That is the picture that polywell advocates have in mind. Let's take a closer look at it. (I've done this before, but I admit it is easy to lose pieces of the argument.) Let's assume your picture of the plasma potential is correct, and that it applies to a flux tube coming out of a point cusp. We will also use some other conditions, like beta = 1 and flux conservation in the sheath. The question I want to ask, semi-quantitatively, is how high the potential on the magrid would have to be, your red tick mark.

By construction, most of the ions are held back by the potential rise. The electrons, in contrast, will be accelerated into this region. By conservation of electron flux, the increased speed is associated with a decreased density, by a moderate factor. We won't be far off if we assume the electron density near the potential peak is half that in the main plasma.

(We won't be able to keep track of all the factors of 2. That gives us an idea of how far we can trust our quantitative results.) In the standard polywell picture, most of the plasma pressure at the edge is due to electrons, and the average energy of the electrons will be a bit more than the well depth phi. Phi will be chosen to give the ions the energy they need to fuse. If we take D-T fuel, but consider that the energy distribution is (presumably) not Maxwellian, we will need about phi = 30 kV. We can now use the beta = 1 condition to relate the density to the magnetic field.

So far so good. Now I'd like to know how thick this pencil of plasma is. Far from a cusp, the transition from field-free plasma to plasma-free field will take place over a sheath of some thickness. Bussard talks about a sheath with a thickness equal to the electron gyroradius, rho_e = m_e*v / eB, where the electron velocity v is given by (1/2)*m_e*v^2 = e*phi. The end result is

rho_e^2 = 2*m_e*e*phi / (e^2*B^2)

It is really just about impossible to imagine a thinner sheath, and I'm afraid I will have to insist on taking this as a lower bound. There is good reason to believe that the sheath will actually be several times thicker than this, but I will give it the benefit of the doubt for now. Now draw a circle around the point cusp as large as you dare, i.e. close to the line cusps or at least halfway between the point cusp and the line cusp. This circle will have a circumference a bit less than 2pi*R. Call it 4*R among friends. Make a stubby cylinder out of this circle by extending it through the sheath. It will have an area of

Because of pressure balance and beta = 1, the magnetic field will be constant along the surface of the palsma, so the field lines passing through the stubby cylinder will map to a pencil pointing along the point cusp with the same area. The radius of this pencil will be roughly

Notice that the radius of the point cusp must be significantly greater than rho_e, no matter what Bussard says.

Let's start putting this together. What I want is lambda, the line density of charge in the pencil.

lambda = n*e*A

= n*e*(4*R*rho_e)

= 2*n_0*e*(R/rho_e)*rho_e^2

= 2*(R/rho_e)*e*( B^2/(2mu_0)/(e*phi) )*( 2*m_e*e*phi / (e^2*B^2) )

= 2 * (R/rho_e) * m_e / (e*mu_0)

= 2 * (R/rho_e) * (m_e*c^2/e) * epsilon_0

= (1 MV) * (R/rho_e) * epsilon_0

The last step follows from the fact that the rest mass of the electron in 500 keV in energy units. (If you get a queasy feeling at this point, seeing 1 MV showing up as a characteristic voltage, I understand completely.)

Why do I want this? Because I want to calculate the potential that the magrid would have to have in order to achieve the assumed potential profile. To get that, I use the fact that a cylinder with line charge density lambda and radius s produces a potential at radius R equal to

V_magrid = (2*lambda/epsilon_0)*ln(R/s)

= (1 MeV) * (R/rho_e) * (m_e*c^2/e) *ln(R/s)

= (1 MeV) * (R/s)^2 * ln(R/s)

That should be enough math. We will not want to choose R/s too small, otherwise our cusps will be bigger than our plasma, our electrons won't be confined by the magnetic field, and probably a slew of other problems. The ln term shouldn't bother us too much, but we expect (R/s)^2 to have to be at least 10, so that our magrid voltage relative to the plasma will have to be on the order of 10 MV. Since phi will probably be around 30 kV for D-T fuel, we see that the red tick mark in the potential diagramm will be way off the map.

The analysis of the line cusps is very similar and yields a very similar result, giving me confidence in the robustness of the result.

That is what I call a serious inconsistency in the usual way of looking at the polywell. Please feel free to try to modify the picture to get it to make some sense.