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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 =...
View Full Document

This note was uploaded on 09/11/2012 for the course PHYSICS 002 taught by Professor Welsch during the Spring '12 term at Berkeley.

Ask a homework question - tutors are online