Week 2: Continuous Charge and Gauss’s Law 75
The second integral is also easy enough, at least if you you remember how tointegrate by
parts: ∫
udv=uv−∫
vdu (80)Our chore, then, is to identify auand advin the integral:
∫ 1
− 1rcos(θ)d(cosθ)
(R^2 +r^2 − 2 rRcos(θ))^3 /^2 =(
1
− 2 R
)∫ 1
− 1− 2 rRcos(θ)d(cosθ)
(R^2 +r^2 − 2 rRcos(θ))^3 /^2 (81)(where I’ve gone ahead and multiplied and divided by− 2 R, thinking ahead).
Let’s let:
u= cos(θ) (82)and
ζ=R^2 +r^2 − 2 rRcos(θ) (83)
so that:
dv=− 2 rR d(cosθ)
(R^2 +r^2 − 2 rRcos(θ))^3 /^2=ζ−^3 /^2 dζ (84)We integrate this to get:v=∫
dv=− 2 ζ−^1 /^2 =− 2
(R^2 +r^2 − 2 rRcos(θ))^1 /^2(85)
Note that this isjust the first integralbefore we plugged in the limits!
So let’s dig into the algebra. This bit isn’t exactly trivial – be patient andtry to understand
each step.
(
1
− 2 R
)∫ 1
− 1− 2 rR d(cosθ) cos(θ)
(R^2 +r^2 − 2 rRcos(θ))^3 /^2=
(
1
− 2 R
){
−2 cos(θ)
(R^2 +r^2 − 2 rRcos(θ))^1 /^2∣∣
∣∣
1− 1−∫ 1
− 1− 2 dcos(θ)
(R^2 +r^2 − 2 rRcos(θ))^1 /^2}
=
(
1
− 2 R
){(
− 2
R−r−
2
R+r)
−
(
1
rR)∫ 1
− 1− 2 Rr dcos(θ)
(R^2 +r^2 − 2 rRcos(θ))^1 /^2}
=
(
1
− 2 R
){(
− 4 R
R^2 −r^2)
−
(
2
rR)
(R^2 +r^2 − 2 rRcos(θ))^1 /^2 )∣∣
∣
1
− 1}
=
(
1
− 2 R
){
− 4 R
R^2 −r^2−(
2
rR)
[(R−r)−(R+r)]}
=
(
1
− 2 R
){
− 4 R
R^2 −r^2+
4
R
}
=
{
2
R^2 −r^2−
2
R^2
}
(86)
Putting it all together we get:Ez = 2πr^2 σke{
2
R^2 −r^2−
2
R^2 −r^2+
2
R^2
}
= 2πr^2 σke2
R^2
=
ke(
4 πr^2 σ)
R^2
=
keQ
R^2