Discussion
A coaxial cable has an inner conductor of radius 1 mm carrying 2 nC/m. The field at a radius of 2 mm is about
Answer: C
Using \( E = 2 \times 9 \times 10^{9}\rho_L/\rho \) gives \( 2 \times 9 \times 10^{9} \times 2 \times 10^{-9}/(2 \times 10^{-3}) = 1.8 \times 10^{4} \) V/m. Fields of tens of kilovolts per metre are entirely ordinary inside a cable.
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