W9_parallel_resonance.eps

74 Week 2: Continuous Charge and Gauss’s Law

dEz

dE

dE

α s

α

θ

z

r

r sin

r cos

θ

θ

R − r cosθ

y

x

P R dq

Figure 14: Geometry for the vector decomposition ofdE~intodEz.

Piece of cake, right? Well, not quite. Sadly,sand cos(α) depend onP,randθvia e.g. the law
of cosines^34 forsand the geometry of the triangle with sidess,R−rcos(θ) andrsin(θ) for the
other. On the other hand, the result still has azimuthal symmetry, which is good! This means we
can immediately do the (trivial)φintegral and rearrange the result so we can tackle it:

Ez = 2πr^2 σke

∫ 1

− 1

(R−rcos(θ))d(cosθ)

s^3

= 2πr^2 σke

∫ 1

− 1

(R−rcos(θ))d(cosθ)

(R^2 +r^2 − 2 rRcos(θ))^3 /^2

= 2πr^2 σke

{∫ 1

− 1

R d(cosθ)

(R^2 +r^2 − 2 rRcos(θ))^3 /^2

−

∫ 1

− 1

rcos(θ)d(cosθ)

(R^2 +r^2 − 2 rRcos(θ))^3 /^2

}

(78)

This integral looks difficult, and perhaps it is, but it isn’tthatdifficult. The worst thing about
it is that we have to integrate the second piece of it by parts. Let’s start with the first (fairly easy)
piece:

∫ 1

− 1

R d(cosθ)

(R^2 +r^2 − 2 rRcos(θ))^3 /^2

= −

1

2 r

∫ 1

− 1

(R^2 +r^2 − 2 rRcos(θ))−^3 /^2 (− 2 rR d(cosθ))

=

1

r

1

(R^2 +r^2 − 2 rRcos(θ))^1 /^2

∣∣

∣∣

1

− 1

=

1

r

{

1

(R^2 +r^2 − 2 rR)^1 /^2

−

1

(R^2 +r^2 + 2rR)^1 /^2

}

=

1

r

{

1

(R−r)

−

1

(R+r)

}

=

1

r

{

2 r

(R^2 −r^2 )

}

=

2

(R^2 −r^2 ) (79)

That’s not so horrible. All I had to do is multiply by− 2 r^1 × 2 r= 1 to get it set up foru-substitution
as

∫

u−^3 /^2 du(easy), and the rest is all algebra.

(^34) Wikipedia: http://www.wikipedia.org/wiki/Law of Cosines.