Homework+Ch+21

# For one wire the vertical component of the eld is e

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Unformatted text preview: is distance from that wire: x= (l/2)2 + z 2 So the electric ﬁeld due to one wire is: E1 = l λ 1 2π 0 ((l/2)2 + z 2 )1/2 (l2 + 4((l/2)2 + z 2 )) 2 Pointing in the direction of a vector from the midpoint of that wire to z. By symmetry, horizontal components of the ﬁelds from the 4 wires will cancel each other out. The net ﬁeld will be in the z direction. For one wire, the vertical component of the ﬁeld is E cos θ where θ is the angle between the E vector and the z direction. Then z cos θ = (l/2)2 + z 2 So the net ﬁeld is: l cos θ λ 2π 0 ((l/2)2 + z 2 )1/2 (l2 + 4((l/2)2 + z 2 )) 1 2 √ z2 2 l λ (l/2) +z =4 2π 0 ((l/2)2 + z 2 )1/2 (l2 + 4((l/2)2 + z 2 )) 1 2 E = Ez = 4E1,z = 4 = π 0 (l2 8λlz + 4z 2 )(2l2 + 4z 2 )1/2 21.49 For an inﬁnitesimal piece of the...
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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 University of California, Berkeley.

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