Lecture 8 - Gauss’s Law A Puzzle... E x a m p l e Calculate the potential energy, per ion, for an infinite 1D ionic crystal with separation a; that is, a row of equally spaced charges of magnitude e and alternating sign. Hint: The power-series expansion of Log[1 + x] may be of use. S o l u t i o n Suppose the array is built inwards from the left (that is, from negative infinity) as far as a particular ion. To add the next positive ion on the right, the amount of external work required equals - ke 2 a + ke 2 2 a - ke 2 3 a +···=- ke 2 a 1 - 1 2 + 1 3 +···(1) The terms in parenthesis look remarkably similar to the Taylor series of Log[1 + x] when x = 1. (What is the Taylor series of Log[1 + x]? ⅆ ⅆx Log[1 + x]= 1 1+x = 1 - x + x 2 -···, and integrating both sides yields Log[1 + x]= x - x 2 2 + x 3 3 -···, since the right hand side is a polynomial expansion, it is also the Taylor series of Log[1 + x]. This Taylor series is converging for -1 < x ≤ 1 (with convergence on -1 < x < 1 assured by the alternating series test)). Therefore the work required to bring this charge in equals - ke 2 a Log[2], which equals the energy of the infinite chain per ion. The addition of further particles on the right doesn’t affect the energy involved in assembling the previous ones, so this result is indeed the energy per ion in the entire infinite (in both directions) chain. The result is negative, which means that it requires energy to move the ions away from each other. This makes sense, because the two nearest neighbors are of opposite sign. Note that this is an exact result! It does not assume that a is small. Getting such a nice closed form is much more difficult in 2 and 3 dimensions. □ Theory Visualizing the Electric Field (Extended) Electric Field Lines Advanced Section: Escaping Field Lines Math Background: What is a Surface Integral? We are already familiar with surface integrals of scalar functions. For example, the surface area of a sphere with radius R centered at the origin is given by ∫ surface ⅆ a = ∫ 0 2 π ∫ 0 π R 2 Sin[θ] ⅆθⅆϕ = 4 π R 2 (4) Although we use spherical coordinates in this particular problem, we could also have used Cartesian coordinates Printed by Wolfram Mathematica Student Edition
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Lecture 8 - Gauss’s LawA Puzzle...
Example
Calculate the potential energy, per ion, for an infinite 1D ionic crystal with separation a; that is, a row of equally
spaced charges of magnitude e and alternating sign.
Hint: The power-series expansion of Log[1 + x] may be of use.
Solution
Suppose the array is built inwards from the left (that is, from negative infinity) as far as a particular ion. To add
the next positive ion on the right, the amount of external work required equals
- k e2
a+ k e2
2 a- k e2
3 a+ · · · = - k e2
a1- 1
2 + 13 + · · · (1)
The terms in parenthesis look remarkably similar to the Taylor series of Log[1 + x] when x = 1. (What is the
Taylor series of Log[1 + x]? ⅆⅆx
Log[1 + x] = 11+x
= 1 - x + x2 - · · ·, and integrating both sides yields
Log[1 + x] = x - x2
2+ x3
3- · · ·, since the right hand side is a polynomial expansion, it is also the Taylor series of
Log[1 + x]. This Taylor series is converging for -1 < x ≤ 1 (with convergence on -1 < x < 1 assured by the
alternating series test)). Therefore the work required to bring this charge in equals - k e2
aLog[2], which equals the
energy of the infinite chain per ion.
The addition of further particles on the right doesn’t affect the energy involved in assembling the previous ones, so
this result is indeed the energy per ion in the entire infinite (in both directions) chain. The result is negative, which
means that it requires energy to move the ions away from each other. This makes sense, because the two nearest
neighbors are of opposite sign.
Note that this is an exact result! It does not assume that a is small. Getting such a nice closed form is much more
difficult in 2 and 3 dimensions. □
Theory
Visualizing the Electric Field (Extended)
Electric Field Lines
Advanced Section: Escaping Field Lines
Math Background: What is a Surface Integral?
We are already familiar with surface integrals of scalar functions. For example, the surface area of a sphere with
radius R centered at the origin is given by
∫surfaceⅆa = ∫0
2 π∫0πR2 Sin[θ] ⅆθ ⅆϕ = 4 π R2 (4)
Although we use spherical coordinates in this particular problem, we could also have used Cartesian coordinates