Mathematical Tools for Physics

2—Infinite Series 55

This is an elementary integral. Letθ= (c/ 2 a) tanφ.

∫Λ

0

dθ

[

c^2 + 4a^2 θ^2

] 3 / 2 =

∫Λ′

0

(c/ 2 a) sec^2 φdφ

[c^2 +c^2 tan^2 φ]^3 /^2

=

1

2 ac^2

∫Λ′

0

cosφ=

1

2 ac^2

sin Λ′

The limitΛ′comes fromΛ = (c/ 2 a) tan Λ′, so this impliestan Λ′= 2aΛ/c. Now given the tangent of an angle,
I want the sine — that’s the first page of chapter one.

sin Λ′=

2 aΛ/c

√

1 + (2aΛ/c)^2

=

2 aΛ

√

c^2 + 4a^2 Λ^2

Asc→ 0 , this approaches one. Put all of this together and you have the behavior of the integral in Eq. ( 25 ) for
smallc. ∫
π/ 2

0

dθ

[

c^2 + 4a^2 sin^2 θ

] 3 / 2 ∼

1

2 ac^2

+a constant

Insert this into Eq. ( 25 ) to get

Fz∼

Q 1 Q 2 c

2 π^2  0

.^1

2 ac^2

=

Q 1 Q 2

4 π^2  0 ac
Now why should I believe this any more than I believed the original integral? When you are very close to one
of the rings, it will look like a long, straight line charge and the linear charge density on it is thenλ=Q 1 / 2 πa.
What is the electric field of an infinitely long uniform line charge?Er=λ/ 2 π 0 r. So now at the distancecfrom
this line charge you know theE-field and to get the force onQ 2 you simply multiply this field byQ 2.

Fz should be

λ

2 π 0 c

Q 2 =

Q 1 / 2 πa

2 π 0 c

Q 2

and that’s exactly what I found in the preceding equation. After all these checks I think that I may believe the
result, and more than that you begin to get an intuitive idea of what the result ought to look like. That’s at least
as valuable. It’s what makes the difference between understanding the physics underlying a subject and simply
learning how to manipulate the mathematics.