# Charge distribution on a conductor

1. Oct 6, 2007

### fantispug

Why does the charge on a conductor accumulate at sharp points? I've read one or two explanations, but I don't follow them, and when I try to think about it I reach the opposite conclusion.

Halliday and Resnick has an example where a large conducting sphere and a small conducting sphere are attached by a long wire (which we neglect the effects of). Since the spheres are a long way away if we charge the system both the spheres become positively charged, the larger sphere has more charge (proportional to radius) but a lower electric field at the surface (inversely proportional to radius).

To my this implies that having a smaller radius of curvature can cause something to have less charge. Why doesn't it?

(Incidentally in Griffiths I found a formula for the charge density of an ellipsoid, but I found it difficult to integrate to find the charge on a part of the ellipsoid with small/large radius of curvature - in this case there is more charge density near points of small radius of curvature, but since they have a smaller area I think it could be like the Halliday and Resnick case where there is less actual charge).

2. Oct 6, 2007

### Staff: Mentor

Let's talk about electrons in a negatively charged conductor. The electrons all exert force on each other by Culomb's law. Since they are in a conductor they will move (f=ma) until all of the forces on each electron are balanced. When they are in the interior of the charged conductor the net force is towards the surface. However, once they get to the surface they cannot leave (unless they are heated and have a lot of KE), all they can do is move around the surface.

So, consider the force on one surface electron due to the neighboring surface electrons. If there is any curvature to the surface then the net force on the electron will not be parallel to the surface. However, since the electron is constrained to not leave the conductor, only the component of the net force which is parallel to the surface can cause motion of the electron. In an area with small radius of curvature the component of the force parallel to the surface is small, and therefore more electrons are required to exert the same force. This results in an accumulation of charges around regions with a small radius of curvature.

3. Oct 6, 2007

### Meir Achuz

"Halliday and Resnick has an example where a large conducting sphere and a small conducting sphere are attached by a long wire (which we neglect the effects of). Since the spheres are a long way away if we charge the system both the spheres become positively charged, the larger sphere has more charge (proportional to radius) but a lower electric field at the surface (inversely proportional to radius)."

In this H&R example, each sphere has the same potential, so that the charge Q on each sphere is proportional to its radius R. The surface charge density is proportional to
Q/R^2~V/R, and thus is larger for the smaller radius.

4. Oct 6, 2007

### fantispug

The force is inversely proportional to the square of the distance, and an area with a small radius of curvature the charges are more bunched together (since circumference is proportional to radius), and every charge exerts a force, so I don't find it obvious whether the force would increase or decrease overall.
Moreover the force on each charge need not be the same, the total force just needs to cancel.

Again using the two sphere example, the charge on the smaller sphere is less. But in general it should be that on an area with less radius of curvature the charge is greater. Why is it different here?

I'm happy to take it as given that the electric field is greater near sharp points; I've seen much experimental evidence to this effect, and it can be shown this is true for an ellipsoid. This implies that the charge density is greater near points of less radius of curvature. But this doesn't necessarily imply the total charge is greater - again as shown in the instructive two sphere example.

This should really be simple and obvious but I just can't understand it.

5. Oct 7, 2007

### Shooting Star

The charge density should be more, not the total charge.

The argument given by Dalespam not only shows why the charge density should be more at regions with high curvature, but also why the charge should accumulate at the surface.

6. Oct 7, 2007

### Staff: Mentor

OK, first what you are trying to understand here is why the charges are more bunched together, so you should not assume the conclusion you are trying to explain.

Correct, the force is inversely proportional to the square of the distance, but only the component of the force parallel to the surface will cause motion. As the radius of curvature decreases so does the parallel component of the force. Since that decreases either you need more charges or smaller distances, both of which result in an increase of charge density.

As you said, the force on each charge need not be the same, and in general it will not be, but the total force due to the other charges also need not cancel, only the component of that force parallel to the surface need cancel. In general, the charges in a region with a small radius of curvature will have a much greater net force due to the other charges, but that net force will always be normal to the surface.

7. Oct 7, 2007

### fantispug

Sorry DaleSpam, I knew it was only the components parallel to the surface, I just thought about it implicitly.

When I was talking about bunching up I meant that on a surface of small radius of curvature the points are closer together than on a surface of large radius of curvature. (e.g. on a sphere the furthest point is 2*Radius away - the smaller the radius the closer the other points are).

(Incidentally, that the force normal to the surface at regions of high radius of curvature is greater is interesting - I wonder if you could measure the greater electromagnetic pressure somehow.)

I still don't follow the logic in:
Shooting star: That's what I keep thinking. Maybe the charge density is greater, but since the surface area goes as R^2 this doesn't necessarily imply that the total charge in this region is more.
But there are numerous things telling me otherwise (including the textbook that originally provoked this question), e.g. this (which gives a 1D argument, but it's not intuitive to me that this scales up to 2D) and this demo which is (supposedly) actually measuring the charge.

It seems unlikely that they'd all be wrong, but it makes sense to me if they are.

It was given in Griffiths (and derived somewhere else, I'll have to check out my library) that on a conductor in the shape of an ellipsoid
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$
with total charge Q, the charge distribution is
$$\sigma=\frac{Q}{4 \pi a b c} \left(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\right)^{-\frac{1}{2}}$$
So, for simplicity set a=b=R, c=2R, write x^2+y^2=r.
Then
$$\sigma=\frac{Q}{8 \pi R^3} \left(\frac{r^2}{R^4}+\frac{z^2}{(2R)^4}\right)^{-\frac{1}{2}}$$
So at r=0,z=2R
$$\sigma=\frac{Q}{4 \pi R^2}$$
Locally the area is like $$\pi R^2$$ (this is easier to see with a diagram), so
$$q\approx\frac{Q}{4}$$

At r=R, z=0:
$$\sigma=\frac{Q}{8 \pi R^2}$$
Area goes like $$\pi (2R)^2$$
so charge goes as
$$q\approx\frac{Q}{2}$$

So the charge density at the sharp point (z=2R) is double the charge density at the round point (r=R), but the charge is half. (This is very similar to the two spheres example).

So this would imply that charge does not necessarily accumulate at sharp points, but charge density does.

Is there a flaw in my reasoning, or are all these other things wrong?

If I had the equipment I'd try to actually measure it, but I don't.

Last edited: Oct 7, 2007
8. Oct 7, 2007

### Staff: Mentor

Yes, there are two effects that show this. First, the E field can be extremely high just outside the surface of a conductor near a sharp point (the "electromagnetic pressure" you are speaking of). And second, when a conductor is heated up to the point that charges can leave the surface they tend to start leaving primarily from sharp points where the large normal force can help push the charges off.
Correct.

9. May 4, 2008

### Surf74

Electron ejection into a vacuum

I have a follow-on question:

Suppose that two equal negatively-charged metal spheres, suspended in a vacuum, are in tangential contact. Does the high curvature at the point of contact act to aid the ejection of electrons into the vacuum? This seems like the inverse of the example above: here the insulating vacuum is the space with the sharp point.

10. May 18, 2008

### Shooting Star

(Sorry for the delayed response.)

When the two conducting spheres are just touching each so that there is only one point of contact, the curvature is infinite or undefined there. The surface charge density $\sigma$ is infinite at that point, and things as usual become intractable.

A better and realistic way would be to consider the slight flattening that occurs when you press two objects, and the circle of intersection is not a strictly 2-d thing, but also a bit smoothed out. In this case, the $\sigma$ is highest in that circular region, and leakage of charge will be the most from this region.

This leads to an interesting result. Suppose two conducting spheres combined have non-zero charge. Now they are connected by a very long wire such that the surface area of the wire is much more than the two spheres. Then the lion’s share of the charge of the whole system will be on the surface of the wire, because the curvature of the wire is much greater than that of the spheres. (Obviously, I am not talking of mini-spheres as small as the wire.)

11. May 18, 2008

### Shooting Star

Correction:

Please read "The surface charge density $\sigma$ is zero at that point."

Just to clarify the last post a bit, it is a well known fact that the charge on the surface of a conductor tends to accumulate at a sharp convexity and tends to vanish at a sharp concavity. If we join the two spheres, and press a bit, and then smooth out this 2-d circle of intersection, it becomes a sharply concave surface in that region, and the $\sigma$ will be very low in this region.

What I had wanted to say in the last post was basically to join the two spheres by a cylinder, and make the joins smooth. Then $\sigma$ may be high on the cylinder, if it's radius is small compared to its height. But on second thoughts, in practice probably the join between two spheres should be considered to be a concave surface, if we just press and flatten the point where they are touching.

We are talking about extrinsic curvatures of the surfaces here, not intrinsic or Gaussian curvatures.