Electrostatic fields with straight field lines This question is motivated by a student of mine. Is there a general characterization, or classification, of electrostatic fields whose field lines are straight? There are some obvious examples:

*

*Uniform field; immediately generalizes to the field of any charge distribution that varies only in one direction, but is constant in the other two directions.

*Field of a point charge; immediately generalizes to the field of any spherically symmetric charge distribution.

*Field of a straight charged wire; immediately generalizes to the field of any axially symmetric charge distribution.

An alternative formulation of the question might be: find all (Newtonian) gravitational fields such that any particle, initially at rest, will freely fall along a straight trajectory.
 A: I do not know about existing classifications. But it seems that the field should satisfy in general the following equation (btw it reminds me of the conformal Killing equation):
\begin{equation}
\vec E(\vec x + \epsilon \vec E) = F(x) \vec E(\vec x),
\end{equation}
with $F(x)$ being a scalar function. The infinitesimal form is
\begin{equation}
f(x) E_i(x) = E_j (x) \partial_j E_i(x).
\end{equation}
Multiplying by $E_i$ both sides fixes $f(x)$ as
\begin{equation}
f(x) = \frac{1}{2E^2(x)} \partial _j \left ( E^2(x) E_j(x) \right ),
\end{equation}
where I assume that $\partial _i E_i(x)=0$. As a result we get
\begin{equation}
E_i(x) E_j (x)\partial_j E^2(x) = 2 E^2(x) E_j(x)\partial_j E_i(x).
\end{equation}
Maybe it would be easier to analyze this equation using scalar potential
\begin{equation}
E_i(x)=-\partial _i \varphi(x).
\end{equation}
A: As a partial answer, it's true that there exist charge distributions with straight field lines that aren't symmetric in any way. For example, suppose a charge is placed outside a grounded conducting sphere; then charges are induced on its surface. We know from the method of images that the electric field produced by those induced charges, outside of the sphere, is precisely the same as the electric field produced by an "image" charge. Therefore, the field of the induced charges has perfectly straight field lines, even though the induced charges don't have any symmetry.
This makes it unlikely that there's a simple general way to describe the charge distributions. But the equipotential surfaces in this case are still spheres; you might have better luck constraining those.
A: Here is an attempt at a partial answer, extending the argument in the previous answer by nwolijin. Let us start with some vector calculus. For any vector fields $\vec u,\vec v$ and scalar field $\phi$ we have the identities
\begin{align}
\tag{1}
\vec\nabla(\vec u\cdot\vec v)&=(\vec u\cdot\vec\nabla)\vec v+(\vec v\cdot\vec\nabla)\vec u+\vec u\times(\vec\nabla\times\vec v)+\vec v\times(\vec\nabla\times\vec u),\\
\tag{2}
\vec\nabla\times(\phi\vec v)&=\phi\vec\nabla\times\vec v-\vec v\times\vec\nabla\phi.
\end{align}
Now the condition that the field lines of $\vec E$ are straight is equivalent to the condition that the directional derivative of $\vec E$ along itself, that is $(\vec E\cdot\vec\nabla)\vec E$, is parallel to $\vec E$. Since electrostatic field is conservative, it follows from (1) that $(\vec E\cdot\vec\nabla)\vec E=\frac12\vec\nabla(\vec E^2)$. Hence $\vec\nabla(\vec E^2)$ must be parallel to $\vec E$.
As the next step, use (2) with $\vec E$ in place of $\vec v$ and any function $f(\vec E^2)$ in place of $\phi$. Since $\vec\nabla(\vec E^2)$ is parallel to $\vec E$, so is $\vec\nabla f(\vec E^2)$, and (2) then tells us that $\vec E'\equiv f(\vec E^2)\vec E$ is a conservative field. At the same time, $\vec E'$ is parallel to $\vec E$ everywhere, and thus also has straight field lines. We arrive at the conclusion that the electric field maintains the straight field line property (and remains conservative) if we rescale it by an arbitrary function of $\vec E^2$. We can in particular replace the field $\vec E$ with a unit vector of the same direction everywhere in space, except for points where $\vec E=\vec 0$; at such points the direction of the field is ill-defined.
So if we are only interested in the topology of the field lines, not in the magnitude of the electric field, we may as well assume that the field is a unit vector everywhere where it is nonzero. It appears that it should be possible to characterize such unit vector fields just by their singularities, i.e. by specifying the set of points in space where the electric field vanishes.
Here is where some hand-waving starts. The singularities can be either isolated points, curves, surfaces or three-dimensional domains. In case of point singularities, the field $\vec E'$ will point radially outwards (or inwards) in their immediate neighborhood. In the other cases, the field lines will enter the singularity from directions perpendicular to it. This also applies to the surface of three-dimensional domains. We have thus reformulated our problem in geometric terms. We look for sets of points, curves, surfaces and domains in three-dimensional Euclidean space such that any straight line starting from a point, or any straight line starting along a normal direction to a curve or surface, will either stretch to infinity or end up on another singularity, entering it along a normal direction. Any two such lines are not allowed to cross anywhere outside of the given set of singularities.
Some obvious possibilities for the set of singularities are:

*

*A set of parallel planes. This corresponds to a one-dimensional electric field.

*A set of concentric spherical surfaces. This corresponds to a spherically symmetric electric field.

*A set of coaxial cylindric surfaces. This corresponds to an axially symmetric electric field.

*Special limits thereof, including a sphere of zero radius (point singularity) or a cylinder of zero radius (straight line singularity), a slab (stack of planes with infinitesimal spacing), a solid cylinder or sphere (the same for cylinders and spheres), and the like.

Is there some elegant way to solve this geometric reformulation of the problem?
A: What is commong among all the three cases is the presence of a symmetry:

*

*reflection symmetry in the case of a charged plane

*spherical symmetry in the case of a point charge

*a symmetry axis in the case of a straight wire or cylinder

The electric field is then directed along the coordinate curves of the coordinate system natural for each symmetry.
So there are two main ingredients for a generalization are:

*

*The system has to possess a point symmetry (not only in intuitive, but in a rather general group theoretical sense).

*Electric field, being a harmonic function (i.e., a solution of the Laplace equation), serves as possible curvilinear coordinate curves compatible with this symmetry (see also orthogonal coordinates).

Note that in general case these coordinate curves are not straight lines in the sense implied in the question - i.e., being straight lines in Euclidean coordinates.
Finally, similar claims could be made about the magnetic field.
A: This is my second attempt at an answer; criticism is welcomed. This could all be incredibly off base, because I've never taken a differential geometry class, and I strongly suspect the problem has a simple formulation in that language.
To incorporate the curl-free condition, we characterize the field in terms of its equipotential surfaces. Consider a specific equipotential surface $S$. The electric field on this surface must all be pointing directly inward or outward; let's suppose it's outward for concreteness, and also that it has a single connected component.
I have a hunch (but no proof) that in order for the field lines to be straight, the electric field magnitude on each surface $S$ must be a constant $E_S$. This is at least true in two dimensions, but I can't really tell in three.
Suppose this is true. Then consider two infinitesimally distant surfaces $S$ and $S'$. By the divergence-free condition, we can apply Gauss's law to the volume bounded by $S$ and $S'$, getting $E_S A_S = E_{S'} A_{S'}$ where $A_S$ is the area of $S$. On the other hand, we can consider a small patch $dA_S$ of $S$, which expands to a small patch $dA_{S'}$ of $S'$. Applying Gauss's law to the volume bounded by these surfaces and the field lines gives $E_S dA_S = E_{S'} dA_{S'}$. Therefore,
$$\frac{dA_S}{dA_{S'}} = \frac{A_S}{A_{S'}} = \text{const.}$$
In other words, as the equipotential surfaces expand outward uniformly, the area of each piece of the surface has to change at the same rate.
Now consider a patch of the surface $S$, and place it at the origin with the normal vector pointing along the $\hat{\mathbf{z}}$ direction. If we orient the $xy$ axes appropriately, the equation of the surface in a small neighborhood is
$$z \approx \frac{a}{2} x^2 + \frac{b}{2} y^2$$
where $a$ and $b$ are the principal curvatures. The normal vector of the surface is $\hat{\mathbf{n}} \propto (ax, yb, 1)$. Thus, after transporting a distance $\epsilon$ along the normal vector, any patch on the initial surface is stretched by a factor $1 + \epsilon a$ in the $x$ direction, and $1 + \epsilon b$ in the $y$ direction, so the area scales as
$$\frac{dA_{S'}}{dA_S} = (1 + \epsilon a)(1 + \epsilon b) = 1 + \epsilon (a + b) + O(\epsilon^2).$$
That is, the rate of increase of area is characterized entirely by the sum of the principal curvatures. This implies the equipotential surfaces must be constant mean curvature surfaces, and there are many examples besides the plane, cylinder, and sphere. The question reduces then to whether these other surfaces continue to have constant mean curvature as they are expanded outward.
A: 
Is there a general characterization, or classification, of electrostatic fields whose field lines are straight?

I had a previous answer (and some other still present answers) that related to the curvature with the expression depending upon something like $\left( \mathbf{E} \cdot \nabla \right) \mathbf{E}$ which only works for divergenceless fields.
The part I had missed is that I left the expression in terms of the magnitude of the field instead of using the unit vectors.  So let's define a unit vector as $\hat{u} = \tfrac{ \mathbf{V} }{ \lvert \mathbf{V} \rvert }$ then we can define curvature in terms of this as:
$$
\boldsymbol{\kappa}_{c} = \left( \hat{u} \cdot \nabla \right) \hat{u} \tag{0}
$$
The radius of curvature is just the inverse of the magnitude of the curvature given by:
$$
\rho_{c} = \frac{ 1 }{ \lvert \boldsymbol{\kappa}_{c} \rvert } \tag{1}
$$
A field line that is straight will satisfy $\rho_{c} \rightarrow \infty$, i.e., a zero magnitude curvature.
Example: Point Charge
For a point charge, the electric field $\hat{u} = (1,0,0)$ in spherical coordinates and so that Equation 0 goes to zero.  If we use the full electric field vector $\mathbf{E} \propto r^{-2} \hat{r}$ in place of $\hat{u}$ then Equation 0 would not be zero, which was my the with my previous answer and the issue with some of the other answers herein.
Example: Linear Gradient
Suppose we a linearly increasing (in magnitude) electric field given by:
$$
\mathbf{E} = E_{o} \left( 1 + z \right) \hat{x} \tag{2}
$$
Again if we use $\mathbf{E}$ instead of its unit vector, Equation 0 will not have a zero magnitude but it will if we use the unit vector.
Example: Dipole
Suppose we have a dipole made of two point charges separated by some vector $\mathbf{d}$ and we observe said field at some displacement $\mathbf{r}$ in the $\phi = 0$ plane.  Then the electric field will have an $r$ and $\theta$ component given by:
$$
\mathbf{E} \simeq \frac{ k \ d }{ r^{3} } \left[ \left( 2 \ \cos{\theta} \right) \hat{r} + \sin{\theta} \ \hat{\theta} \right] \tag{3}
$$
We know the fields are not straight here and indeed the expression for Equation 0 is not zero but a complex function of $r$, $d$, and $\theta$ given by:
$$
\begin{align}
  \left( \left( \hat{e} \cdot \nabla \right) \hat{e} \right)_{r} & = - \frac{ 8 \sin^{2}{\theta} }{ r \left( 5 + 3 \cos{2 \theta} \right)^{2} } \tag{4a} \\
  \left( \left( \hat{e} \cdot \nabla \right) \hat{e} \right)_{\theta} & = \frac{ 8 \sin{2 \theta} }{ r \left( 5 + 3 \cos{2 \theta} \right)^{2} } \tag{4b} \\
  \left( \left( \hat{e} \cdot \nabla \right) \hat{e} \right)_{\phi} & = 0 \tag{4c}
\end{align}
$$
where $\hat{e} = \tfrac{ \mathbf{E} }{ \lvert \mathbf{E} \rvert }$ is the electric field unit vector given by:
$$
\begin{align}
  e_{r} & = \frac{ 2 \cos{\theta} }{ \sqrt{ 4 \cos^{2}{\theta} + \sin^{2}{\theta} } } \tag{5a} \\
  e_{\theta} & = \frac{ \sin{\theta} }{ \sqrt{ 4 \cos^{2}{\theta} + \sin^{2}{\theta} } } \tag{5b} \\
  e_{\phi} & = 0 \tag{5c}
\end{align}
$$
Answer
Therefore, the field line curvature must be calculated in terms of the unit vector of the field, not the field with its magnitude included.  In the limit of $\rho_{c} \rightarrow \infty$, the field lines will always be straight (ignoring critical points where the magnitude is zero).
