9—Vector Calculus 1 273positions of the charges, and the non-primed ones are the position of the point where you are evaluating the
potential, the field point.
For a simple example, what is the graviational potential from a uniform thin rod? Place its center at the
origin and its length= 2Lalong thez-axis. The potential is
φ(~r) =−∫
Gdm
r=−G
∫
λdz′
√
x^2 +y^2 + (z−z′)^2whereλ=M/ 2 Lis its linear mass density. This is an elementary integral. Letu=z′−z, anda=
√
x^2 +y^2.φ=−Gλ∫L−z−L−zdu
√
a^2 +u^2=−Gλ∫
dθ=−Gλθ∣
∣
∣
∣
u=L−zu=−L−zwhereu=asinhθ. Put this back into the original variables and you have
φ=−Gλ[
sinh−^1(
L−z
√
x^2 +y^2)
+ sinh−^1(
L+z
√
x^2 +y^2)]
(45)
The inverse hyperbolic function is a logarithm as in Eq. (1.4), so this can be rearranged and the terms combined
into the logarithm of a function ofx,y, andz, but thesinh−^1 s are easier to work with so there’s not much point.
This is not too complicated a result, and it is far easier to handle than the vector field you get if you take its
gradient. It’s still necessary to analyze it in order to understand it and to check for errors. See problem 49.