Posted by ike on Saturday, March 5, 2011 at 3:32pm.
I suspect a typo here.
Yes, use Gauss's law if the length is great compared to the distance r
here L = 5 and r = 0.1
Area of cylinder around wire at distance r is 2 pi r L
integral of E coming out of surface = charge inside /eo
so
2 pi r L E = 5*10^-8 L /eo
so
E = 5*10^-8 /(2 pi eo) = 10*10^-8 /(4 pi eo)
= 10^-7 k
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