Calculating the Electric field inside an infinite planar slab using Gauss' Law

[Jaccobtw]

Homework Statement

An infinite planar "slab" of charge sits in the x-y plane. It has a thickness of 10cm and a volume charge density ρ=0.005z^2, where the constant 0.005 has units C/m^5 and z=0 is the center of the slab. What is the electric field magnitude in N/C at z=2cm, inside the slab?

Relevant Equations

$$\oint_{}^{} E \cdot dA = \frac {q_e}{\epsilon_o}$$

Draw a Gaussian pill box that starts from 0 (half way between the slab) and extends towards 2 cm.$$A \times \int_{0}^{0.02} \rho dz$$

I'm not sure if I should multiply the integral by A (area) or V (volume)

And if area would I multiply by 0.02^2?

I'm confused here. Thanks for your help.

 

[Orodruin]

The first thing you should ask yourself is not what the integral is, but why you are using that particular integral and how you are going to use the result to conclude something about the electric field.

Once you have done that, the flux out of the closed surface is equal to the volume integral of the charge density divided by $\epsilon_0$ and from this you should be able to conclude what the form of the integral is.

 

[Jaccobtw]

The first thing you should ask yourself is not what the integral is, but why you are using that particular integral and how you are going to use the result to conclude something about the electric field.

Once you have done that, the flux out of the closed surface is equal to the volume integral of the charge density divided by $\epsilon_0$ and from this you should be able to conclude what the form of the integral is.

I end up getting EA = 6.67 x 10^-7/($\epsilon_o$)

I got this number by taking the integral of the charge density from 0 to 0.02. Do you know how I find the area?

 

[Orodruin]

I end up getting EA = 6.67 x 10^-7/(ϵo)

You still have not provided your reasoning. Reasoning with regards to your choice of Gaussian surface and the resulting field is absolutely essential. You also need to put units in your answer.

Please write down your detailed argumentation from start to end.

 

[Jaccobtw]

You still have not provided your reasoning. Reasoning with regards to your choice of Gaussian surface and the resulting field is absolutely essential. You also need to put units in your answer.

Please write down your detailed argumentation from start to end.

The electric field at the center of of the 10cm thick slab must be zero because the charge above cancels with the charge below. According to the charge density $\rho = 0.005z^2$, the charge density increases as I go from the center to the outside by a squared factor. I can use a gaussian surface to enclose the slab up to 2 cm because the question asks for what the electric field is at 2 cm. The surface I will use is a pill box because 4 of the 6 faces have an electric flux of zero because the faces are perpendicular to the electric field. That leaves 2 faces who's areas are parallel with the electric flux. If we integrate the charge density from 0 to 2cm it will give us the total charge enclosed from 0 to 2 cm. Now we have q/$\epsilon_o$. We have 2 areas to divide by. and now we have the electric field. Anything wrong with my reasoning?

 

[Orodruin]

According to the charge density ρ=0.005z2,

Please use units. They will help you make consistency checks.

because the faces are perpendicular to the electric field.

Here you need to solidify your argument by showing that the field is perpendicular to those sides (ie, you must provide an argument for why the field must be in the z-direction).

That leaves 2 faces who's areas are parallel with the electric flux.

The surface normals are parallel to the field, so the field is orthogonal to the surfaces.

If we integrate the charge density from 0 to 2cm it will give us the total charge enclosed from 0 to 2 cm.

Please write down the integral you refer to.

Now we have q/ϵo. We have 2 areas to divide by. and now we have the electric field.

This is not a precise wording. You need to argue for the flux value on both sides. Also note that if you use the electric field multiplied by area, then you must argue that the field is constant on the entire surface.

Write down the value you obtain for the flux integral, defining any quantities involved.

 

[Jaccobtw]

Please use units. They will help you make consistency checks.

The charge density is ρ=0.005z^2 C/m^3 because $\rho$ is always the volume charge density. Because a function is given for the charge density it tells us that as z increases the volume charge density increases.

Here you need to solidify your argument by showing that the field is perpendicular to those sides (ie, you must provide an argument for why the field must be in the z-direction).

The field is in the z direction because the plane is infinite in the x and y directions. The infinite charge in those directions cancel out. So the field points up and down (z direction).

The surface normals are parallel to the field, so the field is orthogonal to the surfaces.

yeah I meant that the area vectors are parallel. Or at least that's how prof taught us. And because of this I can use: $$EA = \frac{q}{\epsilon_o}$$

Because we have Area on the left side we need it to cancel out since we're dealing with volume. To get area on the right side we split the charge into sigma times Area. To get sigma we take the integral of rho. Now the area will cancel on both sides and we're left with $\sigma/2 \epsilon_o$. From there you just plug in numbers

Please write down the integral you refer to.

$$\int_{-0.02}^{0.02} \rho dz$$

which gives 2.67 x 10e-8 for sigma

plug into $\sigma/2 \epsilon_o$

and you get ~1508 N/C which works as the right answer but I'm concerned with the quality of my reasoning.

 

[Orodruin]

The charge density is ρ=0.005z^2 C/m^3 because ρ is always the volume charge density.

No. This would be true if z were dimensionless, but z has dimensions of length.

And because of this I can use: EA=qϵo

You need to be much more careful in your argument here. What is A? Where is the field defined? Why is it constant across the surface?

To get area on the right side we split the charge into sigma times Area. To get sigma we take the integral of rho.

I would recommend against doing this as it is not needed and it is possibly confusing. Just go with the definition of the charge inside your surface. The area will come out automatically.

which gives 2.67 x 10e-8 for sigma

Units!

 

[Jaccobtw]

Hi I wanted to revisit this and finish the problem to see if my reasoning skills have improved since.

So to start I think it's important to define $q_{enc}$. We are given $\rho$ as a function of z. $\rho$ has the constant in it with units of C/m^5 with z^2 over it which has two dimensions of length so the "net" units are C/m^3.

This tells us that to get $q_{enc}$ we need to multiply $\rho$ by dV because the charge enclosed varies based on the volume of the function.
$$q_{enc} = \rho dV$$
The charge density is constant in the two infinite dimensions in the infinite planar slab and varies in the finite dimension from z = 0 to z = 5cm and z= 0 to z = -5cm. I chose a Gaussian pillbox as my surface centered at z = 0 where two dimensions are constant (so for simplicity sake I make them 1m each in length) and the third dimension, $dz$, extends in the finite dimension (from z = 0 to z = 5cm and z= 0 to z = -5cm):
$$ dV = (1m)(1m)dz$$
$$dV = dz$$
The reason we want it centered a z = 0 is so that we can have symmetry for dz. This way the electric field is constant at the top and bottom surfaces of the pillbox (the other four surfaces facing the infinite planar directions have no electric flux because the electric field cancels out ). We will want to now set up an integral from -2cm to 2cm to add up the electric flux of the enclosed charge. Divide both sides by the two 1 by 1 square areas which gives me the electric field:

$$ E = \frac{1}{2\epsilon_o} \int_{-0.02}^{0.02} \rho dz$$

I got about 1507 N/C as my final answer.

 

[vanhees71]

I think the first thing you have to think about is, how the electric field should look, given the symmetries of the problem.

It's of course much simpler to use the differential form of the electrostatic equations, i.e.,
$$\vec{E}(\vec{x})=-\vec{\nabla} \Phi(\vec{x}), \quad -\Delta \Phi(\vec{x})=\frac{1}{\epsilon_0} \rho(\vec{x}).$$
Also here, first think, what the symmetries of the problem imply for $\Phi$.

 

[kuruman]

I got about 1507 N/C as my final answer.

That is correct. However you could have reached that answer more simply if you pulled together what you already know without using numbers.

1. Choosing one of the two Gaussian pillbox faces that are perpendicular to the field at $z=0$ gives only one face to consider at $z_0=2~$cm because the electric field is zero at $z=0$.
2. The left side of the integral form of Gauss's law is always $EA$ where $A$ is the area of the face.
3. If you choose a pillbox of uniform area, the charge enclosed by a slab of thickness $dz$ is $dq=A \rho dz$.

With all this in mind, the integral form of Gauss's law becomes $$E\cancel{A}=\cancel{A}\frac{1}{\epsilon_0}\int_{0}^{z_0}\rho~dz.$$ N.B. Your method of taking half the integral from (-0.02 cm) to (+0,.02 cm) will not work if the volume charge distribution were an odd function of $z$, e.g. $\rho=0.005 z^3$. Can you see why?

 

[vanhees71]

Hm, but as was correctly written in the OP, Gauß's Law in integral form correctly reads
$$\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E}=\frac{1}{\epsilon_0} \int_V \mathrm{d}^3 x \rho.$$
It's not clear to me, how you get your (correct) solution!

 

[Jaccobtw]

Your method of taking half the integral from (-0.02 cm) to (+0,.02 cm) will not work if the volume charge distribution were an odd function of $z$, e.g. $\rho=0.005 z^3$. Can you see why?

Thanks for showing me a much simpler way. I think it's because they would cancel each other out. For example, taking the integral you'd end up with some constants times $z^4 - z^4$ which would give you zero. I think.

 

[kuruman]

Thanks for showing me a much simpler way. I think it's because they would cancel each other out. For example, taking the integral you'd end up with some constants times $z^4 - z^4$ which would give you zero. I think.

That is correct thinking. More generally, a function $f(z)$ is

1. even if $f(-z)=f(z)$. Example: $f(z)=z^2+ \cos\! z.$
2. odd if $f(-z)=-f(z)$ Example: $f(z)=z^5+ \sin\! z.$

An even function integrated over symmetric limits, e.g. from $-z_0$ to $+z_0$, will give you twice the integral from $0$ to $+z_0$ because the area under the curve on the negative side of the $z$-axis is equal to the area under the curve on the positive side of the $z$-axis.

An odd function integrated over symmetric limits will give you zero because the area under the curve on the negative side of the $z$-axis is the negative of the area under the curve on the positive side of the $z$-axis.

 

[vanhees71]

As I tried to say, it's much simpler to use the local form of electrostatic laws and a symmetry argument on the electrostatic potential.

The equations read
$$\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \rho.$$
From the first equation it's clear that (at least locally)
$$\vec{E}=-\vec{\nabla} \Phi,$$
and the 2nd equation implies that
$$\Delta \Phi=-\frac{1}{\epsilon_0} \rho.$$
By symmetry, i.e., because $\rho$ depends only on $z$, it's clear that an ansatz with $\Phi=\Phi(z)$ should be a good idea.

This simplifies the Poisson equation to
$$\Phi''(z)=-\frac{1}{\epsilon_0} \rho(z).$$
Now
$$\rho(z)=\begin{cases} \alpha z^2 &\text{for} \quad z \in (-a,a),\\ 0 & \text{for} z \notin (-a,a). \end{cases}$$
where $a=5 \; \text{cm}$ and $\alpha=5 \cdot 10^{-3} \text{C}{\text{m}^5}$ (though I don't care about these values, which you can plug in at the very end of the calculation).

So for $-a<z<a$ we have
$$\Phi''(z)=-\frac{\alpha}{\epsilon_0} z^2 \; \Rightarrow \; \Phi(z)=-\frac{\alpha}{12 \epsilon_0} z^4 + C_1 + C_2 z.$$
For $z>a$
$$\Phi''(z)=0\; \Rightarrow \; \Phi(z)=C_1'+C_2' z.$$
Because $\rho(z)=\rho(-z)$ we can also assume that $\Phi(z)=\Phi(-z)$. This leads to $C_2=0$. Further $\Phi$ is determined only up to an additive constant. So we can also set $C_1=0$.

Since there are no surface-charge densities, $\Phi$ as well as $\Phi'$ must be continuous, which means that
$$\Phi(a-0^+)=-\frac{\alpha}{12 \epsilon_0} a^4=\Phi(a+0^+)=C_1' + C_2' a \qquad (1)$$
and
$$\Phi'(a-0^+)=-\frac{\alpha}{3 \epsilon_0} a^3 = \Phi'(a+0^+)=C_2'. \qquad (2)$$
Plugging this into (1) leads to
$$C_1'-\frac{\alpha}{3 \epsilon_0} a^4=-\frac{\alpha}{12 \epsilon_0}a^4 \; \Rightarrow \; C_1'=\frac{\alpha}{4 \epsilon_0}a^4.$$
For $z<a$ we can continue the solution for $z>a$ as an even function:
$$\Phi(z)=C_1'-C_2' z=C_1' + C_2'|z|.$$
So finally we have
$$\Phi(z) = \begin{cases}
-\frac{1}{12 \epsilon_0} z^4 &\text{for} \quad |z|<a, \\
-\frac{\alpha}{3 \epsilon_0} a^3 |z| + \frac{\alpha}{4 \epsilon_0}a^4 &\text{for} \quad |z| \geq a
\end{cases}$$
and
$$\vec{E}=-\vec{\nabla} \Phi = \begin{cases}\frac{1}{3 \epsilon_0} z^3 \vec{e}_z&\text{for} \quad |z|<a, \\
\frac{\alpha}{3 \epsilon_0} a^3 \text{sign} z \vec{e}_z &\text{for} \quad |z| \geq a.\end{cases}$$

 

[nasu]

This is the simpler way?

 

[vanhees71]

It's simpler, because it gives all the arguments, why the field must come out this way without much handwaving.

 

[kuruman]

It's simpler, because it gives all the arguments, why the field must come out this way without much handwaving.

Solving Poisson's equation, which is normally done at the intermediate level, is certainly more rigorous than using the integral form of Gauss's law which is taught at the introductory level.

 

[nasu]

It's simpler, because it gives all the arguments, why the field must come out this way without much handwaving.

Using Gauss's law and symmetry arguments is handwaving? Don't you use symmetry too?

 

[vanhees71]

Of course I use symmetry to find the right ansatz for the potential. In this thread at least there was no clear statement, how the electric field must look given the symmetries. Of course you can argue from the symmetry that $\Phi=\Phi(z)$ and thus $\vec{E}=E(z) \vec{e}_z$ and then use Gauss's Law in integral form.