Resistor made from two materials

[RoboNerd]

Hi everyone, I am having problems figuring out how to solve a particular problem.

Because I was not able to access PhysicsForums recently, I typed everything about my issues into a pdf file that I have attached below.

I am grateful for your help in advance, and make it a great day!

Everything in the attached file matches the given template.

Homework Statement

Everything is described in the pdf file attached.

Homework Equations

Everything is described in the pdf file attached.

The Attempt at a Solution

Everything is described in the pdf file attached.

 

[haruspex]

What did you get for the two electric fields? They're different, right? If the field changes abruptly across the boundary, what can you deduce?

 

[RoboNerd]

For my two magnetic fields, I have the following results:

E1 = rho * I / A
E2 = 3 * rho * I /A

Now, I do not have any idea what I need to do if the field changes across a boundary abruptly. If the boundary was a conductor, then there would be a separation of charges on the conductor (boundary) and the electric field would be zero inside it, but would remain the same around it.

Apart from that, I do not know.

The only thing that could be similar to this is the concept of a dielectric that reduces the electric field inside the capacitor. Maybe the boundary is a dielectric, and thus reduces the electric field across the boundary?Thanks for being willing to help.

 

[haruspex]

For my two magnetic fields, I have the following results:

E1 = rho * I / A
E2 = 3 * rho * I /A

Now, I do not have any idea what I need to do if the field changes across a boundary abruptly. If the boundary was a conductor, then there would be a separation of charges on the conductor (boundary) and the electric field would be zero inside it, but would remain the same around it.

Apart from that, I do not know.

The only thing that could be similar to this is the concept of a dielectric that reduces the electric field inside the capacitor. Maybe the boundary is a dielectric, and thus reduces the electric field across the boundary?Thanks for being willing to help.

The question implies there is a constant layer of charge at the boundary, in addition to the charge flowing through. What field is generated by a sheet of charge? (You can treat it as wide since the wire will tend to corral the field lines into being parallel.)

 

[RoboNerd]

The question implies there is a constant layer of charge at the boundary, in addition to the charge flowing through. What field is generated by a sheet of charge? (You can treat it as wide since the wire will tend to corral the field lines into being parallel.)

How it imply that there is a constant layer of charge at the boundary? Even if there is constant charge on the boundary, how is that even possible since the charge is flowing through.

Well, per Gauss's Law for a sheet of charge, the electric field generated by a sheet of charge is going to be sigma/(2* e0).
The electric fields #'s 1 and 2 are thus different due to the effect of the additional electric field created by the charge.

In this case, the Electric Field caused by the charge is going to be 2* rho * I /A, right?

 

[haruspex]

Even if there is constant charge on the boundary, how is that even possible since the charge is flowing through.

A river flows into the upstream end of a lake and out at the downstream end. Water flows through, but there is also a constant volume pool of water.

the Electric Field caused by the charge is going to be 2* rho * I /A, right?

The electric field due to the charge will create that difference. But remember the field from the sheet points both ways.

 

[RoboNerd]

A river flows into the upstream end of a lake and out at the downstream end. Water flows through, but there is also a constant volume pool of water.

That is right. Thanks for fixing this misconception of mine.

But remember the field from the sheet points both ways.

Since the current through section 1 is going towards the slab of charge, this means that the electric field E1 is directed away from the slab of charge. Thus the electric field of the charge is ADDED to the existing electric field in section 1 to get the final field E1.

Since the current going through section 2 is going away from the slab of charge, it means that E2 is directed towards the slab of charge. Thus the electric field of the slab of charge opposes the existing electric field in section 2 to get the final field E2. Right?

Thanks for the help!

 

[haruspex]

That is right. Thanks for fixing this misconception of mine.Since the current through section 1 is going towards the slab of charge, this means that the electric field E1 is directed away from the slab of charge. Thus the electric field of the charge is ADDED to the existing electric field in section 1 to get the final field E1.

Since the current going through section 2 is going away from the slab of charge, it means that E2 is directed towards the slab of charge. Thus the electric field of the slab of charge opposes the existing electric field in section 2 to get the final field E2. Right?

Thanks for the help!

Yes.. not completely sure about the signs/directions, but the basic idea is right: |E1-E2| will be |sigma|/(e0).

 

[RoboNerd]

And then substitute sigma for Q/A and solve for Q!

I get it. Thanks a lot for the help!

The question implies there is a constant layer of charge at the boundary

The final problem I have is understanding how the question implies that there is a constant layer of charge at the boundary.

 

[haruspex]

The final problem I have is understanding how the question implies that there is a constant layer of charge at the boundary.

It asks you to calculate the charge at the boundary, so presumably there is one.