# Why is electric field strong at sharp edges?

I learned about the coronal discharge, and the common explanation is because the electric field is strong where radius of curvature is small. But I haven't found anything yet that explains why electrons like to crowd at the peaks, and escape from the holes.

My intuition suggests electrons try to distribute on the surface as uniform as they can, but they don't. Why?

• Roughly, the surfaces you have in mind are equipotentials and electric fields are the derivatives. Oct 30, 2012 at 18:29
• Electric field is proportional to the density of electrons, rather than the number of electrons. Roughly speaking, at the sharp edges you have a small space, and thus the charge density is larger there. Apr 11, 2016 at 19:39
• Please see: 8.02x - Lect 6 - High-voltage Breakdown, Lightning, Sparks, St-Elmo's Fire by Walter Lewin (MIT) URL: youtube.com/watch?v=ww0XJUqFHXU Jul 9, 2017 at 1:13
• – user4552
Sep 18, 2017 at 15:51

Electrons just don't like each other, a point captured by the phrase that "like charges repel." So, imagine a gymnasium full of students pretending to be electrons, staying as far away from others as possible. Anyone near the center of the crowd will feel badly pressed and will try to work there way towards the edge of the gym, where at least one side will no longer have fellow students milling about. The result? Most of the students will gravitate towards the edge of the gym and hover there, to take advantage of that lack of other students on the wall side of the gym.

Now imagine a narrow corridor leading out of the gym. Even better! Students in that corridor will only have fellow students on their left and right.

Now imagine the very end of that corridor, a sort of point. Even better! Now, the student who finds that spot will benefit from having only one student nearby. But somewhat ironically, that same effect will cause other students to pack themselves into the long, narrow corridor more tightly, since pretty much anywhere in the corridor makes them less exposed to the full set of students than being in the gym does.

This is the kind of effect that makes edges, wires, and points more attractive to electrons, which similarly just don't want other electrons too nearby.

The electric field gradient is the rate at which the electric field falls off, and it is strongest on such edges and lines and points. You can use the gym analogy to see why that is. Imagine the mutual disdain of the students for each other as behaving like spooky spiky hair that extends ghost-like for many meters out from each student. The strands extend easily through ordinary bricks and such, but like Star Wars light sabers they absolutely refuse to move through each other. What happens?

For students lined up against a straight wall, the spooky spikes push against each other and wind up extending almost straight through the gym wall. The gradient in that case is actually quite small, since you end up with about the same number of spooky spikes per square meter far outside the gym wall as right up against it.

But what about the opposite case of that one student at the very end of the long, narrow dead-end corridor? Her spooky spikes are free to expand outward almost like giant ball, very quickly becoming quite sparse even a few meters beyond the corridor. That's a very steep gradient, and with electrons it's what leads to all sorts of interesting effects.

One effect in particular that I should note is that because the only repulsion that the student at the end of the long corridor is from other electrons in that corridor, her desire to move away from that corridor becomes far more directed and acute. She wants to escape! And if there is any weakness in the wall at the end of that corridor, she will succeed, escaping out into free space. And others will then follow! This is why electrons can escape from very sharp points, even at room temperature. The repulsion of electrons for each other is so strong that even the strong binding force of metals may fail if the electrons manage to find a point sufficiently isolated from the main body of free electrons in the metal.

Finally, I should point out that these two perspectives -- mutual avoidance and spooky spikes -- are really just two ways of describing the same thing, which is the way the repulsion of electrons falls off with distance. Calculus provides the machinery needed to make precise predictions from such models, but it's still important to keep in mind that the mechanisms by which such effects occur are by themselves not nearly as exotic as you might think. They have real analogies in events as simple as students following "electron rules" in a gymnasium.

• Though I agree with the rest of the comment, the part about fingers (first sentence) is inaccurate. The mechanism causing fingers to cool before the bulk of the body to cool is entirely unlike the mechanism that ensures the same potential energy across the surface of a conductor. Oct 31, 2012 at 12:18
• Concur! I've removed the sentence. It was part of an originally much longer analogy that dealt with just the issue you mentioned, but the gym analogy worked better. I should have trimmed all of the earlier attempt out. Thanks, good catch! Oct 31, 2012 at 17:39

Consider a charged conductor made out of two spheres of radii $R_1$ and $R_2$, connected with a conducting wire. Assume that $R_1<R_2$, and that the spheres are far apart so that effects of electrostatic interactions between the spheres can be neglected. Then, the surface charge density, the quantity that describes how crowded the charges are, is higher at the smaller sphere.

To see why, remember that, since this system is a conductor, its surface is an equipotential. In particular, the electric potential on the surface of two spheres is the same, $V_1=V_2$, which implies that $$\frac{q_1}{R_1}=\frac{q_2}{R_2}\Rightarrow \frac{q_1}{q_2}=\frac{R_1}{R_2}<1,$$ i.e. that most of the charge is in the bigger sphere. However, the ratio of the surface charge densities behaves the opposite way: $$\frac{\sigma_1}{\sigma_2}=\frac{q_1/4\pi R_1^2}{q_2/4\pi R_2^2}=\frac{q_1}{q_2}\frac{R_2^2}{R_1^2}=\frac{R_2}{R_1}>1.$$ This means that the surface charge density of the smaller sphere is larger, i.e. that the charge is more crowded (you find more charge per unit area) on the smaller sphere.

A generalization of this argument shows that charges are more crowded at pointy parts of a conductor, as opposed to the more gently-curved parts.

• The answer is ok, but I just feel that this is the type of answers that states "this works this way because maths says so". It should be the other way around: have an intuitive understanding of the subject (like the most voted answer), and then use maths to model. IMO, math should be used just to model a phenomenon, but never to explain it, because it can't, math can only describe. You cannot explain something based on its working description alone (although mathematical description often helps complementing). Apr 15, 2019 at 15:04

These are all very good explanations, at an intrinsic and extrinsic level. However, there is a very succinct answer if you need to quote it quickly.

Because the surface of a conductor is always a surface of constant potential, the electric field E = −∇φ, must be perpendicular to the surface at every point on the surface. So if the geometry of your conductor is very sharp, the field lines will diverge at large angles respective to each other: thus a large field gradient at that sharp edge.

Simples.

• why does a high rate of change of electric field imply large magnitude? Jul 27, 2020 at 1:58
• Furthermore, why does a high rate of change of electric field have to do with anything? Sure, the voltage is the gradient of the electric field ($\Delta V=-\nabla\mathbf{E}$), but it's the electric field magnitude that determines when the air breaks down - and that's usually what people are interested in. Furthermore, the question specifically asked about how strong the field was, not the gradient of it. Oct 20, 2020 at 14:04

Consider a pear shaped cathode. In this cathode equipotential lines will be very closed near sharp end and more spaced near round shape of the pear.

We know $E= V/d$. Near sharp end of pear, $d$ (distance between equipotential lines) is smaller as compared to round shape of the pear, therefore $E$ will be more near sharp end as compared to round shape.

Therefore more sharpen the cathode, smaller is $d$, higher is the Electric field. This principle is used in Carbon Nano Tube technology.

If we have two charges Q1 and Q2 separated by a distance r in a vacuum. the force between them is given by equation #1.

Where ϵ_o is the permittivity of vacuum.

If we put the charges in another medium, the force between the two charges is given by equation #2

the force between the two charges will be decreased since the permittivity of any medium is larger than the permittivity of vacuum.

If we put the two charges on the surface of a metallic conductor which has an irregular shape the force between the two charges depends on whether the surface is flat or curved.

Assume we put the two charges at the same distance on fig1, fig2, and fig 3.

If the surface is flat as we see in fig. 1 the conductor shield almost half the electric field lines which means the force between the two charges approximately equals half its value in a vacuum.

In fig.2 the surface gets a little curved and shields more electric field lines than in fig. 1, which means the force between the two charges is less than in the case of fig. 1.

In fig.3 the surface of the conductor gets sharper and shields more electric field lines than in fig2 and fig1, and the force between the two charges decreases more than in the Case of fig.2.

The charges leave the flat edges which have the higher repulsive force and accumulated at sharp edges where the force between them is minimum.

In conclusion as much the surface gets curved the electric field lines get shielded more and more, which means the force between the two charges at sharp edges is less than the flat edge.

So, if we charge a metallic conductor that has an irregular Shape the charges density will be large on sharp edges than on flat ones. We should also mention that a large charge density at sharp edges means a large electric field.