# The charge distribution over a surface

Charge gathers at points and is
also attracted to points. This is why a lightning conductor works, why charge leaks from the
ends of hair or thread fixed to a Van de Graaff generator and why the hairs on the back of
your hand rise when you go to touch a charged object.

But why does this happen
and why is the distribution of charge over a surface with differing curvature different at
different places?

First of all we must assume that the potential is equal over the
surface. This will be true when equilibrium is reached – a very short time after the conducting
solid has been charged.

Let the charge density of a surface be denoted by

s(charge
density is measured in Coulombs per metre squared, and these units give you a good idea
what charge density is).

## Consider an isolated sphere.

The potential at
a distance r from a sphere carrying a charge Q is given by:

Potential (V) = [1/4

pe_{o}]Q/r

Therefore the charge density on the
sphere will be given by the equation:

Charge density of the sphere (s) =
Q/4pr^{2}
However: Potential (V) = [1/4

pe_{o}]4

pr

^{2}/r = [1/

e_{o}]sr and so

Charge density (s = Ve_{o}/r

Therefore the charge density at a point on the surface of radius r is
inversely proportional to r. In other words a small r (a sharply curved surface) has a greater
charge density than a surface with large R.

Another way of looking at the problem is
to start by thinking of two spheres of different radii charged to the same potential. (See
Figure 1(a)).

If they are now
joined together by a wire (See Figure 1(b) no charge will flow between them (they are at the
same potential) but the charge density on each will be different.

Now instead of a wire fit a charged conducting surface (with
the same potential as the spheres) between them to form the shape shown in the following
figure.

Again there will be no charge flow and the charge distribution over the
conducting surface will therefore be different at different points on the surface (Figure
2).

Note that the lines of the electric field will always be at right angles to the
surface. (Figure 3)

## Rough and smooth charged spheres

Consider two metallic spheres with the same radius.
Only difference is that one has a smooth surface while the other has a rough surface.
Which one will have a higher maximum electrical potential before the air break down?

The really important thing when considering electrical breakdown of the air is the
field. For a point conductor the field at its surface is much greater than a larger sphere even if
both have the same charge. You could think of the point as a sphere with a very small radius.
The formula for the electric field of a spherical charged conductor depends on the inverse
square of the radius.

A rough surface is like a large number of small spheres with
small radii and so the field at all the points will be larger than a smooth sphere and so
electrical breakdown will occur at a lower potential.

The old school, experiment with
the Van de Graaff generator shows that if you attach cotton threads to a charged dome the
charge leaks away from the ends of the cotton (electrical breakdown has occurred in the air).
The ends of the cotton act rather like the rough points on your sphere.