Electric field in an infinite layer

[Bestfrog]

Homework Statement

In the layer $0<z<a$ there is a uniform and constant density of charge $\rho>0$. In the layer $-a<z<0$ the density of charge is $\rho<0$.
What is the total electric field in the space?

The Attempt at a Solution

By the Gauss law I find that if $z \geq a$ or $z \leq a$ the electric field is 0. But inside the total layer how can I procede? I rebember that there must be a term $a^2$(!), but I'm not sure. If I integrate I have $E_+=\int_0^a{\frac{\rho}{2 \epsilon_0}}$, the same with the $E_-$. The total is $E=\frac{\rho a}{\epsilon_0}$.

 

[TSny]

By the Gauss law I find that if $z \geq a$ or $z \leq a$ the electric field is 0.

OK. If you want us to check your reasoning for this result, please post your work.

But inside the total layer how can I procede?

Let P be an arbitrary point inside the positively charged portion of the slab. Choose an appropriate Gaussian surface that goes through this point and also allows you to make use of the fact that E = 0 outside the slab.

 

[Bestfrog]

OK. If you want us to check your reasoning for this result, please post your work.

Ok. By the Gauss' law I have that $\int \vec{E} \vec{dA} = \frac{q}{\epsilon_0}$. If I take a Gaussian surface. (a cylinder for example) of height $H>2a$, the net charge in the cylinder is 0, so the electric field is 0.
Inside my thought was to find the electric field due to a infinite layer of thickness $dz$, so $dE= \frac{\sigma}{2 \epsilon_0}= \frac{\rho dz}{2 \epsilon_0}$ and then integrate, but I think that is not correct.
About your hint, I have no idea how to choose an appropriate Gaussian surface... If I work only in the positive layer (considering that there is no negative layer) I can state that in $z=a/2$ the electric field is 0..

 

[TSny]

Ok. By the Gauss' law I have that $\int \vec{E} \vec{dA} = \frac{q}{\epsilon_0}$. If I take a Gaussian surface. (a cylinder for example) of height $H>2a$, the net charge in the cylinder is 0, so the electric field is 0.

If the net charge is zero inside a Gaussian surface, then the total flux through the surface is zero. But that doesn't necessarily imply that the field E is zero on the surface. So, there must be more to the argument.

Inside my thought was to find the electric field due to a infinite layer of thickness $dz$, so $dE= \frac{\sigma}{2 \epsilon_0}= \frac{\rho dz}{2 \epsilon_0}$ and then integrate, but I think that is not correct.

Yes, you could approach it this way. Actually, I think this is a good approach for showing that E is zero at points outside the charge distribution.

About your hint, I have no idea how to choose an appropriate Gaussian surface... If I work only in the positive layer (considering that there is no negative layer) I can state that in $z=a/2$ the electric field is 0..

Suppose you want to find E at a point P located as shown. Try to use Gauss's law. Think about the shape of a Gaussian surface that will get the job done. Assume that E = 0 outside the charge distribution (even though you still need to prove that).

 

[Bestfrog]

Suppose you want to find E at a point P located as shown. Try to use Gauss's law. Think about the shape of a Gaussian surface that will get the job done. Assume that E = 0 outside the charge distribution (even though you still need to prove that).

If I take a cube of side a, at height z. The surface I consider is the down face of the cube. I have $E \cdot a^2=\frac{\rho a^2 \cdot (a-z)}{\epsilon_0}$(?)

 

[TSny]

If I take a cube of side a, at height z. The surface I consider is the down face of the cube. I have $E \cdot a^2=\frac{\rho a^2 \cdot (a-z)}{\epsilon_0}$(?)

Yes, good. Be sure to note the direction of E at P.

 

[Bestfrog]

If the net charge is zero inside a Gaussian surface, then the total flux through the surface is zero. But that doesn't necessarily imply that the field E is zero on the surface. So, there must be more to the argument.

Maybe I can imagine to collect all the charge in a plane with no thickness, so I have 2 planes, one at height $z=a$ with positive charge and the other at $z=-a$ with negative charge, then the electric field if $z>a, z<-a$ is $E=\frac{\sigma'}{2 \epsilon} + \frac{- \sigma'}{2 \epsilon}=0$

 

[TSny]

OK. But how do you justify collecting the charge that way?

 

[Bestfrog]

OK. But how do you justify collecting the charge that way?

My first thought was to "collect" it at the "centre line" of each layer (so at $z=a/2$ and $z=-a/2$) but it would have been wrong, because I have to mantain the symmetry of the problem, so I collected them in the line where the electric field is maximum (negative or positive). It's more an intuitive explanation than a logical one.

 

[Bestfrog]

Yes, good. Be sure to note the direction of E at P.

The vector is $\vec{E} =- (a-z) \frac{\rho}{\epsilon_0} \hat{z}$ ($\hat{z}$ is the positive versor along the z-axys)

 

[TSny]

The vector is $\vec{E} =- (a-z) \frac{\rho}{\epsilon_0} \hat{z}$

Looks good. Does this expression also hold for points in the negatively charged slab?

 

[Bestfrog]

Looks good. Does this expression also hold for points in the negatively charged slab?

Yes, because it would be $a+|z|$ but z is negative so $a-z \rightarrow a+z$ (if z is negative)

 

[TSny]

My first thought was to "collect" it at the "centre line" of each layer (so at $z=a/2$ and $z=-a/2$) but it would have been wrong, because I have to mantain the symmetry of the problem, so I collected them in the line where the electric field is maximum (negative or positive). It's more an intuitive explanation than a logical one.

Suppose you had just two infinite planes of charge separated by a distance d as shown. Top plane has uniform charge density σ₁ and the bottom plane has charge density σ₂,

Find E at point P located a distance x above the upper plane. Does the answer depend on d or x?

 

[Bestfrog]

Find E at point P located a distance x above the upper plane. Does the answer depend on d or x?

Actually, it doesn't depend on d neither on x, because the field generated by an infinite plane is uniform in every point of the space.

 

[TSny]

Yes, because it would be $a+|z|$ but z is negative so $a-z \rightarrow a+z$ (if z is negative)

I'm not sure I follow. Shouldn't the field at $z = -a/2$ be the same as the field at $z = +a/2$? Also, shouldn't E be zero at both $z = a$ and $z = -a$?

 

[Bestfrog]

I'm not sure I follow. Shouldn't the field at $z = -a/2$ be the same as the field at $z = +a/2$? Also, shouldn't E be zero at both $z = a$ and $z = -a$?

Yeah, I'm wrong, the final expression of the field is $\vec{E}=(|z| -a)\frac{\rho}{\epsilon} \hat{z}$

 

[TSny]

Yeah, I'm wrong, the final expression of the field is $\vec{E}=(|z| -a)\frac{\rho}{\epsilon} \hat{z}$

Yes, nice.

 

[Bestfrog]

Yes, nice.

Then, integrating $$E=\int_{a}^{-a} (|z| -a)\frac{\rho}{\epsilon} = \frac{2 \rho}{\epsilon} \int_{a}^{0} (z-a) = \frac{\rho a^2}{\epsilon}$$
Thank you!

 

[TSny]

Then, integrating $$E=\int_{a}^{-a} (|z| -a)\frac{\rho}{\epsilon} = \frac{2 \rho}{\epsilon} \int_{a}^{0} (z-a) = \frac{\rho a^2}{\epsilon}$$

I don't follow why you are integrating.

The expression $\vec{E}=(|z| -a)\frac{\rho}{\epsilon} \hat{z}$ is already the total electric field at $z$.

 

[Bestfrog]

I don't follow why you are integrating.

The expression $\vec{E}=(|z| -a)\frac{\rho}{\epsilon} \hat{z}$ is already the total electric field at $z$.

I wanted to find the total electric field in the zone $-a \leq z \leq a$ (maybe I didn't write it in the statements).

 

[TSny]

I wanted to find the total electric field in the zone $-a \leq z \leq a$ (maybe I didn't write it in the statements).

$\vec{E}=(|z| -a)\frac{\rho}{\epsilon} \hat{z}$ is already the total electric field at a point inside the slab. It doesn't make sense to add together electric fields at different points. So, there is no meaning to trying to add together the electric field at all the points within the slab.

I'm wondering if you might be thinking of using the expression for E to get the potential difference $\Delta V$ between two points. Then you would integrate E along a path between the two points.

 

[Bestfrog]

$\vec{E}=(|z| -a)\frac{\rho}{\epsilon} \hat{z}$ is already the total electric field at a point inside the slab. It doesn't make sense to add together electric fields at different points. So, there is no meaning to trying to add together the electric field at all the points within the slab.

I'm wondering if you might be thinking of using the expression for E to get the potential difference $\Delta V$ between two points. Then you would integrate E along a path between the two points.

Yes, the original problem asks if a point of charge $q$, mass $m$ and a speed $v$ can go through this layer and asks what is the final speed (so I integrate to find $\Delta V$ and then the work done against the particle)

 

[TSny]

Ahhhh. So, if the particle goes all the way through the layer, you will need to find $\Delta V$ for going all the way through. How would you do that?

 

[Bestfrog]

Ahhhh. So, if the particle goes all the way through the layer, you will need to find $\Delta V$ for going all the way through. How would you do that?

I know that $dV=-E dr$, then the work done against the particle is $L=q \Delta V=q \int E dr$(?)

 

[TSny]

I know that $dV=-E dr$, then the work done against the particle is $L=q \Delta V=q \int E dr$(?)

Yes, that's the general idea.