Answer:
E(0.1m)=-16.53.10^6 V.m
E(0.465m)=-3.55.10^6 V.m
E(1.3m)=-1.27^6 V.m
Explanation:
You can find the field using Gauss's Law:
[tex]\int\ E.} \, dS = \frac{Qin}{\epsilon }[/tex]
the surface S is an "infinite long" cylinderr of radio r.
[tex]\int\ {E.} \, dS = E\int\ dS=E.S=E2\pi rL[/tex]
[tex]Qin=\lambda L[/tex]
E(r)=[tex]\frac{\lambda }{2\pi \epsilon} . \frac{1}{r}[/tex]
λ=-92.0 μC/m, ε=8.85.10^-12
E(0.1m)=-16.53.10^6 V.m
E(0.465m)=-3.55.10^6 V.m
E(1.3m)=-1.27^6 V.m