Energy density due to infinite uniform line charges

[Karl86]

Homework Statement

Two infinitely long rods parallel to the $z$-axis have uniform charge density $\lambda$. I need to calculate the energy density $U$ per unit length along the $z$-axis.

Relevant Equations

$\frac{\epsilon_0}{2} \int E^2 d\tau$?

Let $(x_1,y_1)$ and $(x_2,y_2)$ be the point where the rods intersect the $x,y$ plane. I know that on any given point there will be the superpositions of $E_1=\frac{2\lambda}{4\pi \epsilon_0}\frac{1}{(x-x_1)^2+(y-y_1)^2}\hat{r}_1$ and $E_2=\frac{2\lambda}{4\pi \epsilon_0}\frac{1}{(x-x_2)^2+(y-y_2)^2}\hat{r}_2$, which would mean $E^2=\frac{4\lambda^2}{(4\pi\epsilon_0)^2}\frac{1}{(x-x_1)^2+(y-y_1)^2}\frac{1}{(x-x_2)^2+(y-y_2)^2}$. So my thinking was to sum them as vectors and then integrate the square of this sum, which will be a scalar, over the $x,y$ plane, and I though this would give me the energy density per unit length along the $z$-axis. Can you criticize my reasoning? The result should involve a logarithm of the distance between the two rods and that really seems a far cry from what my calculation points to.

 

[PeroK]

Are you sure your relevant equation is valid in this case?

 

[Karl86]

I am not sure it is valid in this case, it can certainly be part of the debate.

 

[PeroK]

I am not sure it is valid in this case, it can certainly be part of the debate.

To get that equation - the integral of $E^2$ - you need to neglect a boundary surface integral, which is not valid if the charge distribution is not bounded.

Do you know another equation for electrostatic energy ?

 

[Karl86]

$\frac{1}{2}\int \lambda V dl?$

 

[Karl86]

My problem with this formula is that I would measure the total energy density due to the rods, not the energy density per unit length along the $z$ axis, as far as I can tell.

 

[PeroK]

My problem with this formula is that I would measure the total energy density due to the rods, not the energy density per unit length along the $z$ axis, as far as I can tell.

The total energy will be infinite. How is energy density related to the integral?

 

[Karl86]

I did not find a paragraph in Griffiths that stated it clearly, but I guess the energy density is actually the integrand?

 

[Karl86]

This was my first, instinctive, idea, but the potential of that configuration is something like
$$ -\frac{\lambda}{2\pi\epsilon_0}\left(\log\frac{r_1}{a}+\log\frac{r_2}{a}\right) $$ whereas the expression I need is something like $C \cdot \log (|r_2-r_1|)$.
Edit: actually I could take $a$ to be #r_2# and that would give $\log (|r_2-r_1|)$. Does it make sense?

 

[PeroK]

This was my first, instinctive, idea, but the potential of that configuration is something like
$$ -\frac{\lambda}{2\pi\epsilon_0}\left(\log\frac{r_1}{a}+\log\frac{r_2}{a}\right) $$ whereas the expression I need is something like $C \cdot \log (|r_2-r_1|)$.
Edit: actually I could take $a$ to be #r_2# and that would give $\log (|r_2-r_1|)$. Does it make sense?

Are you sure that isn't $|\vec{r_2} - \vec{r_1}|$, where these vectors represent the points where the wires cross the $xy$ plane?

PS You could interpret this problem two ways. The way I interpreted it was that you want the energy stored per unit length of wire. The answer you are quoting is consistent with this, as the energy of the configuration is dependent only on the distance between the wires. Note: in this case you need to think about what the variables in the equation for potential of an infinite straight wire actually mean.

Another interpretation would be the energy density of the EM field at points along the z-axis. This may be how you've interpreted the problem. The answer then would depend on the angle between the wires. For example, if the wires were on opposite sides of the z-axis, then the field along the z-axis would be 0.

 

[Karl86]

Are you sure that isn't $|\vec{r_2} - \vec{r_1}|$, where these vectors represent the points where the wires cross the $xy$ plane?

I think it's exactly that, what made you think otherwise? My notation without arrows? I apologise if that's the case. By the way: this expression for the potential confuses me anyway, because my next step is to calculate the electric field of this configuration, assuming the rods are moving in the $x,y$ plane. Should I take the gradient with respect to the $r_1$ or to the $r_2$ coordinates?

 

[PeroK]

I think it's exactly that, what made you think otherwise? My notation without arrows? I apologise if that's the case. By the way: this expression for the potential confuses me anyway, because my next step is to calculate the electric field of this configuration, assuming the rods are moving in the $x,y$ plane. Should I take the gradient with respect to the $r_1$ or to the $r_2$ coordinates?

Moving in the x-y plane?

In any case the field of two rods is just the sum of the two fields by the superposition principle.

 

[Karl86]

Moving in the x-y plane?

In any case the field of two rods is just the sum of the two fields by the superposition principle.

Yes, meaning the two rods move (while staying perpendicular to the $x,y$ plane) and thus the intersection points move. I have to find differential equations for $E_x,E_y$ and show that the component $E_z$ of the electric field satisfies:
$$ \left(-\frac{\partial}{\partial x^2}-\frac{\partial}{\partial y^2}+\epsilon_0 \mu_0 \frac{\partial}{\partial t^2}\right)E_z = 0$$
but $E_z$ should be zero, because the potential $C\cdot \log(|r_2-r_1|)$ does not depend on z. It just seems too trivial. The field I get by taking the gradient wrt both coordinates and adding is:
$$ \mathbf{E}=\left( \frac{x_1-x_2}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}}, \frac{y_1-y_2}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}},\\ \frac{z_1-z_2}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}}\right)$$
but we are on the $x,y$ plane so the last component is $0$.
P.S. The energy density should be understood as the energy in the field between the planes $z=a$ and $z=a+1$ which means our interpretation was correct, right?

 

[PeroK]

The potential you calculated was only valid when the rods were static.

You need to look for the potential formulation for electrodynamics!

Hint: Lorentz Gauge.

 

[Karl86]

We have not done those yet, so I believe there must be a way to prove that $E_z$ obeys that equation and to find equations for $E_x,E_y$ without calculating them explicitly. Do you think that's possible? If you think I should start another thread for this, let me know.

 

[Karl86]

Are you sure that isn't $|\vec{r_2} - \vec{r_1}|$, where these vectors represent the points where the wires cross the $xy$ plane?

PS You could interpret this problem two ways. The way I interpreted it was that you want the energy stored per unit length of wire. The answer you are quoting is consistent with this, as the energy of the configuration is dependent only on the distance between the wires. Note: in this case you need to think about what the variables in the equation for potential of an infinite straight wire actually mean.

Another interpretation would be the energy density of the EM field at points along the z-axis. This may be how you've interpreted the problem. The answer then would depend on the angle between the wires. For example, if the wires were on opposite sides of the z-axis, then the field along the z-axis would be 0.

Actually, the fact that it should be interpreted as the energy stored in the field between the planes of equations $z=a$ and $z=a+1$ for any $a$ makes me think that there needs to be an integration, $\frac{1}{2} \lambda U$ alone is a local quantity. But then I don't understand how it can be so close to the result, this is truly confusing.

 

[PeroK]

Actually, the fact that it should be interpreted as the energy stored in the field between the planes of equations $z=a$ and $z=a+1$ for any $a$ makes me think that there needs to be an integration, $\frac{1}{2} \lambda U$ alone is a local quantity. But then I don't understand how it can be so close to the result, this is truly confusing.

I thought we had solved the first question?

I'm sorry I have not had much time to look at this. In the second part are the rods moving in the z direction?

 

[Karl86]

hi, no, just $x,y$ and they stay parallel to the $z$ axis. That generates an E and a B which I don't need to write explicitly, just to write equations for. For the $z$ component of $E$ I just want to show that
$$
\left(-\frac{\partial}{\partial x^2}-\frac{\partial}{\partial y^2}+\epsilon_0 \mu_0 \frac{\partial}{\partial t^2}\right)E_z = 0
$$
so basically I already have the equation.

 

[PeroK]

hi, no, just $x,y$ and they stay parallel to the $z$ axis. That generates an E and a B which I don't need to write explicitly, just to write equations for. For the $z$ component of $E$ I just want to show that
$$
\left(-\frac{\partial}{\partial x^2}-\frac{\partial}{\partial y^2}+\epsilon_0 \mu_0 \frac{\partial}{\partial t^2}\right)E_z = 0
$$

The problem I see is that symmetry in the z direction should result in $E_z =0$.

 

[Karl86]

That is weird for me too, although the equation would be satisfied (but in a completely trivial way). And if they were indeed moving in the $z$ direction?

 

[Karl86]

Actually: why does the fact that it doesn't move in the z direction imply that $E_z=0$? It should imply that $E_z$ doesn't depend on $z$.

 

[PeroK]

Actually: why does the fact that it doesn't move in the z direction imply that $E_z=0$? It should imply that $E_z$ doesn't depend on $z$.

Because the rod is infinite, we can take any point to have $z=0$. I.e. there's an infinite line of charges above the point mirrored by an infinite line of charges below the point.

We can pair these charges off and study the fields for a pair of charges, the same distance above and below the x-y plane.

If the field at any point has $E_z > 0$, then why is this? What distinguishes the "up" direction? Someone looking from below might see the charges moving to their right, say, and the field point away from the nearer charge. But, someone looking from above (and "upside down") would also see the charges moving to the right, but the field pointing towards the nearer charge.

This is physically incompatible. Hence the field must have $E_z = 0$ everywhere, for every pair of charges, hence for the whole rod.

If we take an example of a magnetic field, say, that points clockwise about the direction of motion for a single charge, then the field for two charges moving in the same direction and the same distance above and below a plane will cancel on the plane.

 

[Karl86]

Because the rod is infinite, we can take any point to have $z=0$. I.e. there's an infinite line of charges above the point mirrored by an infinite line of charges below the point.

We can pair these charges off and study the fields for a pair of charges, the same distance above and below the x-y plane.

If the field at any point has $E_z > 0$, then why is this? What distinguishes the "up" direction? Someone looking from below might see the charges moving to their right, say, and the field point away from the nearer charge. But, someone looking from above (and "upside down") would also see the charges moving to the right, but the field pointing towards the nearer charge.

This is physically incompatible. Hence the field must have $E_z = 0$ everywhere, for every pair of charges, hence for the whole rod.

If we take an example of a magnetic field, say, that points clockwise about the direction of motion for a single charge, then the field for two charges moving in the same direction and the same distance above and below a plane will cancel on the plane.

Ok, and equation-wise what can one say? Taking the curl of one of Maxwell's equations I get
$$ \nabla \times \nabla \times \mathbf{E} = -\frac{\partial}{\partial t} \nabla \times \mathbf{B}$$
from which
$$ \nabla(\nabla \cdot \mathbf{E}) - \nabla^2 E = - \mu_0 \frac{\partial \mathbf{J}}{\partial t} -\mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} $$
but $\mathbf{J}$ should be zero (correct me if I'm wrong) and the gradient of $\nabla \cdot \mathbf{E}=\frac{\rho}{\varepsilon_0}$ should also be zero (although I'm not sure about this, perhaps you can tell me if it is true). So would $E_x, E_y$ satisfy the same equation as $E_z$ or different ones?

 

[PeroK]

Where did you get this problem?

You can, of course, get differential equations for the fields in a vacuum. But, these are always the same equations, independent of the charges and currents elsewhere.

I assume the equations you are looking for are for the specific case of charged rods?

Maybe not?

I'm going offline now. I might have a chance to help later.

 

[Karl86]

Where did you get this problem?

You can, of course, get differential equations for the fields in a vacuum. But, these are always the same equations, independent of the charges and currents elsewhere.

I assume the equations you are looking for are for the specific case of charged rods?

Maybe not?

I'm going offline now. I might have a chance to help later.

Thank, appreciate your help. At this point I am convinced that it's just about getting those differential equations for the generic fields. But I think that this question could also shed light on the matter: if we double the charge density on the rods ($2\lambda$ instead of $\lambda$) how do $E$ and $B$ change?

 

[PeroK]

Ok, and equation-wise what can one say? Taking the curl of one of Maxwell's equations I get
$$ \nabla \times \nabla \times \mathbf{E} = -\frac{\partial}{\partial t} \nabla \times \mathbf{B}$$
from which
$$ \nabla(\nabla \cdot \mathbf{E}) - \nabla^2 E = - \mu_0 \frac{\partial \mathbf{J}}{\partial t} -\mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} $$
but $\mathbf{J}$ should be zero (correct me if I'm wrong) and the gradient of $\nabla \cdot \mathbf{E}=\frac{\rho}{\varepsilon_0}$ should also be zero (although I'm not sure about this, perhaps you can tell me if it is true). So would $E_x, E_y$ satisfy the same equation as $E_z$ or different ones?

You have derived a general equation for $\vec{E}$ in any region where there is no charge or current. In this case, we can see that $\vec{E}$ is independent of $z$ - although, in this case, we can also see that $E_z = 0$.

That's about all there is to it.

 

[Karl86]

This was my first, instinctive, idea, but the potential of that configuration is something like
$$ -\frac{\lambda}{2\pi\epsilon_0}\left(\log\frac{r_1}{a}+\log\frac{r_2}{a}\right) $$ whereas the expression I need is something like $C \cdot \log (|r_2-r_1|)$.
Edit: actually I could take $a$ to be #r_2# and that would give $\log (|r_2-r_1|)$. Does it make sense?

Actually, one last thing. Here I somehow convinced myself I would get $\log (|r_2-r_1|)$ but I actually don't, do I? $$ -\frac{\lambda}{2\pi\epsilon_0}\left(\log\frac{r_1}{a}+\log\frac{r_2}{a}\right) = -\frac{\lambda}{2\pi\epsilon_0}\left(\log(r_2) - \log(r_1) \right)$$ which is quite different from $\log{|\vec{r_2}-\vec{r_1}|}$. Can you see how I would get $\log{|\vec{r_2}-\vec{r_1}|}$ or did you just trust me when I wrote it :D?

 

[PeroK]

Actually, one last thing. Here I somehow convinced myself I would get $\log (|r_2-r_1|)$ but I actually don't, do I? $$ -\frac{\lambda}{2\pi\epsilon_0}\left(\log\frac{r_1}{a}+\log\frac{r_2}{a}\right) = -\frac{\lambda}{2\pi\epsilon_0}\left(\log(r_2) - \log(r_1) \right)$$ which is quite different from $\log{|\vec{r_2}-\vec{r_1}|}$. Can you see how I would get $\log{|\vec{r_2}-\vec{r_1}|}$ or did you just trust me when I wrote it :D?

The variable, $r$, in the potential for a charged rod is the distance from the rod.
$$V(r) = -\frac{\lambda}{2\pi \epsilon_0} \ln(\frac{r}{a})$$

This potential, however, does not go to $0$ at $\infty$. You need to be careful, therefore, if you use this potential to find the energy of an electrostatic configuration. What you need is the potential difference between $\infty$ and the point at which you want to evaluate the potential. That's one issue.

But, this potential difference is infinite for an infinite wire. What you want is the energy per unit length. So, you need to find the energy per unit length for a finite wire and take the limit as the length of the wire goes to infinity.

There might be a neat trick to use the logarithmic potential directly, but I don't see it.

 

[Karl86]

The variable, $r$, in the potential for a charged rod is the distance from the rod.
$$V(r) = -\frac{\lambda}{2\pi \epsilon_0} \ln(\frac{r}{a})$$

This potential, however, does not go to $0$ at $\infty$. You need to be careful, therefore, if you use this potential to find the energy of an electrostatic configuration. What you need is the potential difference between $\infty$ and the point at which you want to evaluate the potential. That's one issue.

But, this potential difference is infinite for an infinite wire. What you want is the energy per unit length. So, you need to find the energy per unit length for a finite wire and take the limit as the length of the wire goes to infinity.

There might be a neat trick to use the logarithmic potential directly, but I don't see it.

But that gives something like $C\cdot \log\frac{a + \sqrt{x^2+a^2}}{-a+\sqrt{x^2+a^2}}$ assuming a symmetric rod, times some constant, $1/2 \lambda$ and that limit is still infinite for $a \rightarrow \infty$...

 

[PeroK]

But that gives something like $C\cdot \log\frac{a + \sqrt{x^2+a^2}}{-a+\sqrt{x^2+a^2}}$ assuming a symmetric rod, times some constant, $1/2 \lambda$ and that limit is still infinite...

Yes, it's tricky. One idea is to go back to your original idea of using the electric field!