When we're dealing with problems in electrostatics (especially when we use Gauss' law) we often refer to the density of electric field lines, which is inversely proportional to the radius in the case of a single point charge (all field lines are directed radially).

My question may sound dumb, despite the fact that this concept is quite intuitive, but if you think about it there is actually an infinite amount of field lines which we can draw everywhere (one through every point), so speaking of an area where field lines are more "dense" or "sparse" doesn't make much sense to me.

With this in mind, why can we still use such a concept? Why does it really work?

  • $\begingroup$ field lines are integral curves to the system of differential equations, en.wikipedia.org/wiki/Integral_curve You will need initial conditions to fix a solution. That means the lines you see in a plot are solutions with the chosen initial conditions. You can make them appear denser by choosing more initial points to draw the lines. The demonstrating diagram is a bit like a sampling of field lines. In areas where the "density of the sampling field lines" is larger, the field there is stronger. $\endgroup$
    – chichi
    Commented May 17 at 4:00

6 Answers 6


Field lines draw all of their validity from Gauss's law for the electrostatic field, $$ \nabla\cdot \mathbf{E}=\frac1{\epsilon_0}\rho,\ \text{or equivalently}\ \oint_{\partial\Omega}\mathbf{E}\cdot\text d\mathbf{S}=\frac1{\epsilon_0}Q_\Omega, $$ where $Q_\Omega=\int_\Omega\rho\,\text d\mathbf{r}$ is the electric charge in a volume $\Omega$ whose surface is $\partial \Omega$, and the fact that it can be mapped exactly into the continuity equation for some fluid in the steady state, which states that $$ \nabla\cdot \mathbf{j}=\sigma,\ \text{or equivalently}\ \oint_{\partial\Omega}\mathbf{j}\cdot\text d\mathbf{S}=\Sigma_\Omega $$ where $\mathbf{j}$ is the current density (i.e. the amount of fluid that crosses a unit area per unit time, which is equal to the fluid density times the local velocity), $\sigma$ is the amount of fluid being created per unit time per unit volume, and $\Sigma_\Omega=\int_\Omega\sigma\,\text d\mathbf{r}$ is the total fluid created inside $\Omega$ per unit time.

The exactly identical form of these two equations means that we can fruitfully interpret the electric field as the velocity of a fictional "fluid", which is conserved or not depending on whether charges are present. More specifically, we can bring in a useful way of describing fluids - the use of streamline diagrams - to help us visualize electric fields (and, indeed, magnetostatic and gravitational fields, to which all of this post also applies).

Let's use, then, this mapping: we can map your question, "how and why do electric field line diagrams work?" into a corresponding question in the dual problem:

"how and why do streamline diagrams work?"

Streamline diagrams depict a finite number of lines which are tangential to the current $\mathbf{j}=\rho_\text{fluid}\mathbf{v}$ at all points. That means a particle that starts on a streamline will stay on it as the fluid flows.

enter image description here (Image source)

The fun starts when streamlines suddenly get closer to each other. This means that either the fluid has been compressed, or that the unit volume of fluid has been stretched so it is longer along the streamlines (as in e.g. this diagram), and therefore the fluid is going faster. Either way, the fluid current density $|\mathbf{j}|$ increases.

This can be made more precise. In the image of the airfoil above, for example, the streamline 'seeds' (the points on the left at which the streamlines start) have been chosen so that the fluid flowing per unit time into the image from the left in between any two adjacent streamlines is constant. Because no fluid flows across a streamline, this rate of fluid flow must remain constant. Because of that, when the streamlines are closer together, the rate of fluid flow per unit area (i.e. the current density) is bigger.

(In addition to that, if there are sources or sinks where the fluid is created or destroyed, then you may need to introduce or remove some streamlines, because in such a situation the rate of fluid flow between the original streamlines is of course no longer conserved.)

Because of this fact, if you are given a pictorial representation of the streamlines of a fluid flow, then you can use it to reconstruct an approximation of its current density field. Its direction is given by the streamlines, and its magnitude is dictated by the spacing to the neighbouring streamlines. This is of course only approximate: you can try and interpolate the values between two streamlines, but you're missing information. You can try and improve your approximation by taking a finer grid, i.e. decreasing the flow rate needed to mark the next streamline, and that will give you better precision (but of course it will still be an approximation).

Conversely, to produce a streamline diagram you must first agree on the flow rate which should be taken until the next streamline. After that, you start off your "official" streamlines at one end of the flow, with the appropriate spacing, and you follow them through to the end. (Note, though, that because of the conservation of rate-of-flow-between-adjacent-streamlines, you can also start off your streamlines in the middle of the flow, as long as you space them appropriately). There will then be a finite number of "official" streamlines, as well as an infinite number of potential streamlines you could draw in between them. It is of course the "official" ones that you should use when estimating the current density from the diagram.

You can then make the spacing between "official" streamlines smaller, and it will get you a more accurate representation of the flow field. Even better, if your streamlines are really close together, it begins to make sense to speak of the local density of official streamlines. If the fluid slows down by a half from one region into another, and your streamlines are still very close together, then you'll find that there are twice as many in a given finite-and-small-ish area where the fluid is fast compared to where it is slow.

In the limit where you take infinitely many streamlines, equally spaced at a correspondingly vanishingly small flow rate, then you get the sort of thing that gets platonically talked about. It doesn't make sense, though, since you have to work at a finite spacing, but if your spacings are very small then you can both (1) have streamlines essentially through every point, and (2) speak meaningfully about the local density of official streamlines, and have that change consistently from one region to another depending on the local current density.

One useful tool to visualize this limit is Paul Falstad's 3-D Vector Fields Applet. For one, it lets you visualize electric fields as the velocity fields of fictitious particles, and it lets you play around in three dimensions with many of the standard electrostatics configurations, but most importantly it lets you increase and decrease the field line density:

enter image description here

It works exactly like this, too, in electrostatics. If you have an electrostatic field (or a gravitational field), then you start off by picking a suitably small unit of electrostatic flux (i.e. $\int_S\mathbf{E}\cdot\text d\mathbf{S}$ to separate your field lines. Once you fix this, the field line diagram is essentially completely determined: starting at a given point, you can draw its field line, then move off one unit of flux up and down from it and draw those field lines, and repeat that until you've covered all the region of interest. And, of course, if you meet regions with charges, then you need to thin down your streamline density or introduce some new ones. Gauss's law then guarantees that your diagram will be consistent: if you pick your spacings so that the local density of field lines reflects the electric field strength $|\mathbf{E}|$ across a given surface, then it will do that throughout the entire diagram. That is why you can use such a concept, and the reason it really works.

I would also like to set down here some thoughts on representation. Through the procedure I outlined above, it is indeed possible to reconstruct an approximation to the electric field in a region from a diagram with a finite number of field lines. Thus, these two representations of the same dipolar electric field are essentially equivalent*:

Mathematica graphics

However, just because they contain the same information, it doesn't mean that they are equally useful. Field line diagrams are very useful for forming an intuitive idea of what direction the field points in and its relative strength in different regions, but it is a terrible tool for estimating the superposition of two fields (for which a vector field representation is better), and it creates several incorrect intuitions. At the end of the day, even though we can indeed 'explain the force of gravity from the field representation without using the mathematical equation', attempting to do so turns out to be absolutely not worth the trouble. You should really see the field lines as merely a representational tool, and stick to the mathematics of the vector field for 'what the field really is'.

Finally, let me make explicit here a warning. The language in this post is primarily suited for two-dimensional situations, and it needs to be slightly modified for 3D scenarios. The reason for this is that in three dimensions there is no "next streamline", and you need to start off with an appropriate spattering of points, with the correct local density, in your 'seed' surface when drawing field lines. The correct generalization of 2D streamlines to 3D is still streamlines, and not "streamsurfaces" (which you can indeed draw, but everything becomes a whole lot messier), and this complicates things. Nevertheless, it is indeed possible to do this properly, and the analogy between a static fluid flow and electrostatics remains.

In fact, this seemingly harmless 2D/3D divide can be the cause of many subtle mistakes. Take, for example, this common representation of the field lines of a point charge, from anupam's duplicate question:

electric field lines of a point charge - an incorrect depiction.

For a point charge, it is the electric field flux through a sphere centred on the charge that is constant, which means that the electric field must go down as $1/r^2$. In this image, though (or in any 2D depiction in which field lines don't vanish), the spacing between the lines goes down as $1/r$ (since the number of field lines which cross a circle is constant), which means that it is seriously overestimating the field strength. This diagram is more suited for depicting the electric field of a line of charge.

Thus, to correctly estimate the field from such a diagram, you need to work from a 3D diagram like the ones produced by Paul Falstad's simulator, which produced the graphics above, ( or this one, if you have the cash), or you should work in 2D systems (i.e. 3D systems with a translational symmetry).

This sort of mistake, though, is exclusive to a field line representation - you wouldn't make it when using a vector field representation. You can use them to build some intuition, but beware of their many breakdowns and always complement them with other ways to visualize the fields.

* Note that if you directly attempt to plot this in Mathematica, StreamPlot will produce an incorrect plot, with field lines appearing out of nowhere; it is surprisingly hard to produce proper streamline plots both in Mathematica and elsewhere. For the code that produced this ones, use Import["http://goo.gl/NaH6rM"]["https://i.sstatic.net/CMlWz.png"]. Also, note that I cheated by reducing the contrast in the vector plot somewhat to make the arrows more visible.


You are absolutely right that the amount of filed lines is infinite, nevertheless thinking about "the density of field lines" is still useful in many applications. Here is an approach that hopefully will clarify things. Let's think of the electric field of a single charge. We choose to draw n lines of it just for illustration. In the figure, $n=8$.

The electric field of a point particle with 8 field lines drawn.

Now imagine a sphere around the charge wih radius $r_1=1\:\mathrm m$. What is the number density per surface area of the fields line we have chosen on that surface? Obviously $$\rho_1=\frac{n}{4\pi r_1^2}=\frac{8}{4\pi} \:\mathrm m^{-2}.$$ For a sphere with radius $r_2=10m$ this quantity is $$\rho_2=\frac{n}{4\pi r_2^2}=\frac{8}{400\pi} \:\mathrm m^{-2}.$$

How do these number densities compare? Which one is bigger and how much bigger is it? Note that $$\frac{\rho_1}{\rho_2}=\frac{\frac{n}{4\pi r_1^2}}{\frac{n}{4\pi r_2^2}}=\frac{r_2^2}{r_1^2}$$ which tells us that $\rho_1$ is bigger than $\rho_2$ by a factor of $\frac{r_2^2}{r_1^2}$ ($=100$ in the example). Now comes the crucial part of the argument:

We note that the above ratio is independent of $n$ so if we had chosen to draw 10, 100 or 10 billion lines it wouldn't have made a difference. Even if we chose to take n to infinity the result still wouldn't change because it doesn't depend on $n$! In that sense, it makes sense to say that the density of the field lines decreases, even if the number of field lines is infinite.


there is an infinite amount of lines which we can draw everywhere, so we can't really think of an area with denser lines.

Despite there being infinite lines there is absolutely no problem with saying one density is larger than another. Mathematically we can write:

$$ \mathop {\lim }\limits_{x \to \infty } \left( x \right) $$

So we are not talking about true infinities but the limit where the number of field lines is arbitrarily large. This has the implicit assumption that the larger the number of field lines the more accurate the model is.

Then why can we still use such a concept?

There are a finite number of water molecules but most models of water assume that it is an infinitely divisible continuous "fluid". This is the opposite situation where we have an infinite number of lines but we can assume a finitely large number and get the approximate right answer. The more field lines used the better the approximation.

Really it all boils down to Gauss's Law:

$$ \nabla \cdot E = \frac {\rho}{\epsilon} $$

Which means that if there are no charges then there is no divergence of the field. Which means that Field lines can only start and end at a charge. Which is what makes this visualisation so useful, (you can't just start and end field lines anywhere) if you draw a field line it must be a loop, or have its ends at charges or at infinity.

Another useful property is that a field line is a locus. This means that every point on the line satisfies a certain condition. Eg. a topographical (mountaineering) map there are lines on the map.

These lines represent lines of equal height, so if you walk along the line you will neither rise nor fall in altitude. And if you travel perpendicular to the lines you will be climbing in the steepest direction. You can also tell how far you climbed by counting the number of lines you passed.

If each line represents an assent of ten meters and you pass four of them on the map you have climbed forty vertical meters. However to be more accurate we should have ten times as many lines each representing one meter, or an infinite number of lines representing an infinitesimal increase in height. And if I am to use the latter map with infinite many lines, does it invalidate the idea that some slopes are steeper than others since there is infinite lines everywhere. Of course not.

Other field lines represent other useful information, like electric field lines represents the motion of a free charge if it were placed on that line (if no other forces were to act on it).

Or gravitational field lines representing lines of equal gravitational potential.

But really field lines are not used in models only in visualisations since they are good at conveying qualitatively information. Most modern models are entirely mathematical, as that is quantitative by nature.

there is an infinite amount of lines

Are there really? (I honestly don't know for electric fields, can someone leave a link in comments if you do know.)

However for magnetic fields there are a finite number of field lines in a given space.

Gauss's Law applied to magnetic fields:

$$ \nabla \cdot B = 0 $$

This states the same as for electric fields, except there are no magnetic monopoles for the lines to begin and end. I.E. Magnetic field lines must be closed loops.

A superconducting quantum interference device takes advantage of this by trapping individual loops of magnetic field through its superconducting loop. (Magnetic fields cannot penetrate superconductors.) There is a weak point manufactured into the superconducting loop. It is possible, through interference to detect every time a single field line passes through the weak point. And since magnetic field lines are quantised there will be a finite number of individual magnetic field lines.


A positively charged point object in space has field lines emanating outwards of the charge, this should be imagined in the 3D space. The figure so obtained is a sphere.

Now suppose I have a sphere of radius $r$ with mass $M$ has a volume $V$ has an expression $\frac{4}{3}\pi r^3$.

The density of the sphere is $M/V$ which has $1/r^3$ dependence. If you increase the radius of sphere, the density decreases.

Similarly, we can account for the geometry of the field line of the point charge. It is less dense as the distance increases.

  • $\begingroup$ Similar to the way a surface charge density $\sigma (r)= \frac{q}{4\pi r^{2}}$, with $q$ constant, distributed over the surface of a sphere with radius $r$, would fall if $r$ is increased. $\endgroup$
    – Hilbert
    Commented Jul 27, 2019 at 10:49

My previous answer got away from the point asked in question, my apologies to the OP and readers.

A complete description of the geometry of all the field lines of a vector field is sufficient to completely specify the direction of the vector field everywhere. In order to also depict the magnitude, a selection of field lines is drawn such that the density of field lines (number of field lines per unit perpendicular area) at any location is proportional to the magnitude of the vector field at that point.
Source : Wikipedia (field lines)

To explain in plain words, field line density depicts the strength of the electric field at a distance $r$ from its source/sink. It is a comparitive study of field lines at different radial distances but would not give us the exact field strength.

It would not give us the exact field strength because it is calculated by relative density of field lines which are arbitrarily drawn and can vary in number from 1 to infinite.

Lets study the field line densities now at radial distances $r_1$ and $r_2$
Let $n$ field lines are drawn from the source/sink of field

Field line density at distance "$r_1$" is given as $$ \rho_{r_1} = \frac{n}{4\pi r_1^2} $$

Field line density at distance $r_2$ is given as $$ \rho_{r_2} = \frac{n}{4\pi r_2^2} $$

Comparing field strengths at the two distances $$\frac{\rho_{r_1}}{\rho_{r_2}} = \frac{n}{4\pi r_1^2} \frac{4\pi r_2^2}{n}$$ $$\frac{\rho_{r_1}}{\rho_{r_2}} = \frac{r_2^2}{r_1^2} $$

It does not matter if you make $n$ infinite, tending to infinite or any other number as taking limits or otherwise it gets cut off and we get an inverse square relation of field line density/strength with radial distance which is all we want to show with the field line density picture. This is why this picture works mathematically.
If you try to draw infinite lines actually you will need to draw lines of $0$ cross sectional area which is not possible by any known man/software. However if you draw very thin lines then you will start getting a complete sphere with protruding hair(spikes really) as the distance between these tentricles keep on increasing density decreases inversely with radius squared.

Note actual field strength at any distance will be $$ E = k \frac{Q}{r^2} $$ while the field line density is $$\rho = \frac{n}{4\pi r^2} $$ Since $k$ is a fixed constant to get actual field strength from field line density we will have to fix $n$ and in that case we won't be able to make it infinite.

Addendum :
If you need to mathematically observe $\frac{\infty}{\infty} $ then try it as $\frac{0}{0}$ both of which give undefined as answer and that is expected as you can see when you draw $0$ lines you can not possibly define field line density much lese relative field line density!

You should rather treat $\infty$ as a big Big number, bigger than all poasible numbers and in that case you can cancel the two ! Just the same way we represent a very long wire or very large sheet as infinitely big, as it represents a HUGE number for our purpose (near vicinity calculations).

  • 2
    $\begingroup$ Give any reference where $\dfrac{\infty}{\infty}$ is mentioned as $1$. Or prove by quoting the definition of infinity. PS: By infinite i mean the amount of points in any physical finite volume. $\endgroup$
    – user31782
    Commented Feb 4, 2014 at 11:46
  • $\begingroup$ @anupam :Let $f(x) = x$ and $g(x) = f(x)/f(x)$ Now mathematically it is true that $g(x) = 1$ it has no dependency on $x$ putting values of x is done after the function is completely defined in terms of $x$ and not before ! If that was the case then $\tan x/ \sec x \neq \sin x$ at various values of $x$. $\endgroup$ Commented Feb 4, 2014 at 12:35
  • $\begingroup$ corrected my comment a bit .Read your mathematics books again. any real valued function $g(x)$ defined by $g(x)=f(x)/f(x)$ where $f(x)$ is also a real function is equal to $1$ except at the point where $f(x)=0$(say $x_0$). For $f(x)=x$ we will have $g(x)=1∀x∈(R−0).$ $\endgroup$
    – user31782
    Commented Feb 4, 2014 at 13:18
  • $\begingroup$ @anupam : I agree I did not think much before writing the first example, it is indeed 1 only when values are other than 0. However the 2nd example has no exception and is true for all values even though individual functions do reach $\infty$ at $x=\pi/2$ $\endgroup$ Commented Feb 4, 2014 at 13:24
  • $\begingroup$ Give me a link it would be $\lim_{x \to {(\pi/2)}}$. $\endgroup$
    – user31782
    Commented Feb 4, 2014 at 13:27

As someone pointed out the actual distribution is continuous. This model is exactly that. A model. Actual infinites don't exist in reality so using them in a model is only going to destroy the model's usefulness. But whether we have 10 lines or 10 billion lines, the density decreases with increasing radius. So the concept of this model works just fine.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.