I'm currently working on an extension to modify the built-in symbolic calculator to display its stack entries in rendered LaTeX--it already has various language modes for displaying its s-expression based formulas, including LaTeX, so this honestly isn't that big of a lift.
Don't fall for what? The fact that some people move between percentiles during their life doesn't change the fact that the distribution has grown more unequal. What the number means is that, while those who are poor are poorer now in relative terms than 1989, the rich have gotten richer in relative terms.
Taking on debt to finance a public good like high speed rail is perfectly reasonable. Privatization isn't a natural outcome of a public corporation running a deficit--it's the result of a political decision to disinvest in that project.
Good question. Under the influence of a magnetic field, charged particles do all sorts of counterintuitive things due to the Lorentz force, which acts at right angles to the particle velocity and the field.
One example is a sideways precession due to the interaction between gravitational acceleration g and the Lorentz force. Fortunately this effect is basically negligible, because gravity is so much weaker than the electromagnetic forces operating on the particles.
The difficulties tend to come from the geometry of the magnetic field itself. For example, in a torus shape, there is an unavoidable drift in the vertical direction due to the combination of centrifugal force in the big circumference, and the force implied by the gradient of the magnetic field.
You're right, my derivation is mistaken for failing to take the z component of the force. EQ. 19 is more like it although you'll note, also goes as 1/z^3.
You're also right that all that is moot from sheathing. But an ion still begins it's journey out at the bottom of a large potential well; one which is particularly steep because of the debye length, but still just as deep.
If you read the exposition you linked, equation 14 gives an expression for the field which is linear in the area. Again, it's pretty important that the capacitor be infinite in extent, otherwise it behaves differently.
What's the dimension that you're proposing to increase of the capacitor? The total work done across the capacitor will be fixed regardless of distance across.
Well, before breaking out a higher order analysis, I'd like to at least see a real first order analysis. The argument from potential at infinity is dispositive, but let's do some practice anyway:
Let A be the area of the capacitor, and dr the distance between the plates. Let c be the appropriate electrostatic constant for the coulomb force between a proton and the charge density on the plate.
At a point a distance r from the capacitor, the field effect from the negative side is, ignoring curvature effects, about
cA/r^2
The repelling charge from the other plate will be about
So the net force drops off approximately as the third power of the distance, to a first order approximation. Integrating over the radius, we have that the potential goes as -1/r^2, with the approximation breaking down near r=0.
Actually inserting appropriate constants of integration would make this argument robust, but would also just reduce to the argument from potential at infinity. Either way it's clear that the effect can't just be ignored out of hand.
The electric field outside an infinite capacitor is zero. For a finite capacitor, there is a nonzero field. The importance of the infiniteness assumption can't be understated--such a capacitor cuts the universe in half, and every point of one half has the same electric potential.
On the other hand, if the capacitor is finite, then the surface integrals over the plates are not equal.
This won't work. The basic idea about the protons gaining more momentum than the electrons is valid. But the dipole creates an opposite field outside of the charged plates. Protons will be decelerated until they pass the first, positively charged plate, then accelerated through the plates, then decelerated back towards the negative plate.
This is all clear if you consider the ions falling through a potential field. The potential is 0 at infinity, positive at the first plate and negative at the second. An incoming ion starts at 0 potential, climbs a big hill to get through the first plate, then falls down below 0. Then on the way out it has to climb back to 0 potential at infinity. So the ions gain energy inside the plates but lose it all back on either side.
It's the inverse square (not a root), because the surface of the planet is two-dimensional. Think of moving a projector screen back and forth. For a fixed aperture angle, the area projected upon will grow quadratically with distance, so the intensity (inverse area) grows inverse-quadratically.