I E-field around charged finite wire - intractable or not?

[Anachronist]

TL;DR Summary

Trying to find field intensity at a point some distance from a finite charged wire with a different charge density on each end. Not exactly electrostatics due to falloff not being inverse square.

I have a problem I'm trying to solve for metaball graphics application. I'm going to use an electric field analogy here, but it isn't actually an electric field because the density falloff is $1/r^s$ rather than $1/r^2$, where $s$ is some arbitrary constant like 3.6 that determines the falloff rate. I am also not concerned with units and physical constants; I am interested in the general expression, so constants like $4\pi\epsilon_0$ can be omitted.

Imagine a straight finite wire of length $L$ in 3D space centered on the z axis. The wire has a total charge $q$, but the charge density is different at each end, changing linearly in between (say, $q_1$ on one end and $q_2$ on the other). While the wire has an infinite number of points along its length, in the illustration below, five points along the wire are shown, with the isosurfaces of constant field intensity shown around each point as if each was in isolation.

This is radially symmetric around the z axis so we can work in 2D.

I want to find the sum of e-field contributions from all points along the wire at point $p$ in the figure above. Each point along the wire has a different distance from $p$, and a different contribution to the field due to its distance and its charge density. It seems like some sort of integration problem but I am not sure where to start.

Is this an intractable problem for obtaining a closed form solution? I thought I'd ask here before diving into it more deeply.

 

[Baluncore]

Is this an intractable problem for obtaining a closed form solution?

It is tractable.
Consider a test charge Q, located at P. Sum the forces on the test charge.
You could approximate it numerically by considering several centroids of charge along the wire, or you could solve for a closed form, assuming the wire could be approximated by straight segments.

 

[Anachronist]

Good to know. I want to avoid a numeric solution if possible. Yes, the wire could be approximated by straight segments. As the length of each segment approaches zero, I would have some sort of integral, I think.

 

[renormalize]

As the length of each segment approaches zero, I would have some sort of integral, I think.

Yes, it's easy to set up an integral expression for your problem. The charge-density on your length-$L$ rod located on the $z$-axis increases linearly from $q_1$ at $z=-L/2$ to $q_2$ at $z=+L/2$ and so takes the form:$$q\left(z\right)=\frac{q_{1}+q_{2}}{2}+\frac{\left(q_{2}-q_{1}\right)z}{L}\tag{1}$$Now I assume that your field is derivable from a potential $\psi$ which falls-off like $r^{-p}$ (for example, $p=1$ in the standard electrostatic case that gives rise to an electric field from a point charge that falls like $r^{-2}\,$). At the observation point $x_o,z_o$ the potential due to the rod is:$$\psi\left(x_{o},z_{o}\right)=\intop_{-L/2}^{+L/2}\frac{q\left(z\right)}{\left[x_{o}^{2}+\left(z_{o}-z\right)^{2}\right]^{p/2}}dz\tag{2}$$This integral can be performed analytically for any value of $p$ but the result is a messy expression in terms of Gauss hypergeometric functions $_{2}F_{1}$. Even the $p=1$ electrostatic case yields an involved combination of $\text{arcsinh}$ functions. So I think you're better off evaluating the integral (2) numerically. Note also that, even if your field is not the gradient of a potential, you can still set up integrals analogous to (2) for each individual component of the field.

 

[PhDeezNutz]

I may have misunderstood the problem statement but how can you have discrete charges at one end or the other if it is a continuous distribution?

Do you instead mean

“The top half has total charge q1 and the bottom half has total charge q2”?

 

[renormalize]

I may have misunderstood the problem statement but how can you have discrete charges at one end or the other if it is a continuous distribution?

In his first post the OP says:

The wire has a total charge $q$, but the charge density is different at each end, changing linearly in between (say, $q_1$ on one end and $q_2$ on the other).

I think this is just sloppy language. My post assumes that the OP intended that the total charge, call it $Q$, on the insulating rod (not a conducting wire) is given by $Q=\intop_{-L/2}^{+L/2}q\left(z\right)dz$ where $q(z)$ is the charge-per-unit-length along the rod.

 

[Anachronist]

Yes, that was sloppy language on my part. I erred in calling it a "wire", I meant a 1D line-segment of charge rather than a single point charge, and that would need to be a thin rod made of an insulator.

Because this is a metaball graphics application, I thought a field generator function for rendering an isosurface shaped like a truncated cone with spherical caps would be a thin rod of finite length with a charge distribution across it, such that charge per unit length on each end has a ratio similar to the truncated cone with different radii on each end. A rod with a constant distribution of charge along its length would have an isosurface appearing like a cylinder with rounded ends. That is an easier problem to solve.

 

[Anachronist]

The charge-density on your length-$L$ rod located on the $z$-axis increases linearly from $q_1$ at $z=-L/2$ to $q_2$ at $z=+L/2$ and so takes the form:$$q\left(z\right)=\frac{q_{1}+q_{2}}{2}+\frac{\left(q_{2}-q_{1}\right)z}{L}\tag{1}$$

Yes, that is exactly what I came up with for the charge density with respect to z. And last night while falling asleep I imagined an integral similar to what you describe. I am wondering if there are certain values of p for which this is simpler and closed form.

Because this is for generating an isosurface (metaball graphics application) I need to probe a few hundred thousand points in the bounding volume, evaluating that integral every time. Therefore, avoiding numerical methods would be best, unless there's a simple approximation that could work.

 

[renormalize]

I am wondering if there are certain values of p for which this is simpler and closed form.
Therefore, avoiding numerical methods would be best, unless there's a simple approximation that could work.

Actually, the integral in post #4 evaluates in closed form for any value of $p$ but it's not always simple to find its numerical value. For the electrostatic case $p=1$, the solution only involves the $\text{log}$ function (see below), but for general $p$ you need access to a function library that can efficiently compute the transcendental Gauss hypergeometric function $_2F_1$. So it may ultimately be just as practical to directly compute the value of the integral numerically at each point in space.
Now I want to illustrate the result of performing the integration for the particular value $p=1$. I start by defining $\overline{q}\equiv\frac{q\left(L/2\right)+q\left(-L/2\right)}{2}$ and $\Delta q\equiv q\left(L/2\right)-q\left(-L/2\right)$ so that the charge-density along the rod can be expressed in the simple form: $q\left(z\right)=\overline{q}+\Delta q\frac{\,z}{L}$. Next, I introduce the dimensionless variables $\chi\equiv\frac{x_{o}}{L},\zeta\equiv\frac{z_{o}}{L},\omega\equiv\frac{z}{L}$ into integral (2) in post #4 so that it becomes:
\begin{align}
\psi\left(\chi,\zeta\right) & \equiv\overline{q}\intop_{-1/2}^{+1/2}\frac{\left(1+\frac{\Delta q}{\overline{q}}\omega\right)}{\left[\chi^{2}+\left(\omega-\zeta\right)^{2}\right]^{1/2}}d\omega\nonumber \\
& = \overline{q}\left[\left(1+\frac{\Delta q}{\overline{q}}\zeta\right)\log\left(\frac{2\zeta+1+R_{+}}{2\zeta-1+R_{-}}\right)-\frac{1}{2}\frac{\Delta q}{\overline{q}}\left(R_{+}-R_{-}\right)\right] \nonumber\tag{1a}
\end{align}where$$R_{\pm}\equiv\sqrt{4\chi^{2}+\left(2\zeta\pm1\right)^{2}}\tag{1b}$$This is the electrostatic potential that arises from the the vertical rod due to its linearly-varying charge-density. It's straightforward to plot from (1) the "normalized" equipotentials of $\psi/\overline{q}$ simply by choosing a particular value of the ratio $\frac{\Delta q}{\overline{q}}$. As an example, setting this ratio to $-1$ gives:

Here I've plotted the $\text{log}$ of the potential to compress the dynamic range and I indicate the charged rod in red. Note that near the top of the rod, this plot agrees conceptually with the diagram in your OP: as you move down from the low-charge top, the equipotential lines do diverge in tandem with the increasing charge-density. Nevertheless, the lines ultimately re-converge again at the bottom where the charge-density is highest, as they must since the equipotentials are necessarily closed curves that approach circles at infinity. I suspect that this remains true for any fall-off power $p$. If so, does that impact the utility of this approach to your metaball graphics task?

 

[Anachronist]

Wow, that's incredible, and far more than I expected. Really helpful to see a plot of the isosurfaces, as well as a not-terribly-complicated solution for p=1.

To your last question, yes, that impacts the utility of this approach, because it isn't quite what I expected to see.

I was going for an isosurface that's shaped like a truncated cone capped with a sphere on each end. I already have a "metaball" isosurface shaped like a cylinder capped with hemispheres; I planned to use that as a connector between other shapes, and I thought it would be useful to have something that changes radius along its length. Now that I've seen it, however, due to the mathematics involved, it might make more sense to approximate this with a series of small spherical metaballs arranged in a line.

 

[renormalize]

I was going for an isosurface that's shaped like a truncated cone capped with a sphere on each end.

OK, that makes sense. I would describe those equipotential lines shown above that are near to the charged rod as having a "teardrop" shape; a truncated-cone with hemispherical end caps-looks very much the same and is easier to calculate.