problemset3

# 641 spring 2009 figure 6 the superposition of the

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Unformatted text preview: for r0 = 0.25, 0.5, 1, and 2 meters with 4πε0 = 100 volt−1 -m−2 (Image by MIT OpenCourseWare.) p Φ = .0025 Equipotential Lines 2 1 Φ = .01 Φ = 0.04 Φ = .16 -1 -0.5 0.5 1 Φ=0 -1 -2 p cos Figure 5: Polar plot of equipotential lines Φ = 4πε0 rθ for 0 ≤ θ ≤ π , Φ = 0, ±0.0025, ±0.01, ±0.04, ±0.16, 2 and ±0.64 volts with 4πε0 = 100 volt−1 -m−2 (Image by MIT OpenCourseWare.) p 4 Problem Set 3 6.641, Spring 2009 Figure 6: The superposition of the previous two plots of perpendicular equipotential and ﬁeld lines (Image by MIT OpenCourseWare.) Problem 3.2 A We can think of the bird as a perfectly conducting small sphere. When it lands on the uninsulated wire, it must become the same potential as the wire. This forces it to acquire a charge. When it ﬂies away, the charge stays with it because air is a poor conductor. B, C For B and C, use the method of images. We can use superposition to get the total potential for a charge q at height h moving in the x direction at velocity U . � � q 1 1 Φ= − 1 4πε0 [(x − U t)2 + (y − h)2 + z 2 ] 1 2 [(x − U t)2 +...
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