In a region of space, I have an electric field represented by parallel curves as shown in the diagram.

enter image description here

The curves are equally spaced. We know that a uniform electric field can be represented by lines that are parallel and evenly spaced. Can this definition of uniform electric field be extended to parallel and equally spaced curves?

  • 1
    $\begingroup$ The electric field lines on a conductor all bend in so that they intersect the surface at right angles. Arrange enough conductors in a large uniform electric field, you have your configuration $\endgroup$ Commented Jul 1, 2014 at 12:10

2 Answers 2


Short answer - no. If field lines are curved and parallel, then the path integral along one line or the other will either give a different potential difference, or require different field strength. In either case you cannot call the field "uniform".

A little picture to clarify:

enter image description here

The dotted lines represent equipotential lines (at right angles to the electric field). Now when you start at point A and integrate the field around the path indicated with arrows, you can see that the first and third elements of the integral are zero because they are along an equipotential line; that leaves us with the two curved segments of the path. Now if the path lengths are $s_1$ and $s_2$, and fields $E_1$ and $E_2$ are in general different, then the total loop integral is $E_1 s_1 - E_2 s_2$ where I use the minus sign since the path is in the opposite direction for the last segment. We know this integral should be zero because the electric field is conservative - no work done by a charge after returning to the starting point.

I can rewrite that as an equality:

$$\frac{E_1}{E_2}= \frac{s_2}{s_1}$$

In other words, the only way that $E_1=E_2$ is if $s_1 = s_2$. And if the two paths are the same length, and they connect two equipotential surfaces (which are at right angles to the field lines), then the two paths must be straight and parallel.

Of course the above is using the fundamental result that for a static electric field,

$$\underline{\nabla} \times \underline{E} = 0$$

since the electric field is conservative (Maxwell's equation).

  • $\begingroup$ Another way of looking at it: "uniform" means that the field is everywhere pointing in the same direction with the same magnitude. Curved field lines don't meet this definition. $\endgroup$
    – garyp
    Commented Jul 1, 2014 at 12:21
  • $\begingroup$ @garyp - that is true but I was trying to show explicitly that the field strength (never mind direction) cannot be the same everywhere if the lines are not parallel. It is a weaker definition of "uniform", but I thought it was more intuitive than just saying "no that is by definition not uniform". $\endgroup$
    – Floris
    Commented Jul 1, 2014 at 12:42
  • $\begingroup$ You asked "Can the definition of uniform electric field be extended to parallel and equally spaced curves." The answer to that question is no. Uniform means constant magnitude and constant direction. I'm still not quite sure what you are asking. Perhaps you should edit and clarify the question? $\endgroup$
    – garyp
    Commented Jul 1, 2014 at 13:21
  • 1
    $\begingroup$ That was a very good explanation. In short, the equipotential surfaces in a uniform electric field are equally spaced. This is not true in this case. $\endgroup$
    – R004
    Commented Jul 3, 2014 at 10:08

Equi-spaced field lines are used to denote a uniform electric field. That means the field strength does not vary in the region, there is no potential difference between any two points on a surface normal to ${\vec E}$, no work is done in moving along a closed loop, or between two points on such an equipotential surface. Further, you expect that if you place a charged particle in such a region, its motion will correspond to uniformly accelerated/retarded motion, i.e. parabolic curves and all, with which I assume you are familiar.

(More part regarding first situation, added in edit(v2) - Another perspective on the issue is that, over here ${\vec E}$ is irrotational, i.e. ${\vec \nabla} \times {\vec E} = 0$. Hence, ${\vec E}$ can be written as the gradient of a scalar, ${\vec E} = -{\vec \nabla}\phi$, $\ \ \phi$ being the electrostatic potential. Hence, the line integral $\int_a^b {\vec E}\cdot {\vec {dl}} = \phi(a) - \phi(b)$ only depends on the endpoints and not on the path followed between a and b. Hence, the same, as well as the ''electrostatic'' work done in the process, vanishes for a closed path, where a coincides with b.)

However, the field lines that you show in the diagram, the curvature implies that the field curls, i.e. ${\vec \nabla} \times {\vec E} \ne 0$. Hence, in accordance with Maxwell's third equation, there will be a time-varying magnetic field set up, and the line integral $\int_l {\vec E}\cdot {\vec {dl}}$ shall be non-zero in general (see the equivalent integral form of the eqn in the link). Thus, there is net non-zero work done in moving along the path, which is not what you expect from a uniform field situation. Also, the particle motion here would be very complicated - there has to be a net effect of the (comparatively) simple ${\vec E}$ and the highly non-trivial ${\vec B}$, and you don't associate haphazardness with uniformity.

  • $\begingroup$ "no work is done in moving along the field lines". Is that what you meant to say? Think about a parallel plate capacitor. Is there no work done moving from one plate to the other? $\endgroup$
    – garyp
    Commented Jul 1, 2014 at 13:13
  • $\begingroup$ @garyp - Sorry, I was sprinting and hence ended up writing something gross. I meant no work done in moving along either plate, in the context of your example, and ended up saying ''along field lines'', instead of ''equipotential surface''. You have shamed me enough into revising and expanding my answer. So, thanks :) $\endgroup$
    – 299792458
    Commented Jul 1, 2014 at 18:56

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