2—Infinite Series 602.25 The electric potential from one point charge iskq/r. For two point charges, you add the potentials of
each: kq 1 /r 1 +kq 2 /r 2. Place a charge−q at the origin; place a charge+q at position(x,y,z) = (0, 0 ,a).
Write the total potential from these at an arbitrary positionP with coordinates (x,y,z). Now suppose that
ais small compared to the distance ofP to the origin (r =
√
x^2 +y^2 +z^2 ) and expand your result to the
first non-vanishing power ofa. This is the potential of an electric dipole. Also express your answer in spherical
coordinates. Ans:kqacosθ/r^2
2.26 Do the previous problem, but with charge− 2 q at the origin and charges+qat each of the two points
(0, 0 ,a)and(0, 0 ,−a). Again, you’re looking for the potential at a point far away from the charges, and up
to the lowest non-vanishing power ofa. In effect you’re doing a series expansion ina/rand keeping the first
surviving term. Also express the result in spherical coordinates. The angular dependence should be proportional
toP 2 (cosθ) =^32 cos^2 θ−^12 , a “Legendre polynomial.” This potential is that of a linear quadrupole.
2.27 The combinatorial factor Eq. ( 13 ) is supposed to be the number of different ways of choosingnobjects
out of a set ofmobjects. Explicitly verify that this gives the correct number of ways form= 1, 2 , 3 , 4. and all
nfrom zero tom.
2.28 Pascal’s triangle is a visual way to compute the values ofmCn. Start with the single digit 1 on the top
line. Every new line is computed by adding the two neighboring digits on the line above. (At the end of the line,
treat the empty space as a zero.)
1
1 1
1 2 1
1 3 3 1Write the next couple of lines of the triangle and then prove that this algorithm works, that is that themthrow
is themCnwhere the top row hasm= 0. Mathematical induction is the technique that I recommend.
2.29 Sum the series
1
2!
+
2
3!
+
3
4!
+···
Ans: 1