Homework+Ch+21

# 52 a 1 4 0 4 0 4 0 4 0 b e a b a b a b a dq r2 dy cos

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Unformatted text preview: = Magnitude: E= Direction: θ = arctan( 2 2 Ex + Ey = kQ l2 Ey 1 ) = arctan(− √ ) = 330◦ Ex 3 2 21.52 (a) 1 4π 0 λ = 4π 0 λ = 4π 0 λ = 4π 0 b E= a b a b a b a dQ r2 dy (cos θx − sin θy) ˆ ˆ r2 x sec2 θdθ (cos θx − sin θy) ˆ ˆ r2 x sec2 θdθ (cos θx − sin θy) ˆ ˆ (x/ cos θ)2 b λ (cos θx − sin θy)dθ ˆ ˆ 4π 0 x a λ = (sin θx + cos θy)|b ˆ ˆa 4π 0 x = The limits of integration are: a = arctan(−4/0.25) = −86.42◦ and b = arctan(2.5/0.25) = 84.29◦ E = (9 × 109 ) 3.15 × 10−6 (1.993x + 0.037y) = (34770x + 646y)N/C ˆ ˆ ˆ ˆ (6.5)(0.25) (b) E= λ 2Q (2)(3.15 × 10−6 )(9 × 109 ) = = = 34892N/C 2π 0 x 4π 0 xl (0.25)(6.5) Error: (Ex − E )/E = (34770 − 34892)/34892 = −0.0035 Ey /E = 646/34892 = 0.019 21.88 At rest ⇒ net force is zero. Let’s look at the vertical direction ﬁrst: Fnet,y =...
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## This note was uploaded on 09/11/2012 for the course PHYSICS 002 taught by Professor Welsch during the Spring '12 term at Berkeley.

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