Uneven charge distribution on a conductor

[vcsharp2003]

Homework Statement

Why is the charge density i.e. charge per unit area higher on pointed areas of a conductor as compared to more flatter areas? The uncharged conductor is placed in an electric field as shown in diagram below.

Relevant Equations

$V= k \frac {Q} {r}$, where V is potential on surface of a charged sphere having a total charge of Q on it and a radius of r

All I can say is that where the charge density on surface is higher, we will have a stronger electric field compared to areas where charge density is lower since more charges means greater electrical force on a test charge placed very close to the surface.
Also, the potential on pointed areas must be the same as potential on flatter areas since charges are not moving on the surface.

 

[vcsharp2003]

I think it's got to do something with the potential on a charged sphere. A very small pointed area can be imagined to be part of a sphere of radius $r_p$ and a very small flatter area a part of a sphere of radius $r_f$ so that $r_f > r_p$. Let $Q_p$ and $Q_f$ be charges on a corresponding sphere having the corresponding radius of curvature i.e. spheres corresponding to radius $r_p$ and $r_f$.

Using the formula for potential on surface of a charged sphere and the fact that the potential on pointed area must equal the potential on flatter area, we get

$$V_p = V_f$$
$$k \frac {Q_p} {r_p} = k \frac {Q_f} {r_f}$$

If charge density per unit area on pointed area is $\sigma_p$ and flatter area is $\sigma_f$, then above equation becomes.

$$k \frac {4\pi r_p^2\sigma_p} {r_p} = k \frac {4\pi r_f^2\sigma_f} {r_f}$$
$$\therefore {\sigma_p} {r_p} = {\sigma_f} {r_f}$$

Since $r_f > r_p$ so $\sigma_f < \sigma_p$ so that above equation is always true, which means charge density is LESS on flatter surface.

The above analysis clearly proves that flatter the surface the lesser the charge density per unit area.

 

[haruspex]

if we assume that the flatter area and pointed area are both equal to A,

That may not be valid.
Consider two spheres connected by a wire, long enough that they are not affected much by each other's field. If one has double the radius of the other it has twice the charge but four times the area.

 

[vcsharp2003]

That may not be valid.
Consider two spheres connected by a wire, long enough that they are not affected much by each other's field. If one has double the radius of the other it has twice the charge but four times the area.

You're right.

I found a mistake in my analysis. First I cannot assume that areas of pointed and flatter surfaces are equal and second the terms $Q_p$ and $Q_f$ are NOT charges on the areas considered but charges on a corresponding sphere having the corresponding radius of curvature. Then, my analysis works.

I have just now corrected my post and the expected result logically follows from my analysis.

 

[TSny]

I think we have to be careful not to come to a general conclusion that the surface charge density on a conductor in electrostatic equilibrium will be greatest at points where the curvature of the surface is greatest.

In the figure above we appear to have a conductor with zero net charge that has been placed in a horizontal, uniform external electric field. The surface charge density here is depicted as being greatest in the regions of the surface that have the greatest curvature.

However, suppose we replace this conductor with one that is shaped like an ellipsoid of revolution as shown below.

By symmetry the points $a$ and $b$ will have zero induced surface charge density even though these are the points of highest surface curvature.

 

[kuruman]

By symmetry the points and will have zero induced surface charge density even though these are the points of highest surface curvature.

It seems that you are assuming that there is a left-to-right external electric field. Given that, if you rotate the ellipsoid by 90^o, the surface charge density will be higher at points $a$ and $b$ than $c$ and $d$.

The fact that there is a relatively lower surface charge density on relatively flatter surfaces of a charged conductor is related to the way deposited charge is distributed in the absence of an external electric field. It would be a mistake to draw conclusions when external electric fields are present.

For example, if charge Q is placed on a conducting sphere in a uniform electric field $\vec E=E_0~\hat z$, the induced surface charge density is $\sigma(\theta)=\frac{Q}{4\pi R^2}+3E_0\epsilon_0 \cos\theta$ where $\theta$ is the conventional spherical angle. Although the system is in electrostatic equilibrium and the curvature is everywhere the same, the surface charge density is not.

 

[PhDeezNutz]

I may be simplifying this too much but in this picture coincidentally the flatter areas are further from the plates than the more pointy areas.

Field will be stronger in magnitude the closer we are to the plates therefore more charge per unit area will be induced on areas of the conductor closer to the plates.

I can’t think of any reason why this “flat sharp” conclusion would be true in general.

 

[PhDeezNutz]

Wait…..isn’t $\sigma = \vec{E}_{ext} • \hat{n}$ (modulo a constant). Coincidentally in this picture the flatter areas are on top and bottom and the external field is parallel to the surface i.e. Orthogonal to the surface normal.

 

[TSny]

It seems that you are assuming that there is a left-to-right external electric field.

Yes. I wanted to contrast the ellipsoid shape with the shape of the conductor in the OP where both conductors are in an external field. I was concerned that some students might get the impression from the example in the OP that you would always find a greater induced charge density at points of the surface with greater curvature. Your example of the sphere is another good, simple example that shows that there is not a simple relation between surface charge density and curvature, in general. You can think of many other examples, such as bringing a point charge very near the surface of a conductor so that you get a large induced surface charge density on the conductor even if the conductor is flat in that region. So, I think I was probably unduly worried.

The fact that there is a relatively lower surface charge density on relatively flatter surfaces of a charged conductor is related to the way deposited charge is distributed in the absence of an external electric field. It would be a mistake to draw conclusions when external electric fields are present.

Suppose there are no external electric fields present and we have an isolated conductor with a nonzero net charge. Is it always true that the magnitude of the surface charge density is greatest at points of the surface that have the greatest curvature? It would be nice to have a simple counterexample.

There was a paper published in the American Journal of Physics some years ago that proved that for isolated, charged conductors with shapes of ellipsoids, paraboloids, and hyperboloids, the surface charge density at a point of the surface is proportional to the fourth root of the Gaussian curvature of the surface at that point. At the end of the paper, the author conjectured that this proportionality relation would hold for essentially any conductor whose surface can be expressed as an analytic function. But, it was quickly pointed out by others that this conjecture is not true.

 

[vcsharp2003]

I think we have to be careful not to come to a general conclusion that the surface charge density on a conductor in electrostatic equilibrium will be greatest at points where the curvature of the surface is greatest.

Even in the diagram presented in question, the bottom most area of conductor has no charges even though it's more curved than the side flatter areas which have some charges.

So, it seems, it's the electric field that's causing the distribution so no charges appear in bottom most curved area.

But, if the conductor was charged postively or negatively by some means and there was no applied electric field , then we could say that more curved areas including the bottom most curved area would have a higher charge density. Is that correct?

 

[haruspex]

Even in the diagram presented in question, the bottom most area of conductor has no charges even though it's more curved than the side flatter areas which have some charges.

So, it seems, it's the electric field that's causing the distribution so no charges appear in bottom most curved area.

But, if the conductor was charged postively or negatively by some means and there was no applied electric field , then we could say that more curved areas including the bottom most curved area would have a higher charge density. Is that correct?

With a spherical uncharged conductor in a uniform field, the greatest induced charge density will occur where the applied field's potential is at its maximum and minimum. (Extreme left and right in a horizontal field.) That's why there is high density on the left and right 'wings' in the illustration, but not at the 'tail' at the bottom.
For any symmetric uncharged conductor in a symmetric field, there will be no charges in the plane of symmetry.
For an arbitrary shape, uncharged, uniform horizontal field, we can expect a basic left and right pattern, locally modified to favour high curvatures.

 

[kuruman]

There was a paper published in the American Journal of Physics some years ago that proved that for isolated, charged conductors with shapes of ellipsoids, paraboloids, and hyperboloids, the surface charge $\dots$

You are probably not referring to this paper: David J. Griffiths, and Ye Li, Charge density on a conducting needle, American Journal of Physics 64, 706 (1996); doi: 10.1119/1.18236]. In it the authors try to model the surface charge distribution when charge Q is placed on a conducting needle. Their first model is a prolate ellipsoid, $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$. They write the charge distribution on this as $$\sigma=\frac{Q}{4\pi abc}\left(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\right)^{-1/2}.$$ Then they find that the linear charge density along $x$ is surprisingly constant: $\lambda(x)=\frac{dQ}{dx}=\frac{Q}{2a}$. It is worth quoting the authors' remarks:

"Since this result is independent of $b$ and $c$, it holds in the limit $b,c\rightarrow 0$ when the ellipsoid collapses to a line segment along the $x$ axis. Conclusion: If the needle is the limiting case of an ellipsoid, then the linear charge density is constant. In this case the tapering of the ends exactly cancels the tendency for charge to push out toward the extremities.​

The authors consider other models such as a cylinder for which there is no exact form for the charge distribution and which they solve numerically, a "charged-bead" model and more. The closing paragraph of the article begins with the statement "It is embarrassing to conclude that we still do not know what the charge density on a conducting needle is." They base this conclusion on the suspicion that the problem is ill posed (in the sense that it is model-dependent) and because they cannot "absolutely exclude the counterintuitive possibility that it is in fact a constant."

There you have it. I think the take-away message is that one cannot a priori deduce, simply by looking at curvatures, the surface charge distribution on a charged conductor in the absence of external electric fields. The devil is in the details. That is not the opinion I held when I first posted here.

 

[TSny]

You are probably not referring to this paper: David J. Griffiths, and Ye Li, Charge density on a conducting needle, American Journal of Physics 64, 706 (1996); doi: 10.1119/1.18236].

The paper that I had in mind is this one. The paper by Griffiths is interesting, too. This paper by Jackson also investigates conducting needles.

I think the take-away message is that one cannot a priori deduce, simply by looking at curvatures, the surface charge distribution on a charged conductor in the absence of external electric fields. The devil is in the details.

I agree.

 

[haruspex]

I think the take-away message is that one cannot a priori deduce, simply by looking at curvatures

The expression for $\sigma(x,y,z)$ looks like the fourth root of the Gaussian curvature of an ellipsoid, no?

Also, let me clarify that you didn't mean $\lambda(x)=\frac{dQ}{dx}$. Q is a constant. That Q should be the charge on a ring element, right?

 

[kuruman]

Also, let me clarify that you didn't mean $\lambda(x)=\frac{dQ}{dx}$. Q is a constant. That Q should be the charge on a ring element, right?

Yes and no. It's easier to explain with the relevant screenshots from the article.

 

[haruspex]

Yes and no. It's easier to explain with the relevant screenshots from the article.
View attachment 289736
View attachment 289737

Ok, the abuse of notation is in the original. They define dQ as the charge on a ring element even though Q is a constant, not a function. It would be better to use dq.

Anyway, the more interesting point in post #14 is that the surface charge density function quoted is proportional to the fourth root of the Gaussian curvature, suggesting that curvature is a good guide for charge distribution. But of course that is in the absence of an external field.

 

[kuruman]

Ok, the abuse of notation is in the original. They define dQ as the charge on a ring element even though Q is a constant, not a function. It would be better to use dq.

Yes, I quoted directly from the original. It is curious that the AJP referee didn't pick it up.

Anyway, the more interesting point in post #14 is that the surface charge density function quoted is proportional to the fourth root of the Gaussian curvature, suggesting that curvature is a good guide for charge distribution. But of course that is in the absence of an external field.

As a guide only. See the last paragraph in Post #9 by @TSny.

 

[TSny]

In this paper the following example is given to show that the surface charge density $\sigma$ cannot, in general, be proportional to some power of the curvature $K$.

Here we have a charged, isolated solid conductor in the shape of a cylinder of radius $R$ with hemispherical caps. The surface of the cylinder has zero curvature and the caps have $K = \frac 1 {R^2}$. If $\sigma$ were proportional to some power of $K$, then $\sigma$ would be zero for the cylindrical part and $\sigma$ would be uniform over the caps. But it is not hard to see that this cannot be the case since such a charge distribution would not produce zero electric field throughout the interior of the conductor.