# Gas station without pumps

## 2012 October 2

### Chapter 14 done

Filed under: home school — gasstationwithoutpumps @ 16:21
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We’re a week behind already, because we did not get any of the Chapter 14 homework done during the trip to Boulder, but we did finally get Chapter 14 done today. The twelve problems I assigned for Chapter 14 (plus one program) were too many, because they were almost all just plug-in-numbers-and-turn-the-crank exercises (despite having been given “P” for problem rather than “X” for exercise). I was rather tired last night and this morning when I did the problems, and I made a huge number of copying errors: copying the charge of an electron wrong (and using it for all the exercises that needed charge of proton or electron), copying the wrong line from a problem, copying numbers wrong from my calculator to the paper, … .  I don’t think I’ve ever had so many wrong answers. On only one problem when my son and I compared answers did we get different results that stemmed from his error rather than mine (and that was also a clerical error, not a conceptual one).

There was one problem that wasn’t just an exercise, though my son treated it as one.  That problem is 14P40 part c:

The electric field at a location C points north, and the magnitude is 1×106 N/C.  Give numerical answers to the following questions:

(c) where should you place a proton and an electron, at equal distances from C, to produce this field.

Locus of positions for electron with C at origin. The proton would be symmetrically located at (x,-y) for an electron located at (x,y). Locus plotted by Wolfram Alpha command “plot y/(x^2+y^2)^(3/2)=347.2E12”

The trivial answer places an electron to the north and a proton to the south, and students just compute the distance. But there are an infinite number of solutions and the locus of the solutions is an interesting one.  I don’t know whether the authors were thinking of this infinite set of solutions, or if they had only considered the trivial solution.

Since C is equidistant from the proton and electron, it must be on the perpendicular bisector of the line segment between them. Since the field points north, and we’re on the perpendicular bisector, the electron must be due north of the proton. If we do a 2D plot, putting C at the origin and the electron at (x,y), we get a formula of the form $y/(x^2+y^2)^{3/2}=347.2E12$, which we can ask Wolfram Alpha to plot for us. Note: I’m deliberately not providing the derivation for the number, so that students have to do some work before copying this answer!

The trivial solution is the one on the y-axis (not the origin). I had not expected to see the lower part of the curve, where the locus approaches the origin, but it makes sense. If we look at the formula in polar coordinates, we get $r = \sqrt{\sin(\theta)/347.2E12}$, where r is the distance from C to the electron (and to the proton), and θ is the angle from the horizontal axis (angle north of east). This parameterization also makes it easier to find where the x value is maximized by taking the derivative of $\cos(\theta)\sqrt{\sin(\theta)/347.2E12}$ with respect to θ, and setting it to 0. I get the maximal value for x as about 3.33E-08 m, at about 35.26°.

For very small angles, the electron and proton need to be very close together, though the solutions with them too close together are bogus, because classical electrostatics breaks down once quantum effects become significant.

The polar plot made by giving Wolfram Alpha “polar plot r=sqrt(sin(theta) / 347.2e12)” is cleaner and faster than solving for the x,y values directly.

For next week, we’ll have to read Chapter 15. I should have some problems selected before this weekend.