Required math: vectors, calculus
Required physics: basics
Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problems 2.6 – 2.8.
Now a few examples of surface charge.
First, we consider a circular disk of radius with surface charge density lying in the plane and centred at the origin. Find the electric field at a point on the axis.
To solve this we can make use of the solution to the circular loop. In this case we’re considering a circular ring of circumference and thickness , so the amount of charge in the ring is and from the earlier solution, the field due to this ring is
To get the total field from the disk, we integrate over :
To get the limiting behaviours we can Taylor-expand the result. For , we expand about and find the leading non-zero term is in :
This is correct since the total charge on the disk is so the field is that due to a point charge of that amount.
If we let we get
This is the field due to an infinite plane of charge. Note that the field is independent of so is the same no matter how far away from the plane we are.
We have a spherical shell of charge with radius and surface density , centred at the origin. Again, we seek the field at a point on the axis.
Using spherical coordinates, a point on the sphere has coordinates where is the angle from the positive axis, and is the azimuthal angle. We can use the cosine law to write the distance between a point on the sphere and the field point:
By symmetry, the field will again be in the direction, so we need the component of . To get this, we need the angle between and the axis. To get this, project the point on the sphere onto the axis; this gives a point with coordinate . The remaining distance along the axis to the field point is therefore , but this distance is the projection of onto the axis. The cosine of the angle between and the axis is this projection divided by , so we get
Now for a given value of , we have a ring of charge with radius and thickness at distance from the field point, so we can integrate over to get the total field.
This integral can be done using Maple, but there are two possibilities. First, if so the field point is outside the sphere, we get
Since is the total charge on the sphere, we see that the sphere behaves like a point charge for all field points outside it.
Second, if so we are inside the sphere, we get
So anywhere inside a spherical shell with a uniform charge distribution, we feel no field at all.
The result of the last example can be used to find the field due to a sphere that contains a uniform volume charge density . Since each spherical shell within the sphere behaves as a point charge to all points outside the shell, the field at a point outside the sphere () is just
At a point inside the sphere, all shells outside the field point contribute nothing, so we get, for :
The field thus increases linearly within the sphere and then falls off as an inverse square outside.