Physics 500: Mathematical Methods (first midterm exam)

• Now, keeping in mind that
∞∑
ln(1 + x) = x − x 2 /2 + · · · = (−1) i+1 x i /i, (16)
i=1
show that V (z, 0) can be expanded as
∞∑
V (z, 0) = 2KQ/L (1/i)(L/2z) i [(−1) i−1 − (1) i−1 ] (17)
i=1
∞∑
= 2KQ/L (L/2) 2j+1 z −(2j+1) , (18)
j=1
so that we now have a power series in z.
• Finally, determine the cofficients A l by equating the expansion formula
V (r, θ) = ∑ ∞
l=0 A l r −(l+1) P l (cosθ) at r = z and θ = 0 to the series in
powers of z which you obtained earlier; show that
V (r, θ) =
∞∑
j=0
Hint: keep in mind that P l (1) = 1.
KQ
r (L/2r)2j P 2j (cosθ) (19)
• Let’s carry out the same calculation, only in a more direct manner
this time. The task now is to calculate V off of the z axis. We
will lose no generality by concentrating on the point P = (x, 0, z) =
(r sin θ, 0, r cosθ); show that the electrostatic potential at P is given by
V (x, 0, z) = V (r, θ) = K
∫ L/2
−L/2
λdz ′
√x 2 + z 2 + (z ′ ) 2 − 2zz ′ ; (20)
finally, show that this expression leads to a formula for V (r, θ) identical
to that given in Equation 19. To work toward this goal, it will be useful
to express x and z in terms of spherical polar coordinates as needed.
Recall that the generating function for the legendre polynomials can
be expanded as
∞∑
(1 − 2tx + t 2 ) −1/2 = t l P l (x). (21)
You should obtain an integral of a series, and it may be helpful to
note that you are free to exchange the order of the integration and the
summation of the series; integrating term by term should lead to the
desired result.
l=0
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