P25_067 - 67. From the previous chapter, we know that the...

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Unformatted text preview: 67. 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 + λ 2πε0 rf 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 will be the z axis described in this problem) at a distance rf = a from the wire. In the “end-view” presented below, the wires and the z axis appear as points as they intersect the xy plane. The potential due to the wire on the left (intersecting the plane at x = −a) is Vnegative wire = Vo + (−λ) ln 2πε0 a ( x + a) 2 + y 2 , and the potential due to the wire on the right (intersecting the plane at x = +a) is Vpositive wire = Vo + (+λ) ln 2πε0 a . ( x − a) 2 + y 2 Since potential is a scalar quantity, the net potential at point P is the addition of V−λ and V+λ which simplifies to Vnet = 2Vo + λ 2πε0 ln a ( x − a) 2 + y 2 − ln a ( x + a) 2 + y 2 = λ ( x + a) 2 + y 2 ln 4πε0 ( x − a) 2 + y 2 where we have set the potential along the z axis equal to zero (Vo = 0) in the last step (which we are free to do). This is the expression used to obtain the equipotentials shown below. The center dot in the figure is the intersection of the z axis with the xy plane, and the dots on either side are the intersections of the wires with the plane. ...
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