68 Week 2: Continuous Charge and Gauss’s Law
In these equations, we must evaluate sinφand cosφusing the right triangleazr:
sinφ =
a
r
=
a
(z^2 +a^2 )^1 /^2
(45)
cosφ =
z
r=
z
(z^2 +a^2 )^1 /^2
(46)
so that:
Ez=
∫ 2 π
0
keλz adθ
(z^2 +a^2 )^3 /^2
=
keλ(2πa)z
(z^2 +a^2 )^3 /^2
=
keQ z
(z^2 +a^2 )^3 /^2
(47)
AlthoughEx=Ey= 0 from symmetry as noted, it is pretty easy to actually evaluate them:
Ex=−
∫ 2 π
0
keλa^2 cosθdθ
(z^2 +a^2 )^3 /^2
=− keλa
2
(z^2 +a^2 )^3 /^2
·sinθ|^20 π= 0 (48)
(and ditto, of course, forEy)!