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 π0keλz 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 π0keλa^2 cosθdθ
(z^2 +a^2 )^3 /^2=− keλa2
(z^2 +a^2 )^3 /^2·sinθ|^20 π= 0 (48)(and ditto, of course, forEy)!