10. Appendix

R
-R
A
y
E’-E
y
E’-E
R
x
Solution to Problem 5.1 623
Let us first consider the contour integral:
A
dz ′
(z ′ E ′ E)(z ′ i°)(z ′ i°)
In the limits of R ⇒∞and ‰ ⇒ 0 the
part of the contour integral over the xaxis
will give us the principal value of:
∞
∞
dz ′
(z ′ E ′ E)(z ′ i°)(z ′ i°)
-R
R
R
x
Similarly, when we perform the contour
integral over B in the limits of R ⇒∞
B
and ‰ ⇒ 0, the part of the contour integral
over the x-axis will give us again the
same principal value of the above integral.
On the other hand, the parts of the two contour integrals over the semicircles
(whether of radius R or ‰) will exactly cancel each other since the directions
of the integrations in A and B are opposite. Thus, the principal value
of the integral becomes:
∞
dz
∞
′
(z ′ E ′ E)(z ′ i°)(z ′ i°)
1
dz
2 A
′
(z ′ E ′ E)(z ′ i°)(z ′ i°)
dz
B
′
(z ′ E ′ E)(z ′ i°)(z ′ (4.5q)
i°)
The value of the two contour integrals in (4.5q) can now be obtained by
the Residue Theorem. For example, the contour integral over A is given by
2applei times the residue of the integrand at the only pole: i°, and is equal to
(2applei)(1/2apple°)[i° E ′ E] 1 . Similarly the contour integral over B is given by
the residue at the pole: i° and is equal to (2applei)(1/2i°)[i° E ′ E] 1 .
Substituting these results into (4.5q) one can obtain the same expression as in
(4.5o).
Solution to Problem 5.1
[Note: (5.19) in the first, second and third editions has an error in the sign of
gk. This error has led to errors in the sign of the term qÙkvk · F in Problem 5.1
in those older editions].
In Sect. 5.2.1 it is shown that, when a small enough electric field F is applied
to a charge distribution with a distribution function f 0 k in the absence of