# Charge distribution on a doughnut shaped conductor

In a spherical solid conductor the charge is always on the outer surface. Even if the sphere has a cavity, the surface of the inner cavity can not carry a charge due to Gauss's Law.

1. What would be the charge distribution in case the solid conductor is exactly doughnut shaped?

2. How does one define the outer and inner surface in this case?

3. What is the right way to use Gauss's law in this case to find charge distribution.

Gauss's law can only easily be used in cases with high degrees of symmetry and where you can define surfaces where the E-field is either parallel or perpendicular to the surface vector(s) and is constant or is zero over that surface.

In the case of a doughnut, there is clearly a high degree of symmetry, either along the axis of the doughnut itself or along a line perpendicular to the plane of the doughnut and passing through its centre.

However, the second condition is much harder to satisfy. If you consider a cylinder that encloses the doughnut and joins in on itself, the problem is that although the E-field immediately at the surface of the conductor is parallel to the surface vector, it is not of constant magnitude over the surface. Similarly, if you use a spherical surface, with centre at the centre of the doughnut and which cuts through the doughnut. You know the E-field cutting the surface of the sphere inside the conductor is zero, but it is not zero outside and not parallel to the sphere's surface vector.

In fact I do not think you are going to get very far with Gauss's law (I am of course immediately prepared to withdraw my answer should someone else come up with one; at least a straightforward one) unless you know what the variation of E-field strength is over the surface of the donut (see below). Instead you could construct your donut by superposing the fields from rings at different radii and at different heights above the central donut plane. Each one of these rings has an E-field that has an intrinsic asymmetry with respect to the geometric symmetries of the donut. So the superposed result (which will have a charge distribution that gives zero E-field inside the conductor) will also inherit these asymmetries.

You can see the field of a ring of charge below (taken from Wolfram.com). If you knew the strength of the E-field over the surface then you could use Gauss's law to argue that the charge density followed the E-field strength. This can be done by constructing small Gaussian "pillboxes" that cut across the donut surface. We know that the E-field exits the surface of a conductor parallel to the surface vector, but is zero inside the conductor. The charge inside the pillbox is equal to the charge surface density times the area. Hence by Gauss's law, the charge surface density will be proportional to the E-field strength at that point.

• I saw this lecture in which Gauss's law is used to explain the charge distribution.This is similar to a doughnut shape.I am confused as to which surface is outer and which the inner.Since this is essentially a cylinder which is bent to meet end to end isn't the entire surface an external surface.Where does the external surface begin and where does it end?youtu.be/qaZQzIXv2RQ – Chappy Nov 7 '14 at 3:33
• @Chappy I'm not going to watch all of an undergrad electrostatics lecture. At what time in the film is Gauss's law used to predict the surface charge distribution on a donut conductor? – Rob Jeffries Nov 8 '14 at 9:51
• Sorry Rob.Go to 25:27 approx. – Chappy Nov 8 '14 at 15:13
• @Chappy can you not see this is topologically different? The "heart" shape is more like a hollow spherical shell (although the lecturer does not clearly describe this) i.e. it isn't 2D as drawn on his blackboard - there is an "inside". There is no "inside" surface of a donut. Effectively all of it is "exterior" surface according to the definition in this lecture. Therefore there is charge all over the exterior surface, but it is not uniformly distributed because the geometry is asymmetric and the E-field is not the same magnitude everywhere on the surface. – Rob Jeffries Nov 8 '14 at 15:38
• @Chappy To make something equivalent to the lecture, you would use a hollow tube to make the donut. Then you could say that the E-field in this interior space inside the tube was zero and that there was no charge on the interior surface of the tube (using Gauss's law with a closed surface that cuts entirely through the conductor between the interior and exterior surface of the tube and where there is zero E-field inside a conductor). – Rob Jeffries Nov 8 '14 at 15:40