The electric field of a piecewise uniform 1D charge distribution

[sudera]

Homework Statement

Find the electric field of a line of charge on the x-axis, with charge density $\lambda_1$ (negative) between $x=-a$ and $x=0$ and charge density $\lambda_2$ (positive) between $x=0$ and $x = b$.

Relevant Equations

$\dfrac{dE}{dx} = \dfrac{\lambda}{\epsilon_0}$

This is not really homework, but I'm having trouble understanding it intuitively. I came across this when learning about the space charge layer of a diode. The solution I know simply uses the 1D form of Gauss's law: $\vec{\nabla} \cdot \vec{E}$ = $\dfrac{\rho}{\epsilon_0}$ becomes $\dfrac{dE}{dx} = \dfrac{\lambda}{\epsilon_0}$.
The integral of the left part is an integral of a negative constant, thus will give a declining line. The integral from 0 to the right will give an inclining line (picture below). The electric field is negative everywhere, thus points to the left everywhere. This means that when you placed a positive charge anywhere between $-a$ and $b$, it would accelerate to the left. But my actual questions is this: what if you place it at $x=b-\epsilon$, where $\epsilon$ is a small number? It's much closer to all the positive charges than it is to the negative ones so why doesn't it get repelled to the right?

Thanks in advance.

 

[Orodruin]

The solution is one-dimensional, i.e., it corresponds to infinitely extended charges in the y and z directions. In this scenario, the electric field does not decrease with distance (cf, the electric field of an infinite flat surface with constant charge density).

 

[sudera]

The solution is one-dimensional, i.e., it corresponds to infinitely extended charges in the y and z directions. In this scenario, the electric field does not decrease with distance (cf, the electric field of an infinite flat surface with constant charge density).

But in case of an infinite surface, the electric field is constant perpendicular to the surface, whereas here it is declining. So I don't really understand what you mean.

 

[Orodruin]

whereas here it is declining

No it is not. It is constant. It just is constantly zero since the positive and negative charges are equal and thus contribute with the same magnitude field (but opposite sign).

 

[sudera]

No it is not. It is constant. It just is constantly zero since the positive and negative charges are equal and thus contribute with the same magnitude field (but opposite sign).

What about the electric field "triangle" in the picture then? Perhaps I am interpreting it wrong.

 

[sudera]

The solution is one-dimensional, i.e., it corresponds to infinitely extended charges in the y and z directions. In this scenario, the electric field does not decrease with distance (cf, the electric field of an infinite flat surface with constant charge density).

I think I understand what you meant now. But the question here is the electric field IN or ALONG the line of charge, not perpendicular to it.

 

[archaic]

The solution is one-dimensional, i.e., it corresponds to infinitely extended charges in the y and z directions. In this scenario, the electric field does not decrease with distance (cf, the electric field of an infinite flat surface with constant charge density).

As we get closer to the wire we should consider it as an infinite plane, not an infinite line?

 

[TSny]

Homework Statement:: Find the electric field of a line of charge on the x-axis, with charge density $\lambda_1$ (negative) between $x=-a$ and $x=0$ and charge density $\lambda_2$ (positive) between $x=0$ and $x = b$.

I came across this when learning about the space charge layer of a diode.

Are you sure you are dealing with a line of charge? It seems to me that the space charge would be modeled as negative and positively charged slabs. Then, the $\lambda$'s would be volume charge densities.

 

[sudera]

Are you sure you are dealing with a line of charge? It seems to me that the space charge would be modeled as negative and positively charged slabs. Then, the $\lambda$'s would be volume charge densities.

View attachment 254211

Well, I came across this when learning about diodes, so a slab would be more accurate. However, my textbook solves for the electric field using the equation $\dfrac{dE}{dx} = \dfrac{\lambda}{\epsilon_0}$, and so does this site in section 4.3.3: https://ecee.colorado.edu/~bart/book/book/chapter4/ch4_3.htm
I understand the calculation, but like I said, I don't know why it makes sense intuitively that placing a positive charge at the right border of the positive region wouldn't be repelled to the right by all the positive charges in its proximity to the left.

 

[TSny]

Well, I came across this when learning about diodes, so a slab would be more accurate. However, my textbook solves for the electric field using the equation $\dfrac{dE}{dx} = \dfrac{\lambda}{\epsilon_0}$, and so does this site in section 4.3.3: https://ecee.colorado.edu/~bart/book/book/chapter4/ch4_3.htm
I understand the calculation, but like I said, I don't know why it makes sense intuitively that placing a positive charge at the right border of the positive region wouldn't be repelled to the right by all the positive charges in its proximity to the left.

The slabs of charge can be thought of as made of many thin sheets of charge. So you should consider the nature of the electric field produced by a single thin sheet of charge. I think this is what @Orodruin was alluding to. If a sheet can be approximated as infinite in area, then the E field that it produces does not depend on the distance from the sheet. The magnitude of the force that the sheet would produce on a point charge would be independent of the distance of the point charge from the sheet.

 

[sudera]

The slabs of charge can be thought of as made of many thin sheets of charge. So you should consider the nature of the electric field produced by a single thin sheet of charge. I think this is what @Orodruin was alluding to. If a sheet can be approximated as infinite in area, then the E field that it produces does not depend on the distance from the sheet. The magnitude of the force that the sheet would produce on a point charge would be independent of the distance of the point charge from the sheet.

If it's independent of the distance, then why is the electric field not constant? @Orodruin said it was, in fact, constant and zero, but the electric field in the link I shared and in the picture of the original post isn't zero in the "triangle".

 

[Orodruin]

If it's independent of the distance, then why is the electric field not constant? @Orodruin said it was, in fact, constant and zero, but the electric field in the link I shared and in the picture of the original post isn't zero in the "triangle".

It is not zero there because the total charge on either side is not zero.

 

[TSny]

If it's independent of the distance, then why is the electric field not constant? @Orodruin said it was, in fact, constant and zero, but the electric field in the link I shared and in the picture of the original post isn't zero in the "triangle".

Think of the positive slab of charge as made of many individual sheets, like a deck of cards where each card is a sheet. The magnitude of the field of any particular sheet does not depend on the distance from the sheet. But the direction of the field changes to the opposite direction when you go from one side of the sheet to the other side of the sheet.

Imagine boring a little ways into the deck of cards. The field produced at your location by the cards that you have passed through points oppositely to the field produced by the cards that you have not passed through. So, there's some cancellation of the fields of the cards in the positive slab. The net field of the positive slab no longer cancels the field of the negative slab. So, overall, the field at your location will point in the negative direction.

 

[sudera]

Think of the positive slab of charge as made of many individual sheets, like a deck of cards where each card is a sheet. The magnitude of the field of any particular sheet does not depend on the distance from the sheet. But the direction of the field changes to the opposite direction when you go from one side of the sheet to the other side of the sheet.

Imagine boring a little ways into the deck of cards. The field produced at your location by the cards that you have passed through points oppositely to the field produced by the cards that you have not passed through. So, there's some cancellation of the fields of the cards in the positive slab. The net field of the positive slab no longer cancels the field of the negative slab. So, overall, the field at your location will point in the negative direction.

Oh, this makes a lot of sense, I get it now. Thanks a lot for elaborating on the analogy!