Charge distribution on a torus (doughnut) shaped conductor

Question

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.

Answer 1

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.

Answer 2

Here, Gauss law is a special case of Maxwells electrostatic equation $$\nabla \cdot \vec E =\rho$$ at a conducting surface with constant potential U along the two tangent directions, yielding $$\vec n \times \vec E = \vec n \times \nabla U = 0$$ for a vector $\vec n$ normal to any surface geometry. The normal component $(\vec n \cdot \vec E)(out-in)$ yields the surface charge density, because $E(in)=0$ inside a conductor in the absence of currents.

I order to determine the potential, the electric field and the surface charge, one needs a solution of the Laplace equation with monopole character of a point charge at infinity and constant value on the torus surface.

One may consult Moon/Spence 'A Field Theory Handbook', or use the state of art e.g. in Mathematica, in order to get a separable coordinate system for the Laplacian, that has a family of concentric torus surfaces.

      toroidalTransforms = 
       CoordinateTransformData[ ][[First /@ 
        Position[CoordinateTransformData[ ], "Toroidal", \[Infinity]]]];

       Select[toroidalTransforms, 
            (Cases[#, "Cartesian", \[Infinity]] =!= {} &)]  

       Select[toroidalTransforms, 
          (Cases[#, "Cartesian", \[Infinity]] =!= {} &)] 

        {{"Cartesian" -> {"Toroidal", {\[FormalA]}}, "Euclidean", 3},
 {{"Toroidal", {\[FormalA]}} -> "Cartesian", "Euclidean", 3}}

        CoordinateTransformData[{{"Toroidal", 
          {\[FormalA]}} -> "Cartesian",  "Euclidean", 3} , "Properties"]

    Set @@@ 
  ({ {X[R_, u_, \[Eta]_, \[Phi]_], Y[R_, u_, \[Eta]_, \[Phi]_], 
 Z[R_, u_, \[Eta]_, \[Phi]_]}  , 
CoordinateTransformData[{{"Toroidal", {R}} -> "Cartesian", 
  "Euclidean", 3} , "Mapping", {u, \[Eta], \[Phi]}] }\[Transpose]) 

$$X=\left.\frac{R \sinh (u) \cos (\phi )}{\cosh (u)-\cos (\eta )},\quad Y=\frac{R \sinh (u) \sin (\phi )}{\cosh (u)-\cos (\eta )},\quad Z=\frac{R \sin (\eta )}{\cosh (u)-\cos (\eta )}\right.$$

The Laplacian is, as always in a orthogonal system with scale factors $h_i$,

$$\Delta =\sum_i \frac{1}{\prod_k h_k^2} \ \partial_i \left( \frac{1}{h_i} \prod_k{ h_k^2}\ \frac{1}{h_i}\partial_i\right) $$

yields the following form of the Laplacian, applied to

$$\Delta _{u,\eta ,\phi } \left(H(\eta ) U(u) \sqrt{\cosh (u)-\cos (\eta )}\right)$$

$$\frac{(\cosh (u)-\cos (\eta ))^{-5/2} \Delta _{u,\eta ,\phi } \left(H(\eta ) U(u) \sqrt{\cosh (u)-\cos (\eta )}\right)}{H(\eta ) U(u)}$$

     1/(U[u] H[\[Eta]]) (-Cos[\[Eta]] + Cosh[u])^(-5/2)
 Laplacian[
 Sqrt[Cosh[u] - Cos[\[Eta]]] U[u] H[\[Eta]], {u, \[Eta], \[Phi]}, 
 "Toroidal"   ] /. {\[FormalA] -> R, f_[u, __] -> f} //
  FullSimplify // Numerator 

$$\frac{4 H''(\eta )}{H(\eta )}+\frac{4 \left(U''(u)+\coth (u) U'(u)\right)}{U(u)}+1$$

The $\phi,\theta$-dependence is discarded in order to yield solutions of the separated ODE'S constant over all of a given torus as boundary with Dirichlet condition 1.

The remaining equation yields the solution according to Mathematica $$\sqrt{\frac{1}{\cosh (u)+1}} K\left(\frac{2}{\cosh (u)+1}\right)$$

This is identical with Moon/Spencer by the identity for the elliptic integral $K$ and the Legendre functions of index -1/2

$$ \mathit P_{-1/2}(x) =\frac{2 K\left(\frac{1-x}{2}\right)}{\pi } $$

Resubstitution of the square root factor $1/\sqrt{\cosh u - \cos \eta}$ and determination of the normal derivative at the torus surface is an easy exercise. Its clear, that the result is independent of the surface coordinates $\theta,\ \phi$, but in cartesian coordinates, the surface element yields a $\theta$-dependence of the area density.

This result is general and merely a restatement of Gauss law: In local, tangent space orthonormal, coordinate system, the potential is constant along tangents and linear in normal direction, yielding the the charge density as in the case of a capacitor.