referring to is called induced current actually.

You see the difference in displacement current and real (induced) current ? That is the clue of the story

regards
marlon

**bakshi** · Dec10-05, 12:37 PM

Maybe my question wasn't clear enough. I do know that the displacement current is not a real current but just a changing electric flux through a surface. Let me try to formulate my question again. Let's suppose that our Amperian loop is centered around the wire leading to one of the charging capacitor plates. If the surface delimited by the loop is between the plates, then there is obviously no conduction current through the surface, just a displacement current. It can be shown that this displacement current will be equal to the conduction current I through the wire of the circuit. Now, if our surface is in the plane of the Amperian loop, we say that there is no displacement current through it, just the conduction current I. Thus we say that Ampere-Maxwell's law gives us the same value for the integral for both surfaces.

What I don't see is why we say that there is no displacement current through the surface that is in the plane of the Amperian loop, since there must be a changing electric field in the wire while the capacitor is charging.

**marlon** · Dec11-05, 07:14 AM

Originally Posted by bakshi

What I don't see is why we say that there is no displacement current through the surface that is in the plane of the Amperian loop, since there must be a changing electric field in the wire while the capacitor is charging.

What is the surface in the plane of the Amperian Loop ?

Are you referring to the surface of a circle in the case that the wire was a closed loop ? There will be no current over that surface because apart from the wire, there is nothing to be charged and therefore no displacement current can be generated.

marlon

**heman** · Dec11-05, 08:13 AM

Originally Posted by bakshi

The example of the charging capacitor is often used to show that Ampère's law cannot work in electrodynamics, since the value of the integral (I'm talking about Ampère in integral form here) would depend on the choice of surface delimited by the Amperian loop. It is then shown that since the displacement current between the plates of the capacitor is equal to the conduction current in the wire of the circuit, Ampère-Maxwell's law gives the same result for the integral.
My problem is the following: why isn't there also a displacement current inside the wire of the circuit? The electric field obvously changes inside the wire, since the voltage changes while the capacitor is charging. Since there is a changing electric field, there should be a displacement current, no?
Thank you,
Bakshi

I got yours doubt if i am not mistaken.I would like to try.The net flux through a closed surface is always zero unless there is a source or sink inside the surface.And in the wire you can easily see that since current is flowing not originating nor sinking.
Therefore,if we choose the Flat surface Electric field is Zero and Hence Displacement current is Zero.

**marlon** · Dec11-05, 09:05 AM

Originally Posted by heman

I got yours doubt if i am not mistaken.I would like to try.The net flux through a closed surface is always zero unless there is a source or sink inside the surface.And in the wire you can easily see that since current is flowing not originating nor sinking.
Therefore,if we choose the Flat surface electric field is Zero and Hence Displacement current is Zero.

But, one can never have displacement current in a wire (conductor) since the current present there are just moving electrons. This is a fundamental difference between "normal" current and displacement current.

When you guys look at the Ampère Maxwell equation, you will see that (in the case of a wire) the closed loop MUST encapsulate the current in the wire. Than, ofcourse there is the extra term containing the variation of E-flux with respect to time. Displacement current ONLY arises if there is a change in E-flux BUT no change in current, such as in the region between two charging capacitor plates. So changing the current in the wire cannot be linked with displacement current.

If the current in the wire changes with time, there is a change in B-flux and the induction law predicts an induced EMF.

marlon

**erickalle** · Dec11-05, 06:09 PM

Bakshi:
My problem is the following: why isn't there also a displacement current inside the wire of the circuit? The electric field obvously changes inside the wire, since the voltage changes while the capacitor is charging. Since there is a changing electric field, there should be a displacement current, no?

Marlon:
But, one can never have displacement current in a wire (conductor) since the current present there are just moving electrons. This is a fundamental difference between "normal" current and displacement current.

Marlon, can I try to explain what I think Bakshi means?
Suppose first we have an established current in a wire. We are now going to change this current by altering the field E. I think Bakshi is trying to say that apart from the normal electron current in the wire there must on top of that also be a changing electrical field. I know that this changing field has the speed of signal speed, which is ~20% of the speed of light. Since the drift speed of the electrons is far lower then the signal speed we must be dealing with some other phenomenon. Bakshi (erroneously) calls this changing electrical field in the wire also a displacement current.

Bakshi am I right?
Sorry Marlon for crossing you jet again but we seem to have similar interests. Can you both reply again?

**Meir Achuz** · Dec12-05, 05:28 PM

There is E in a wire, but \partial_t(E) is negligible compared to j,
because j=\sigma E.

**bakshi** · Dec14-05, 09:04 AM

Thank you Erickalle for clarifying my point. This is exactly what my question was.

A "symmetric" problem would arise if the surface delimited by the Amperian loop was between the plates of the capacitor, but didn't contain a whole plate (I would need a figure here...).

Then, there would also be a conduction current (in addition to the displacement current) through the surface while the charge is spreading on the plate.

By the way, I still like to use the term displacement current instead of "changing electric flux through a surface". I know it is wrong.

**erickalle** · Dec15-05, 01:07 PM

There’s a very interesting chapter about this in the Feynmann lectures. (Which one is not interesting?) Its 27-5 in volume 2. There it is shown that for capacitors and wires a travelling E-field (displacement current as you call it) comes from the space surrounding these components. So does the electrical energy which is heating a wire up.