HW5, Problem 3 **Spoiler a Trois**

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HW5, Problem 3 **Spoiler a Trois**

Post by Melissa on Sat Nov 05, 2011 4:28 pm

Damn trig, I defined Phi to be one of the top angles, and R1 is the outside distances, and R2 is the distance straight up the y axis.

E=r ̂ ρ_l/(2πε_0 r) Eq 3.33 p 78 (International Edition)
(r_1 ) ⃑=-x ̂R_1 cosϕ-y ̂R_1 sinϕ |r_1 |= R_1
(r_2 ) ⃑=-y ̂R_2 |r_2 |= R_2
(r_3 ) ⃑=x ̂R_1 cosϕ-y ̂R_1 sinϕ |r_3 |= R_1
E=(ρ_l (-x ̂R_1 cosϕ-y ̂R_1 sinϕ))/(2πε_0 R_1^2 )+(-2ρ_l (-y ̂R_2))/(2πε_0 R_2^2 )+(ρ_l (x ̂R_1 cosϕ-y ̂R_1 sinϕ))/(2πε_0 R_1^2 )
E=-x ̂ (ρ_l cosϕ)/(2πε_0 R_1 )-y ̂ (ρ_l sinϕ)/(2πε_0 R_1 )+y ̂ (2ρ_l)/(2πε_0 R_2 )+x ̂ (ρ_l cosϕ)/(2πε_0 R_1 )-y ̂ (ρ_l sinϕ)/(2πε_0 R_1 )
E=-y ̂ (〖2ρ〗_l sinϕ)/(2πε_0 R_1 )+y ̂ (2ρ_l)/(2πε_0 R_2 )
E=-y ̂ 〖2ρ〗_l/(2πε_0 R_2 )+y ̂ (2ρ_l)/(2πε_0 R_2 )=0


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