Find the density of surface charges at the boundary of two conductors with different resistivities

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SUMMARY

The discussion centers on calculating the density of surface charges at the boundary of two conductors with differing resistivities using Gauss's Law. The equation derived is $$(\frac{\rho_{2} I}{A} - \frac{\rho_{1} I}{A})A = \frac{\sigma A}{\epsilon_{0}}$$, which is valid but requires knowledge of the cross-sectional area, ##A##, to determine the surface charge density, ##\sigma##. The participants confirm that the boundary condition for the electric field's normal component at the interface is correctly derived, though there is confusion regarding the units of the answer provided in the key, which should refer to charge rather than charge density.

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tellmesomething
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Homework Statement
Two cylindrical conductors with equal cross sections and different resistivities p1 and p2 are put end to, end as shown in figure-3.11.
Find the density of surface charges at the boundary of the conductors if a current I flows from conductor I to conductor 2 .
Relevant Equations
None
I tried using gauss law. This is how i set it up: I took a gaussian pill box of same area and very small length dx, the diagram is amplified for obvious reasons.

Assuming there's a constant field in the yellow portion ##E_{1}## and a constant field ##E_{2}## in the green portion. I assume the area of cross section as ##A## and surface charge density as ##\sigma##

$$(\frac{\rho_{2} I}{A} - \frac{\rho_{1} I}{A})A = \frac{\sigma A}{\epsilon_{0}}$$

Is this valid? Because it doesnt get me the answer as i do not know what ##A## is.

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Your work looks correct
tellmesomething said:
$$(\frac{\rho_{2} I}{A} - \frac{\rho_{1} I}{A})A = \frac{\sigma A}{\epsilon_{0}}$$
This looks good.

tellmesomething said:
Is this valid? Because it doesnt get me the answer as i do not know what ##A## is.
The answer for ##\sigma## will depend on the cross-sectional area ##A##. The problem statement is a little inconsistent in not providing a symbol for the cross-sectional area while it does provide symbols for the current and resistivities.
 
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You have derived the boundary condition for the discontinuity of the normal component of the electric field at the interface between conductors. That is correct. You are also correct in saying that you need the area. Think of it this way, two setups like this carrying the same current ##I## but having different cross sectional areas will have different surface charge densities at the interface.

On edit:
I see that @TSny has preempted me (again.) I will leave this post up as additional confirmation.
 
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TSny said:
Your work looks correct

This looks good.


The answer for ##\sigma## will depend on the cross-sectional area ##A##. The problem statement is a little inconsistent in not providing a symbol for the cross-sectional area while it does provide symbols for the current and resistivities.
Yes i did a unit check as well and the answer given in the key is ##\epsilon_0(\rho_{2}-\rho_{1})I## , the unit comes out to be Coulumb instead of Coulumb per sqaured meter. I guess they meant charge instead of charge density.
 
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kuruman said:
You have derived the boundary condition for the discontinuity of the normal component of the electric field at the interface between conductors. That is correct. You are also correct in saying that you need the area. Think of it this way, two setups like this carrying the same current ##I## but having different cross sectional areas will have different surface charge densities at the interface.

On edit:
I see that @TSny has preempted me (again.) I will leave this post up as additional confirmation.
Yes thankyou :)
 
tellmesomething said:
I guess they meant charge instead of charge density.
Or current density instead of current?
 
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haruspex said:
Or current density instead of current?
Yes possible
 

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