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Unformatted text preview: 70. From the previous chapter, we know that the radial ﬁeld due to an inﬁnite linesource is
E= λ
2πε0 r which integrates, using Eq. 2518, to obtain
Vi = Vf + rf λ
2πε0 ri λ
rf
dr
ln
= Vf +
r
2πε0
ri . The subscripts i and f are somewhat arbitrary designations, and we let Vi = V be the potential of some
point P at a distance ri = r from the wire and Vf = Vo be the potential along some reference axis (which
intersects the plane of our ﬁgure, shown below, at the xy coordinate origin, placed midway between the
bottom two line charges – that is, the midpoint of the bottom side of the equilateral triangle) at a
√
distance rf = a from each of the bottom wires (and a distance a 3 from the topmost wire). Thus, each
side of the triangle is of length 2a. Skipping some steps, we arrive at an expression for the net potential
created by the three wires (where we have set Vo = 0):
Vnet √
λ
(x2 + (y − a 3)2 )2
=
ln
4πε0
((x + a)2 + y 2 )((x − a)2 + y 2 ) which forms the basis of our contour plot shown below. On the same plot we have shown four electric
ﬁeld lines, which have been sketched (as opposed to rigorously calculated) and are not meant to be as
accurate as the equipotentials. The ±2λ by the top wire in our ﬁgure should be −2λ (the ± typo is an
artifact of our plotting routine). y
±2λ +λ +λ x ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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