HW5, Problem 2 **Another Spoiler**

[Melissa]

Again written poorly, because I don't want to spend the time cleaning it up.

E= 1/4πε ∫_(l^')▒〖R ̂^' (ρ_l dl^')/R^'2 〗 Eq 3.21c p 74 (International Edition)
R ⃑^'= -x ̂rcosϕ-y ̂rsinϕ+z ̂z
|R ⃑'|= √(〖(rcosϕ)〗^2+(rsinϕ)^2+ z^2 )= √(r^2+z^2 )
E=ρ_l/(4πε_0 ) ∫_0^(π⁄4)▒(-x ̂rcosϕ-y ̂rsinϕ+z ̂z)r/(r^2+ z^2 )^((3/2)) dϕ
E=ρ_l/(4πε_0 ) ∫_0^(π⁄4)▒(-x ̂r^2 cosϕ)/(r^2+ z^2 )^((3/2)) dϕ-∫_0^(π⁄4)▒(y ̂r^2 sinϕ)/(r^2+ z^2 )^((3/2)) dϕ+∫_0^(π⁄4)▒(z ̂zr)/(r^2+ z^2 )^((3/2)) dϕ
E=ρ_l/(4πε_0 ) [ (-x ̂r^2 sinϕ)/(r^2+ z^2 )^((3⁄2) ) |■(π⁄4@0)+(y ̂r^2 cosϕ)/(r^2+ z^2 )^((3⁄2) ) | ■(π⁄4@0)+(z ̂zr)/(r^2+ z^2 )^((3⁄2) ) ϕ│■(π⁄4@0)]
E=(5×〖10〗^(-6))/(4π(8.854×〖10〗^(-12))) [(-x ̂〖0.02〗^2 sin⁡(π⁄4))/(2^2+ z^2 )^((3⁄2) ) +(y ̂〖0.02〗^2 (cos π⁄4-cos0))/(2^2+ z^2 )^((3⁄2) ) +(z ̂0.02z(π⁄4))/(2^2+ z^2 )^((3⁄2) ) ]
E=(-x ̂12.71)/(〖0.02〗^2+ z^2 )^((3⁄2) ) +(-y ̂5.265)/(0.02^2+ z^2 )^((3⁄2) ) +(z ̂705.9z)/(〖0.02〗^2+ z^2 )^((3⁄2) ) (V/m)