Nature of charge at the junction of a composite wire

AI Thread Summary
The discussion focuses on the nature of charge accumulation at the junction of copper and iron in a composite wire. It is noted that the current density in copper is higher due to its greater conductivity, leading to negative charge accumulation at the junction. The conversation also touches on the potential for electrochemical reactions due to differing electronegativities of the metals, although some participants argue that such reactions may not occur if the metals are merely in contact. The application of Gauss' law is suggested to understand the charge distribution, particularly in transient states where charge buildup occurs. Overall, the participants emphasize the importance of understanding current density and the effects of material properties on charge behavior at the junction.
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Homework Statement


Please help me answer thew question in the image.[/B]
KVPY doubt.PNG


Homework Equations


Current Density J=conductivity X Electric field[/B]

The Attempt at a Solution


As the current density depends on the conductivity of the material through which the electrons constitutuing it are flowing hence the current density in copper should be more than iron due to greater conductivity of copper. Hence negative charge should accumulate at the junction of copper and iron.
 

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Aishikdesto said:
Hence negative charge should accumulate at the junction of copper and iron.
As the choices indicate, you need to distinguish the two sides of the junction. And don't forget that the conventional current flow is opposite to the electron flow.
 
Could it perhaps be that any charges there are a result of an electrochemical reaction between dissimilar metals?
 
gneill said:
Could it perhaps be that any charges there are a result of an electrochemical reaction between dissimilar metals?
That depends on what the word "joined" connotes. If they have been just kept in contact, should any electrochemical reaction occur?
 
Wrichik Basu said:
That depends on what the word "joined" connotes. If they have been just kept in contact, should any electrochemical reaction occur?
Yes, I'd think so. The metals have different electronegativity, different "standard electrode potentials". Galvanic corrosion comes to mind. I am the first one to admit that I am not an expert in this area, but I remember the terms from high school and college chemistry :smile:
 
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gneill said:
Yes, I'd think so. The metals have different electronegativity, different "standard electrode potentials". Galvanic corrosion comes to mind. I am the first one to admit that I am not an expert in this area, but I remember the terms from high school and college chemistry :smile:
The doubt I have is that the question implies it is related to the applied current. I think I have come across this effect before on this forum.
 
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Sorry guys, as long as the metals just touch each other all this chemical mumbo-jumbo doesn't apply.
 
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gneill said:
Yes, I'd think so. The metals have different electronegativity, different "standard electrode potentials". Galvanic corrosion comes to mind. I am the first one to admit that I am not an expert in this area, but I remember the terms from high school and college chemistry :smile:
It looks like the same principle as this one:
https://www.physicsforums.com/threads/resistor-made-from-two-materials.900569/#post-5670963
Essentially, the voltage gradients in each wire imply different field strengths. Applying Gauss' law to the boundary yields a local static charge.
 
  • #10
Welcome to PF, Aishikdesto.

Aishikdesto said:
...the current density in copper should be more than iron...
This would imply that the current ##I## in the copper is greater than the current ##I## in the iron. This isn't correct under steady-state conditions. Can you see why the currents must be the same? What can you say about the current density ##j##?

##j = \sigma E##
Which material has a greater value of ##\sigma##? A greater value of ##E##?
Apply Gauss' law to an appropriate closed surface.
 
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  • #11
Charles Link said:
Looks to me like the Peltier effect will occur here. See: https://en.wikipedia.org/wiki/Thermoelectric_effect As for excess electrostatic charge build-up, perhaps @Borek has it right when I think he suggests that there isn't any.
I'm reasonably sure my interpretation in post #8 is the right one. TSny seems to agree.
 
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  • #12
For the steady state I totally agree with @TSny post.

But can anyone tell us what happens in the transient state during which the charge at the interface is building up? I think the OP's reasoning at post #1 maybe applies for the transient state.
 
  • #13
haruspex said:
I'm reasonably sure my interpretation in post #8 is the right one. TSny seems to agree.
Yes, I agree. Somehow I overlooked post #8. My post was fairly redundant.
 
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