Physics - Why charges accumulate more at the sharp edges in a conductor

A bit about Lightening Conductor:

It is good to know this fact "why charges accumulate more at the sharp edges", as the lightening conductor uses this fact to prevent lightening strike directly to the building. A lightening conductor is a long thick wire from the top of the building to inside the earth. The wire at the building end is kept sharp. 

So during the stormy weather when the clouds and earth are oppositely charged and creates ideal situation for lightening. Since the lightening conductor is connected to earth also becomes charged, the sharp edge has more charges and therefore instense electric field is created that can break electic conductivity of air. It literally creates the lightening from earth and tempts the lightening from clouds to strike the wire. The building is therefore saved.

Now for the physics of Charge Accumulation at the Sharp Edges 

Let me show the figure of a charged conductor below. You can see where the curvature is small (sharp edges), the accumulation of the charge is more. Correspondingly, where the curvature is large (flatter portion) the charges per unit area  is less.

Defining the Problem:

So let us try to define the problem more concretely so we can prove it. Here is the Problem to prove:  The charge density is more at high curvature then low curvature area.

Strange Properties of Conductor:

Since are are talking about conductor, let me elaborate some interesting properties of the conductor:

1. For a conductor, the potential at any two points in a conductor is the same. (If it is not, the electrons will move quickly to make it so!)
2. In a conductor, there are no charges inside the conductor. All the charges are on the surface! Strange, but it is because if there are, say positive charges, inside the conductor, the electron nearby will come to neutralize. It will do to keep the potential between two points the same)
3. The electric field comming out of surface of the conductor is alway perpendicular to surface. (for positive charged conductor it is comming out, for negatively charged conductor it is going in. But always perpendicular to surface)
4. More charge accumulate where the curvature of the conductor is large (we will prove it this below)

The Proof:

Consider the figure above. A small conducting sphere of radius R1 and  a larger conducting sphere of radius R2. They are connected by a wire, which will ensure the potential of both sphere are the same. Since they are conducting sphere, the charges will be spread out evenly on the surface. We will assume the charge density \(\sigma =Q/(4\pi R^2) \)

The potential of a sphere at the surface is given by: 

$$ \phi=kQ/R, \quad Q=\sigma\times 4\pi R^2 $$ substituting for Q, we get:

$$\phi=4\pi k\sigma R $$

Applying to both conducting spheres and that the potential of the two are same due to wire connecting it:

$$ \phi =4\pi k\sigma_1 R_1 = 4\pi k\sigma_2 R_2 $$  

or  $$\sigma_1 R_1=\sigma_2 R_2$$  

Therefore for small \(R_1 \)  the charge density,  \( \sigma_1 \),  has to be higher. The smaller sphere has high curvature. 

Hence we prove that higher curvature means the charge density will be higher.

THE END