Homework+Ch+21

# 47 from problem 46 we know that for a uniformly

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: eld due to the piece dy’ at the position -y. So the net ﬁeld will be horizontal. The net ﬁeld is: l/2 E = Ex = −l/2 1 E= xλ 2π 0 1 λdy cos θ = 2 + y2 4π 0 x l/2 2 2 −3/2 (x +y ) 0 l/2 −l/2 1 λdy 2 + y2 4π 0 x x2 x2 + y 2 l/2 y λx dy = 2π 0 x2 x + y2 = 0 l/2 =2 λx 2π 0 x2 0 1 λdy 2 + y2 4π 0 x l/2 x2 + l2 /4 = 2π 0 21.47 From problem 46, we know that for a uniformly charged straight wire with length l and point 0 at the midpoint, the ﬁeld at a point P perpendicular distance x from 0 is: λ l 2 + 4x2 )1/2 2π 0 x(l 5 x2 + y 2 λl + 4x2 )1/2 x(l2 Appendix B of the textbook has a formula to evaluate the integral. The direction of the ﬁeld is the +x direction. E= x For our problem, let’s consider one wire. Then the point P at a distance z above the center of the square loop is at th...
View Full Document

Ask a homework question - tutors are online