# Built on Facts

## Launching Electrons

#### July 24th, 2008 · 3 Comments

I love teaching, but I have to say I am not a fan of teaching the summer session. Everything is way too disorganized. Oh well, at least it’s only for another three weeks or so.

The way undergraduate intro physics is taught is usually in two halves, and right now I’m teaching the second. The first course is an introduction to classical mechanics: Newton’s laws, gravity, friction, momentum, energy, the usual. The second course is an introduction to electricity and magnetism. The second course does build on the first to some extent; you can’t very well find the acceleration of a particle in an electric field without knowing how force and acceleration work in the first place.

But the connection between the first and second courses is not always very obvious. To take a tiny example, problems with charged particles used as projectiles almost always ignore gravity. It’s not always clear to a beginner that this is valid. Sure the book says that the electron is going so fast that gravity doesn’t have time to do much in the context of the problem, but it’s hard to have an intuitive understanding without some concrete examples. How about we try one?

Find the electric potential required to accelerate an electron to escape velocity.

The answer to that question should give us some idea of what potential serves as kind of a lower limit for non-negligibility in comparison to gravity. We could find or look up the escape velocity directly, but I prefer to keep everything in terms of energy for conceptual benefit. The potential energy of an object in a gravitational field is the integral of the force with respect to distance, taken from the surface of the earth to infinity.

Where R and M are the earth’s radius and mass, and little m is the mass of the electron. I’ve played a little fast and loose with some negative signs, but it doesn’t affect anything. Now the potential energy a charge gains through a potential difference V is just qV. So set the two potential energies equal:

And solve that for V.

Plugging the mass and radius of the earth, along with the mass and charge of the electron and I get

A third of a millivolt. Tiny. An AA battery is 1.5 volts, thousands of times stronger than that. An electron accelerated through just about any realistic potential is going to be traveling at speeds which render gravity irrelevant for most practical purposes. An electron accelerated through the 1000 V or so of an old TV tube will experience “bullet drop” in the nanometer regime, compared to an actual sniper rifle drop of several inches over a typical 300 yard shooting distance.

Incidentally, are these speeds fast enough to worry about the effects of relativity? The answer is “not really”. As a general rule of thumb, relativistic speeds become important once the kinetic energy starts to become comparable to the mass of the particle. For the case of an electron this is 511 keV, and so even a few thousand volts is small enough to treat classically except in cases requiring very high precision.

### 3 responses so far ↓

• 1 Uncle Al // Jul 24, 2008 at 9:54 am

The Sony Trintron color CRT had 0.125 mm diameter phosphor dots and a 0.015 mm slit aperture grill (not a shadow mask). Relativistic effects were substantial versus target dimension and location. They were adjusted out as a sum.

• 2 Erik Remkus // Aug 2, 2008 at 6:59 am

Wouldn’t it be easier to simply compare the gravitational force and electric force between two electrons?
F = GMm/r^2 and F = KQq/r^2.
That’s 10^-11*10^-31*10^-31/r^2 compared to 10^9*10^-19*10^-19/r^2. That comes out to 10^-73 vs. 10^-21. When you demonstrate to them that the force due to electric field is 52 orders of magnitude larger, that should drive the point home fairly well.

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