P25_070 - 70 From the previous chapter we know that the radial field due to an infinite line-source is E= 2πε0 r which integrates using Eq

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Unformatted text preview: 70. From the previous chapter, we know that the radial field due to an infinite line-source is E= λ 2πε0 r which integrates, using Eq. 25-18, 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.

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